Sadiku Practice Problem Solutionpdf [PDF]

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February 5, 2006



CHAPTER 1



P.P.1.1



A proton has 1.602 x 10-19 C. Hence, 2 million million protons have +1.602 x 10-19 x 2 x 106 = 3.204 x 10–13 C



P.P.1.2



i = dq/dt = -10(–2)e-2t mA At t = 0.5 sec, i = 20e-1 =



P.P.1.3



q=



1



∫ idt = ∫ 2dt + ∫ 0



= 2 + 14/3 =



P.P.1.4



(a)



2



1



7.358 mA



2



2t dt = 2 t



2



1



+ (2 / 3) t 0



3



2 1



6.667 C



Vab  = w/q = -30/2 =



–15 V



The negative sign indicates that point a is at higher potential than point b.



P.P.1.5



(b)



Vab  = w/q = -30/-6 =



(a)



v = 2 i = 10 cos (60 π t)



5V



 p = v i = 50 cos2 (60 π t) At t = 5 ms, ms, p = 50 cos2 (60 π 5x10-3) = 50 cos2 (0.3 π) = (b)



v = 10 + 5



∫



t 0



17.27 watts



idt = 10 +



∫



t 0



25 cos 60 π   t dt = 10+



25 60π  



sin 60



 p = vi = 5 cos (60 πt)[10 + (25/(60 π)) sin (60 π t)] At t = 5 ms, p = 5 cos (0.3π){10 + (25/(60 π)) sin (0.3 π)} =



29.7 watts



π  



t



P.P.1.6



 p = v i = 15 x 120 = 1800 watts; w = p x t therefore, t = w/p = (30x103)/1800 =



P.P.1.7



16.667 seconds



 p1 = 5(-8) = –40w  p2 = 2(8) = 16w  p3 = 0.61(3) = 0.6(5)(3) = 9w  p4 = 3(5) = 15w



P.P.1.8



i=



= e



dn = -1.6 x 10-19 x 1013 = -1.6 x 10-6 A dt



 p = v0 i = 30 x 103 x (1.6 x 10-6) = 48mW



P.P.1.9



Minimum monthly charge



= $12.00



First 100 kWh @ $0.16/kWh



= $16.00



 Next 200 kWh @ $0.10/kWh



= $20.00



Remaining 100 kWh @ $0.06/kWh = $6.00 Total Charge = $54.00 Average cost = $54/[100+200+100] = 13.5 P.P.1.10



cents/kWh



This assigned practice problem is to apply the detailed problem solving technique to some of the more difficult problems of Chapter 1.



P.P.1.6



 p = v i = 15 x 120 = 1800 watts; w = p x t therefore, t = w/p = (30x103)/1800 =



P.P.1.7



16.667 seconds



 p1 = 5(-8) = –40w  p2 = 2(8) = 16w  p3 = 0.61(3) = 0.6(5)(3) = 9w  p4 = 3(5) = 15w



P.P.1.8



i=



= e



dn = -1.6 x 10-19 x 1013 = -1.6 x 10-6 A dt



 p = v0 i = 30 x 103 x (1.6 x 10-6) = 48mW



P.P.1.9



Minimum monthly charge



= $12.00



First 100 kWh @ $0.16/kWh



= $16.00



 Next 200 kWh @ $0.10/kWh



= $20.00



Remaining 100 kWh @ $0.06/kWh = $6.00 Total Charge = $54.00 Average cost = $54/[100+200+100] = 13.5 P.P.1.10



cents/kWh



This assigned practice problem is to apply the detailed problem solving technique to some of the more difficult problems of Chapter 1.



February 5, 2006



CHAPTER 2



P.P.2.1



i = V/R = 110/12 = 9.167 A



P.P.2.2



(a)



v = iR = 2 mA[10 kohms] = 20 V



(b)



G = 1/R = 1/10 kohms = 100 µS



(c)



p = vi = 20 volts[2 mA] = 40 mW



P.P.2.3



 p = vi which leads to i = p/v = [20 cos2 (t) mW]/[10cos(t) mA] or i = 2cos(t) mA R = v/i = 10cos(t)V/2cos(t)mA 10cos(t)V/2cos(t)mA = 5 k 



P.P.2.4



5 branches and 3 nodes. nodes. The 1 ohm and 2 ohm ohm resistors are in parallel. parallel. The 4 ohm resistor and the 10 volt source are also in parallel.



P.P.2.5



Applying KVL to the loop we get: -10 + 4i – 8 + 2i = 0 which leads to i = 3A v1  = 4i = 12V  and v2  = -2i =  –6V



P.P.2.6



Applying KVL to the loop we get: -35 + 10i + 2v x + 5i = 0 But, vx  = 10i and v0  = -5i. Hence, -35 + 10i + 20i + 5i = 0 which leads to i = 1A. Thus, vx = 10V and v0 =  –5V



Applying KCL, 6 = i0 + [i0  /4] + [v0  /8], but i0 = v0/2



P.P.2.7



Which leads to: 6 = (v0/2) + (v0/8) + (v0/8) thus, v 0 = 8V  and i0 = 4A P.P.2.8



2



+ -



5V



i1



+ V1 -



4 + V3 -



i2 + V2 -



Loop 1



i3



+



Loop 2



8



3V



At the top node,



i1 = i2 + i3



(1)



For loop 1 or



-5 + V1 + V2 = 0 V1  = 5 - V2 



(2)



For loop 2 or



- V2 + V3  -3 = 0 V3 = V2 + 3



(3)



Using (1) and Ohm’s law, we get (V1/2) = (V2/8) + (V3/4) and now using (2) and (3) in the above yields [(5- V2)/2] = (V2/8) + (V2+3)/4 or



V2 = 2 V



V1 = 5- V2 = 3V, V3 = 2+3 = 5V, i1 = (5-2)/2 = 1.5A, i2 = 250 mA, i3 = 1.25A 2



P.P.2.9



R eq



6



1



3



4



5



4



3



Combining the 4 ohm, 5 ohm, and 3ohm resistors in series gives 4+3+5 = 12. But, 4 in parallel with 12 produces [4x12]/[4+12] = 48/16 = 3ohm. So that the equivalent circuit is shown below. 2



Req



3



3



6



1



Thus, R eq = 1 + 2 + [6x6]/[6+6] = 6



20



P.P.2.10



8



Req



5



20



18



1 9 2



Combining the 9 ohm resistor and the 18 ohm resistor yields [9x18]/[9+18] = 6 ohms.



Combining the 5 ohm and the 20 ohm resistors in parallel produces [5x20/(5+20)] = 4 ohms We now have the following circuit: 8



4



6



1



20 2



The 4 ohm and 1 ohm resistors can be combined into a 5 ohm resistor in parallel with a 20 ohm resistor. This will result in [5x20/(5+20)] = 4 ohms and the circuit shown below: 8



4



6



2



The 4 ohm and 2 ohm resistors are in series and can be replaced by a 6 ohm resistor. This gives a 6 ohm resistor in parallel with a 6 ohm resistor, [6x6/(6+6)] = 3 ohms. We now have a 3 ohm resistor in series with an 8 ohm resistor or 3 + 8 = 11ohms. Therefore: R eq = 11 ohms



P.P. 2.11 8S



4S



8||4 = 8+4 = 12S 12 S



Geq



Geq 2S



4S



2||4 = 2+4 = 6S



12 S in series with 6 S = {12x6/(12+6)] = 4 or:



Geq = 4 S



6S



12



P.P.2.12



15V



-



+ v



i1



i2 +



6



+ -



+ v1



10



15V



40



v2



+ -



+



4 8



v2



-



-



6||12 = [6x12/(6+12)] = 4 ohm and 10||40 = [10x40/(10+40)] = 8 ohm. Using voltage division we get: v1 = [4/(4+8)] (15) = 5 volts, v2 = [8/12] (15) = 10 volts



i1 = v1/12 = 5/12 = 416.7 mA, i2 = v2/40 = 10/40 = 250 mA P1 = v1 i1 = 5x5/12 = 2.083 watts, P2 = v2 i2 = 10x0.25 = 2.5 watts



P.P.2.13 i1



1k



+ 3k



v1



-



i2



+ 10mA



4k 5k



20k



v2



-



4k



10mA



Using current division, i 1 = i2 = (10 mA)(4 kohm/(4 kohm + 4 kohm)) = 5mA (a)



v1  = (3 kohm)(5 mA) = 15 volts v2  = (4 kohm)(5 mA) = 20 volts



(b)



For the 3k ohm resistor, P1 = v1 x i1  = 15x5 = 75 mw For the 20k ohm resistor, P2 = (v2)2  /20k = 20 mw The total power supplied by the current source is equal to: P = v2 x 10 mA = 20x10 = 200 mw



(c)



P.P.2.14 R a = [R 1 R 2 + R 2 R 3 + R 3 R 1]/ R 1 = [10x20 + 20x40 + 40x10]/10 = 140 ohms R b = [R 1 R 2 + R 2 R 3 + R 3 R 1]/ R 2  = 1400/20 = 70 ohms R c = [R 1 R 2 + R 2 R 3 + R 3 R 1]/ R 3  = 1400/40 = 35 ohms



P.P.2.15



We first find the equivalent resistance, R. We convert the delta sub-network to a wye connected form as shown below: 13



i



a



24



100V + -



a



13 20



30



10



24



10



a’



 b’



6



50



10



n



 b



15 b



c’



R a’n = 20x30/[20 + 30 + 50] = 6 ohms, R  b’n  = 20x50/100 = 10 ohms R c’n  = 30x50/100 = 15 ohms. Thus, R ab = 13 + (24 + 6)||(10 + 10) + 15 = 28 + 30x20/(30 + 20) = 40 ohms. i = 100/ R ab  = 100/40 = 2.5 amps



P.P.2.16



For the parallel case, v = v0  = 110volts. p = vi i = p/v = 40/110 = 364 mA For the series case, v = v0/N = 110/10 = 11 volts i  = p/v = 40/11 = 3.64 amps (a)



We use equation (2.61) -3 -3 R 1  = 50x10 / (1-10 ) = 0.05/999 = 50 mΩ (shunt)



(b)



R 2 = 50x10 /(100x10  – 10 ) = 50/99 = 505 mΩ (shunt)



(c)



R 3  = 50x10 /(10x10 -10 ) = 50/9 = 5.556 Ω (shunt)



P.P.2.17



-3



-3



-3



-3



-3



-3



February 5, 2006



CHAPTER 3



P.P.3.1



1A



6



i1



1



i1



i2



1A



2



4A i3



2



4A



7



At node 1, 1 = i1 + i2



1=



or 6 = 4v1 - v2 



(1)



v1



− v2



+



v1



= 4+



v2



6



−0 2



At node 2, i1



v1



= 4 + i3



− v2 6



−0 7



or 168 = 7v1 - 13v2  (2) Solving (1) and (2) gives v1 = –2 V, v2 = –14 V 2



i1



P.P.3.2



4ix v1



i2



i2 3



10 A



v2



v3 ix 4



i3 6



At node 1, 10 = i1 + i2 =



v1 − v 3



+



− v2



v1



2 or 60 = 5v1 - 2v2 - 3v3 



3 (1)



At node 2,



i2



v1



+ 4i x = i x



− v2 3



+3



v2 4



=0



or 4v1 + 5v2 = 0



(2)



At node 3,



v1



i1 = i3 + 4ix



− v3 2



=



v3



−0 6



+4



v2 4



or -3v1 + 6v2 + 4v3 = 0



(3)



Solving (1) to (3) gives v1 = 80 V, v2 = –64 V, v3 = 156 V



P.P.3.3 4



v



7V



v1



- +



+



3



+



2



v



-



6



-



At the supernode in Fig. (a),



7−v 4



v



v1



3



2



= +



+



or 21 = 7v + 8v1 



v1 6 (1)



Applying KVL to the loop in Fig. (b), - v - 3 + v1 = 0



+



+



v



v1



-



(b)



(a)



v1 = v + 3



(2)



3V



Solving (1) and (2), v = – 200 mV v1 = v + 3 = 2.8, i1 =



v1 2



= 1.4



i1 = 1.4 A P.P.3.4 v1



v2



3V



v3



 + -



-



+



+



+



v1



v2



v3



-



-



-



(a)



(b)



From Fig. (a),



v1



+



v2



+



v3



=0



6v1 + 3v2 + 4v3 = 0



(1)



- v1 + 10 + v2 = 0



v1 = v2 + 10



(2)



- v2 - 5i + v3 = 0



v3 = v2 + 5i



(3)



2



4



3



From Fig. (b),



Solving (1) to (3), we obtain v1 = 3.043V, v2 = –6.956 V, v3 = 652.2 mV P.P.3.5 We apply KVL to the two loops and obtain



- 12 + 18ii - 12i2 = 0 8 + 24i2 - 12i1 = 0 From (1) and (2) we get i1 = 666.7 mA, i2 = 0A



+



3ii - 2i2 = 2



(1)



- 3i1 + 6i2 = -2



(2)



P.P.3.6 For mesh 1,



- 20 + 6i1 – 2i2 - 4i3 = 0



3i1 - i2 - 2i3 = 10



(1)



For mesh 2, 10i2 - 2i1 - 8i3 - 10i0  = 0 = -i1 + 5i2 – 9i3 



(2)



But i0 = i3, 18i3 - 4i1 - 8i2 = 0



- 2i1 - 4i2 + 9i3 = 0



From (1) to (3),



⎡ 3 − 1 − 2⎤ ⎡ i 1 ⎤ ⎢ − 1 5 − 9⎥ ⎢i ⎥ ⎥ ⎢ 2⎥ ⎢ ⎢⎣− 2 − 4 9 ⎥⎦ ⎢⎣i 3 ⎥⎦ 3



−1 Δ = −2 3



−1 10 0



Δ1 =



0 10 0



−1 − 2 5 −9 − 4 9  = 135 - 8 - 18 - 20 - 108 - 9 = - 28 −1 − 2 5 −9 −1 − 2 5 −9 − 4 9 = 450 − 360 = 90 −1 − 2 5 −9



0



−2 −9



0



9



3



10



−1



0



−2 −9



3



−1 Δ2 = − 2



=



⎡10⎤ ⎢0⎥ ⎢ ⎥ ⎢⎣ 0 ⎥⎦



10



= 180 + 90 = 270



(3)



−1



10



−1 5 Δ3 = − 2 − 4 3 −1 −1 5



0



3



i1 =



= 40 + 100 = 140



0 10 0



Δ Δ 270 140 Δ1 90 = −9.643 , i3 = 3 = = −5A = = −3.214, i2 = 2 = Δ − 28 Δ − 28 Δ − 28



i0 = i3 = –5A P.P.3.7 2



i3



2



i1 6V



i1



i3



2



2



4



+



-



4



+



3A



-



3A



i2



8



i2 8



1



i1



0 i2 (a)



(b)



For the supermesh, - 6 + 2i1 - 2i3 + 12i2 - 4i3 = 0



i1 + 6i2 - 3i3 = 3



(1)



For mesh 3, 8i3 - 2i1 - 4i2 = 0



- i1 - 2i2 + 4i3 = 0



At node 0 in Fig. (a), i1 = 3 + i2



i1 - i2 = 3



Solving (1) to (3) yields i1 = 3.474A, i2 = 473.7 mA, i3 = 1.1052A



(2)



P.P.3.8 G11 = 1/(1) + 1/(10) + 1/(5) = 1.3, G12 = -1/(5) = -0.2, G33 = 1/(4) + 1 = 1.25, G44 = 1/(2) + 1/(4) = 0.75, G12 = -1/(5) = - 0.2, G 13 = - 1, G14 = 0, G21 = -0.2, G23 = 0 = G26, G31 = -1, G32 = 0, G34 = - 1/4 = - 0.25, G41 = 0, G42 = 0, G43 = 0.25, i1 = 0, i2 = 2 - 1 = 1, i3 = - 1, i4 = 3.



Hence, 0 ⎤ ⎡ 1.3 − 0.2 − 1 ⎢− 0.2 0.2 ⎥ 0 0 ⎢ ⎥ ⎢ −1 0 1.25 − 0.25⎥ ⎢ ⎥ 0 − 0.25 0.75 ⎦ ⎣ 0 P.P.3.9



⎡ v1 ⎤ ⎡ 0 ⎤ ⎢v ⎥ ⎢ 3 ⎥ ⎢ 2⎥ = ⎢ ⎥ ⎢ v3 ⎥ ⎢− 1⎥ ⎢ ⎥ ⎢ ⎥ ⎣v4 ⎦ ⎣ 3 ⎦



R 11 = 50 + 40 + 80 = 170, R 22 = 40 + 30 + 10 = 80, R 33 = 30 + 20 = 50, R 44 = 10 + 80 = 90, R 55 = 20 + 60 = 80, R 12 = -40, R 13 = 0, R 14 = -80, R 15 = 0, R 21 = -40, R 23 = -30, R 24 = -10, R 25 = 0, R 31 = 0, R 32 = -30, R 34 = 0, R 35 = -20, R 41 = -80, R 42 = -10, R 43 = 0, R 45 = 0, R 51 = 0, R 52 = 0, R 53 = -20, R 54 = 0, v1 = 24, v2 = 0, v3 = -12, v4 = 10, v5 = -10



Hence the mesh-current equations are



⎡ 170 − 40 0 − 80 0 ⎤ ⎢− 40 80 − 30 − 10 0 ⎥ ⎢ ⎥ ⎢ 0 − 30 50 0 − 20⎥ ⎢ ⎥ − − 80 10 0 90 0 ⎢ ⎥ ⎢⎣ 0 − 20 0 0 80 ⎥⎦ P.P.3.10



⎡ i1 ⎤ ⎢i ⎥ ⎢ 2⎥ ⎢i 3 ⎥ ⎢ ⎥ ⎢i 4 ⎥ ⎢⎣i 5 ⎥⎦



=



⎡ 24 ⎤ ⎢ 0 ⎥ ⎢ ⎥ ⎢− 12⎥ ⎢ ⎥ ⎢ 10 ⎥ ⎢⎣− 10⎥⎦



The schematic is shown below. It is saved and simulated by selecting Analysis/Simulate. The results are shown on the viewpoints:



v1 = –40 V, v2 = 57.14 V, v3 = 200 V



-40.0000



P.P.3.11



57.1430



200.0000



The schematic is shown below. After saving it, it is simulated by choosing Analysis/Simulate. The results are shown on the IPROBES. i 1 = –428.6 mA, i2 = 2.286 A, i3 = 2 A



-4.286E-01



2.286E+00 2.000E+00



P.P.3.12



For the input loop, 3



-5 + 10 x 10  IB + VBE + V0 = 0



(1)



For the outer loop, -V0 - VCE - 500 I0 + 12 = 0



(2)



But



V0 = 200 IE 



(3)



Also



IC = βIB = 100 IB, α = β/(1 + β) = 100/(101) IC = αIE



IE = IC/(α) = βIB/(α)



IE = 100 (101/(100)) IR  = 101 IB 



(4)



From (1), (3) and (4), 10,000 IB + 200(101) IR  = 5 - VBE IB =



5 − 0.7 10,000 + 20,000



= 142.38μA



V0 = 200 IE = 20,000 IB = 2.876 V From (2), -6



VCE = 12 - V0 - 500 IC = 9.124 - 500 x 100 x 142.38 x 10 VCE = 1.984 V {often, this is rounded to 2.0 volts}



P.P.3.13



20 k



i1 i0



iC 30 k



iB



+ 20 k



+



1V



+



-



v0



-



VBE -



+



-



22V



First of all, it should be noted that the circuit in the textbook should have a 22V source on the right hand side rather than the 10 V source.



iB =



1 − 0.7



B



30k 



= 10μA,  iC = βiB = 0.8 mA



i1 = iC + i0  Also,



-20ki0 – 20ki1 + 22 = 0



(1) i1 = 1.1 mA – i0 



Equating (1) and (2), 1.1mA – i0 = 0.8 mA + i0 v0 = 20 ki0 = 20 x 0.15 = 3 V



i0 = 150 A



(2)



February 5, 2006



CHAPTER 4



P.P.4.1 6



i2 i1



+ 2



iS



By current division, i 2



2



=



is



2+6+4 2 v 0 = 4i 2 = i s 3 2 When is = 15A, v 0 = (15) = 10V 3 2 When is = 30A, v 0 = (30) = 20V 3



=



1 6



4



vo



is



P.P.4.2 12



v1



+



+



VS = 10 V



Let v0 = 1. Then i =



1 8



5



−



 and v 1



giving vs = 2.5V. If vs = 10V, then v0 = 4V



=



1 8



(12



+



8)



=



2 .5



vo



8



P.P.4.3



Let v0 = v1 + v2, where v1 and v2 are contributions to the 20-V and 8-A sources respectively. 3



5



i



+ v1



+



2



−



(a)



3



i2



i1



5



+ v2



8A



2



(b)



To get v1, consider the curcuit in Fig. (a). (2 + 3 + 5)i = 20 v1 = 2i = 4V



i = 20/(10) = 2A



To get v2, consider the circuit in Fig. (b). i1 = i2 = 4A, v2 = 2i2 = 8V Thus, v = v1 + v2 = 4 + 8 = 12V



20 V



P.P.4.4 Let vx = v1 + v2, where v1 and v2 are due to the 10-V and 2-A sources



respectively. 20



v1



+



10 V



20



(a)



v2



2A



4



(b)



To obtain v1, consider Fig. (a).



0.1v1



+



10 − v1 20



=



v1



v1 = 2.5



4



For v2, consider Fig. (b). 2 + 0.1v2 +



0 − v2 20



=



vx = v1 + v2 = 12.5V



0.1v1



4



−



v2 4



v2 = 10



0.1v2



P.P.4.5



Let i = i1 + i2 + i3



where i1, i2, and i3 are contributions due to the 16-V, 4-A, and 12-V sources respectively.



2 6



2



8



6



8 4A



i1 16V



+



i2



−



(a)



(b) 6



2



8 i3 12V



+ −



(c)



For i1, consider Fig. (a), i 1



=



16 6+2+8



= 1A



For i2, consider Fig. (b). By current division, i 2



For i3, consider Fig. (c), i 3



=



2 2 + 14



( 4 ) = 0 .5



− 12



= −0.75A 16 Thus, i = i1 + i2 + i3 = 1 + 0.5 - 0.75 = 750mA



P.P.4.6



=



Combining the 6-Ω and 3-Ω resistors in parallel gives 6 3 =



6x3



= 2Ω . 9 Adding the 1-Ω and 4-Ω resistors in series gives 1 + 4 = 5 Ω. Transforming the left current source in parallel with the 2-Ω resistor gives the equivalent circuit as shown in Fig. (a).



2



5V −



+



io



+



10V



−



7



5



3A



7



5



3A



(a)



io 7.5A



2



(b) io 10.5A



(10/7)



7



(c)



Adding the 10-V and 5-V voltage sources gives a 15-V voltage source. Transforming the 15-V voltage source in series with the 2-Ω resistor gives the equivalent circuit in Fig. (b). Combining the two current sources and the 2-Ω and 5-Ω resistors leads to the circuit in Fig. (c). Using circuit division, 10 io



=



7 10 7



+7



(10.5) = 1.78 A



P.P.4.7



We transform the dependent voltage source as shown in Fig. (a). We combine the two current sources in Fig. (a) to obtain Fig. (b). By the current division principle, ix



=



5 15



(4 − 0.4i x )



ix = 1.176A



ix 4A



10



0.4ix



5



(a)



ix 4 – 0.4ix A



10



5



(b) P.P.4.8



To find R Th, consider the circuit in Fig. (a). 6



6



4



RTh



(a) 6 + 2A



6



2A



(b)



R Th



=



( 6 + 6) 4 =



12 x 4 18



=3



4



VTh



To find VTh, we use source transformations as shown in Fig. (b) and (c). 6



6 + 4



+



24 V



VTh



−



(c)



Using current division in Fig. (c), VTh



i=



=



4 4 + 12



VTh R Th



+1



( 24) = 6V



=



6 3 +1



=



1.5A



P.P.4.9 To find VTh, consider the circuit in Fig. (a).



5



6V



+ −



3



Ix



a



+



i2 i1



VTh



4 1.5Ix i2



i1 o



b (a)



0.5Ix



5



3



Ix



1.5Ix



i



a



+



4



(b)



−



b



1V



Ix = i2 i2 - i1 = 1.5Ix = 1.5i2



i2 = -2i1 



(1)



For the supermesh, -6 + 5i1 + 7i2 = 0



(2)



From (1) and (2), i2 = 4/(3)A VTh = 4i2 = 5.333V To find R Th, consider the circuit in Fig. (b). Applying KVL around the outer loop, 5(0.5I x ) − 1 − 3I x = 0 1 i = − I x = 2.25 4 1 1 R Th =  = = 444.4 m 2.25 i



P.P.4.10



Ix = -2



Since there are no independent sources, VTh = 0



4vx



10



+



−



+ vx



+ 5



15



vo



io



(a) 4vx



10



+



−



+ vx



15 +



5



vo



i



+ –



15io



(b)



To find R Th, consider Fig.(a). Using source transformation, the circuit is transformed to that in Fig. (b). Applying KVL, ).



But vx = -5i. Hence, 30i - 20i + 15io = 0 vo = (15i + 15io) = 15(-1.5io + io) = -7.5io R Th = vo/(io) = –7.5



10i = -15io



P.P.4.11 3



3



6



RN



(a)



3



5A



3



4A



(b)



From Fig. (a), R  N = (3 + 3) 6 = 3



From Fig. (b), I N =



1 2



(5 + 4) = 4.5A



IN



P.P.4.12



2vx i



+



−



+



+ 6



vx



2



ix



vx



+



1V



−



(a)



2vx



+



−



+ 6



2



10 A



Isc



vx



(b)



To get R  N consider the circuit in Fig. (a). Applying KVL, 6 i x But vx = 1, 6ix = 3 ix = 0.5 v i = i x + x = 0.5 + 0.5 = 1 2 1 R  N = R Th = = 1 i



−



2v x



−



1



=



0



To find I N, consider the circuit in Fig. (b). Because the 2Ω resistor is shorted, v x = 0 and the dependent source is inactive. Hence, I N = isc = 10A. P.P.4.13



Fig. (a).



We first need to find R Th and VTh. To find R Th, we consider the circuit in



vx



+



v0 4



−



+



i



2



vx



4



−



2 1



1



+ −



+



1V



+ −



9V



VTh



io +



3vx



3vx



(a)



(b)



Applying KCL at the top node gives



1 − vo 4



+



3v x



−



vo



1



=



vo 2



But vx = -vo. Hence



1 − vo 4



− 4v o =



1 − vo



1−



vo



vo = 1/(19)



2 1



19 = 9 4 4 38 R Th = 1/i = 38/(9) = 4.222Ω i=



=



To find VTh, consider the circuit in Fig. (b), -9 + 2io + io + 3vx = 0 But vx = 2io. Hence, 9 = 3io + 6io = 9io



io = 1A



VTh = 9 - 2io = 7V R L = R Th = 4.222 Pmax



=



2 v Th



4R L



=



49 4(4.222)



=



2.901W



+



P.P.4.14



We will use PSpice to find Voc and Isc which then can be used to



find VTh and R th.



Clearly Isc = 12 A



Clearly VTh = Ioc = 5.333 volts . R Th  = Voc/Isc = 5.333/12 = 444.4 m-ohms.



P.P.4.15



The schematic is the same as that in Fig. 4.56 except that the 1-k Ω resistor is replaced by 2-k Ω resistor. The plot of the power absorbed by R L is shown in the figure  below. From the plot, it is clear that the maximum power occurs when R L = 2k Ω and it is 125 W.



VTh = 9V, R Th



P.P.4.16



=



(v oc − VL )



R L VL



=



(9 − 1)



20 8



=



2.5Ω



2.5 + 9V



VL



=



10 10 + 2.5



(9) = 7.2V



+ −



VL



10



P.P.4.17



R 1 = R 3 = 1k Ω, R 2 = 3.2k Ω R  R x = 3 R 2 = R 2 = 3.2k R 1



P.P.4.18



We first find R Th and VTh. To get R Th, consider the circuit in Fig. (a). R Th



=



20 30 + 60 40 =



20 x 30 50



+



60 x 40 100



= 12 + 24 = 36Ω



20



30



20



30



a



+



v2



a



+ VTh



RTh



+ v 1



b



60



40



60



b



40



10 V



+



(a)



−



(b)



To find VTh, we use Fig. (b). Using voltage division, v1



But



−



v1



=



+



60 100



v2



(16) = 9.6,



+ v Th =



v2



20



=



50



0



(16) = 6.4



vTh = v1 - v2 = 9.6 - 6.4 = 32V IG



=



VTh R Th



+ R m



=



3.2 3.6 + 1.4



=



64mA



February 5, 2006



CHAPTER 5



The equivalent circuit is shown below:



P.P.5.1



2M



vd vs



+



+



-



40 k



1



+



5k



2



At node 2,



v s − v1 2x10 6



=



Av d  − v 0 50



v1 5x10 3



+



+ Avd



v0



-



-



-



At node 1,



50



+



20 k



v1



i0



+



v1 − v 0 40x10 3



v1 − v 0 40x10 3



=



v1 =



v S  + 50v 0 451



 



(1)



v0 20x133



But vd  = v1 - vS. 5



[2 x 10  (v1 - vS) - v0] 4000/(5) + v1 - v0 = 2v0 1600 x 10  (vS - v1) + 803v0 ≅ 0 5



Substituting v1 in (1) into (2) gives 8



1.5914523 x 10  vS - 17737556v0 = 0 v0 vS



=



1.5964523x10 8 17737556



= 9.00041



If vS = 1 V, v0 = 9.00041 V, v1 = 1.0000455 -5



vd  = vS - v1 = - 4.545 x 10 Av d  − v 0 = 657 A Avd  = - 9.0909, i0 = 50



(2)



P.P.5.2 20 k



i V1



10 k



VS



+



+



V2



-



+ V0



-



At node 1,



v S − v1 10



=



v1 − v 0 20



But v1 = v2 = 0,



vS 10



i0 =



=−



v0



v0



20



vS



0 − v0 20x103



=−



v0 20x10 3



When vs = 2V, v0 = -4, i0 =



P.P.5.3



v0 = −



i=



P.P.5.4



R 2 R 1



0 − v0 15k 



(a) iS =



= –2



vi =



4 x10 −3 20



− 15 5



 = 200 A



(40mV) = –120 mV



= 8 A



0 − v0



v0



R 



iS



= −R 



iS



R1



0V



V1



R2



R3



20 k



+



+



V2



V0



-



(b)



0 − v1



At node 2, iS =



At node 1,



v1 = -iSR 1 



R 1



0 − v1 R 1



=



⎛  1



-v1 ⎜⎜



⎝ R 1



v1 − 0 R 2



+



1 R 2



+



+



v1 − v 0 R 3



1  ⎞



− v0 ⎟⎟ = R 3  ⎠ R 3



⎛  1



v0 = -iSR 1R 3 ⎜⎜



(1)



⎝ R 1



+



1 R 2



+



1  ⎞



⎟



R 3  ⎠⎟



⎛  R  R   ⎞ = − R 1 ⎜⎜1 + 3 + 3 ⎟⎟ iS ⎝  R 1 R 2  ⎠



v0



P.P.5.5



By voltage division v1 =



8 4+8



(3) = 2V



where v1 is the voltage at the top end of the 8k Ω resistor. Using the formula for noninverting amplifier,



⎛  ⎝ 



v0 = ⎜1 +



5 ⎞



⎟( 2) =  7 V



2 ⎠



This is a summer.



P.P.5.6



8 8 ⎡8 ⎤ v 0 = − ⎢ (1.5) + ( 2) + (1.2) ⎥  = –3.8 V 10 6 ⎣ 20 ⎦



i0 =



R 2 R 1 R 2 R 1



8



+



v0 4



=−



3.8 8



−



3.8 4



= –1.425 mA



If the gain is 4, then



P.P.5.7



But



v0



=



=4



R 2 = 4R 1



R 4



R 4 = 4R 3



R 3



If we select R1 = R3 = 10k , then R2 = R4 = 40k P.P.5.8



v0 =



R 2 ⎛  2R 3  ⎞ ⎜1 + ⎟ (v2 - v1) R 1 ⎜⎝  R 4  ⎠⎟



R 3 = 0, R 4 = ∞, R 2 = 40k Ω, R 1 = 20k Ω v0 = i0 = P.P.5.9



40 20 v0



(8.01 − 8) = 0.02



0.02



=



10k  10x10 3



= 2 A



Due to the voltage follower va = 4V



For the noninverting amplifier,



⎛  ⎝ 



v0 = ⎜1 +



i0 =



v b 4



6 ⎞



⎟ va = (1 + 1.5) (4) = 10V



4 ⎠



mA



-



a +



+



vS



+



+



b



-



4k



v0 6k



-



But v b = va = 4 i0 = P.P.5.10



4 4



= 1mA



As a voltage follower, va = v1 = 2V



where va is the voltage at the right end of the 20 k Ω resistor. As an inverter, v b = −



50 10



v 2 = −7.5V



Where v b is the voltage at the right end of the 50k Ω resistor. As a summer 60 ⎤ ⎡ 60 v b v0 = − ⎢ v a + 30 ⎥⎦ ⎣ 20



= [6 - 15] = 9V The schematic is shown below. When it is saved and run, the results are P.P.5.11 displayed on 1PROBE and VIEWPOINT as shown. By making vs = 1V, we obtain v0 = 9.0027V  and i0 = 650.2 µA



6.502E-04



9.0027



P.P.5.12



or



-V0 =



R f  R 1



V1 +



R f  R 2



V2 +



R f 



V3



R 3



V0 = V1 + 0.5V2 + 0.25V3



(a)



If [V1V2V3] = [010], V0 = 0.5V



(b)



If [V1V2V3] = [110], V0 = 1 + 0.5 = 1.5V



(c)



If



V0



= 1.25, then V1 = 1, V2 = 0, V3 = 1, i.e.



[V1V2V3] = [101] (d)



V0



= 1.75, then V1 = 1, V2 = 1, V3 = 1, i.e.



[V1V2V3] = [111]



P.P.5.13



Av = 1 +



2R  R G



R G =



R G =



2R  Av −1



2x 25x10 3 142 − 1



= 354.6



February 5, 2006



CHAPTER 6



q 



v=



P.P.6.1



w



=



=



=  40V



3x10 −6 1 Cv 2 = x 3x10 − 6 x1600 = 2.4mJ 2 2



C 1



i( t ) = C



P.P.6.2



120x10 −6



dv dt



= 10 x10 −6



d  dt



(50 sin 2000 t )



= cos2000t A v=



P.P.6.3



=−



1



∫



t



C 0 500



idt



=



10 −3



t



∫ 50 sin 120πt dt V



0.1x10 −3



cos 120πt 0t



0



50



(1 − cos 120πt )V 120π 12π 50 (1 − cos 0.12π) = 93.14mV v(t = 1ms) = 12π 50 (1 − cos 0.6π) = 1.7361 V v(t = 5ms) = 12π



i(t) =



P.P.6.4



v



=



1



C∫



idt



=



−3



∫ idt ⋅10 1



−3



= ∫ idt



50 t dt = 25t  x 10 C∫ 1 v = ∫ 100dt + v( 2) = (100 t − 0.2 + 0.1) C



For 0< t 0, the current source is cut off and the RL circuit is shown in Fig. (b). L 2 τ= = = 0.5 R eq  = (12 + 8) || 5 = 20 || 5 = 4 Ω , R eq  4 i( t ) = i(0) e - t τ = 2 e -2t A, t



0



For t < 0, the switch is closed. The inductor acts like a short so the equivalent circuit is shown in Fig. (a). P.P.7.5



3 i i



io



1H



io 6A



4



2 4



(a)



i=



4 (6) = 4 A , 4+ 2



2



(b)



io = 2 A ,



v o = 2i = 8 V



For t > 0, the current source is cut off so that the circuit becomes that shown in Fig. (b). The Thevenin equivalent resistance at the inductor terminals is L 1 τ= = R th = (4 + 2) || 3 = 2 Ω , R th 2 3 (-i) - 1 - 4 -2t 8 = i= io = e   and v o = -2i o = e -2t 6+3 3 3 3



Thus, t 0



i= ⎨



P.P.7.6



8V t 0



⎧ 0 t 4,



t



I = 20 − 10t



4 2



t 2



= 40 − 10 t



=0



Thus,



⎧ 0 t 0, 20 u (-t) = 0 so that the voltage source is replaced by a short circuit. Transforming the current source leads to the circuit below. 10



i



10



+ v



0.2 F



+ −



30 V



5 (30) = 10 15 10 Ω, R th = 5 || 10 = 3 v(∞) =



τ = R th C =



10 2 × 0.2 = 3 3



v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ v( t ) = 10 + (20 − 10) e -3t 2 v( t ) = 10 ( 1 + e -1.5 t ) i( t ) = i( t ) =



P.P.7.12



- v( t ) = -2 ( 1 + e -1.5 t ) 5 0



t



0



e -1.5t A



t



0



-2 1



20 V



v( t ) =



10 1



e - 1.5t V



0



t



0



Applying source transformation, the circuit is equivalent to the one below. i



1.5 H



10



t=0



5



+ −



30 V



At t < 0, the switch is closed so that the 5 ohm resistor is short circuited. 30 i ( 0 − ) = i (0) = =3A 10 For t > 0, the switch is open. R th = 10 + 5 = 15 ,



i(∞) =



t



τ=



L 1.5 = = 0.1 R th 15



30 =2A 10 + 5



i( t ) = i(∞) + [ i(0) − i(∞)] e - t τ i( t ) = 2 + (3 − 2) e -10t i ( t ) = 2 e-10t A, t 0



P.P.7.13



For 0 < t < 2, the given circuit is equivalent to that shown below. 10



20 i(t)



6A



5H



15



Since switch S1 is open at t = 0 − , i(0 − ) = 0 . Also, since i cannot jump, i(0) = i(0 − ) = 0 . 90 =2A i(∞) = 15 + 10 + 20 L 5 1 = = R th = 45 Ω , τ = R th 45 9 i( t ) = i(∞) + [ i(0) − i(∞)] e - t τ i( t ) = 2 + (0 − 2) e -9 t i( t ) = 2 (1 − e -9 t ) A When switch S 2 is closed, the 20 ohm resistor is short-circuited. i(2 + ) = i(2 − ) = 2 (1 − e -18 ) ≅ 2 This will be the initial current 90 = 3.6 A i(∞) = 15 + 10 5 1 = R th = 25 Ω , τ = 25 5 i( t ) = i(∞) + [ i(2 + ) − i(∞)] e -( t − 2 ) τ i( t ) = 3.6 + (2 − 3.6) e -5( t − 2 ) i( t ) = 3.6 − 1.6 e -5( t − 2 ) 0



Thus, i ( t ) =



t



2 (1 e -9t ) A 3.6 1.6 e -5( t



2)



0 A



0 t



t



2 2



At t  = 1 ,



i(1) = 2 (1 − e -9 ) = 1.9997 A



At t  = 3 ,



i(3) = 3.6 − 1.6 e -5 = 3.589 A



P.P.7.14



The op amp circuit is shown below. C



+



v



−



Rf  1 2



R1



− +



+ vo



Since nodes 1 and 2 must be at the same potential, there is no potential difference across R 1 . Hence, no current flows through R 1 . Applying KCL at node 1, v dv dv v + C = 0  ⎯   ⎯→ + =0 R f  dt dt CR f  which is similar to Eq. (7.4). Hence, v( t ) = v o e - t τ , τ = R f C



v(0) = v o = 4 ,



τ = (50 × 10 3 )(10 × 10 -6 ) = 0.5



v( t ) = 4 e -2 t V, t > 0 Alternatively, since no current flows through R 1 , the feedback loop forms a first order RC circuit with v(0) = 4 and τ = R f C = 0.5 . Hence, v( t ) = 4 e -2 t V, t > 0 To get to v o from v, we notice that v is the potential difference between node 1 and the output terminal, i.e.  ⎯→ v o = -v 0 − v o = v  ⎯  v o = - 4 e -2t V , t



0



Let v1 be the potential at the inverting terminal. v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ v ( 0) = 0 where τ = RC = 100 × 10 3 × 10 -6 = 0.1 , P.P.7.15



v1  = 0  for all t v1 − v o = v  



(1)



For t > 0, the switch is closed and the op amp circuit is an inverting amplifier with - 100 v o (∞) = ( 4 mV) = -40 mV 10 From (1), v(∞) = 0 − v o (∞) = 40 mV Thus, v( t ) = 40 1 e -10t mV v o = v1 − v = -v v o = 40 e -10t



1 mV



This is a noninverting amplifier so that the output of the op amp is ⎛  R f  ⎞ ⎟ vi v a = ⎜1 + ⎝  R 1  ⎠



P.P.7.16



⎛  R f  ⎞ ⎛  40 ⎞ ⎟ v i = ⎜1 + ⎟ 2 u ( t ) = 6 u ( t ) v th = v a = ⎜1 + ⎝  20 ⎠ ⎝  R 1  ⎠ To get R th , consider the circuit shown in Fig. (a), where R o is the output resistance of the op amp. For an ideal op amp, R o = 0  so that R th = R 3 = 10 k Ω R3



Ro



Rth Rth



R2



(a)



τ = R th C = 10 × 10 3 × 2 × 10 -6 =



Vth



+ −



C



(b)



1 50



The Thevenin equivalent circuit is shown in Fig. (b), which is a first order circuit. Hence,



v o (t) = 6 ( 1 − e - t τ ) u (t)



v o ( t ) = 6 1 e -50 t u(t ) V



The schematic is shown in Fig. (a). Construct and save the schematic. Select Analysis/Setup/Transient to change the Final Time to 5 s. Set the Print Step slightly greater than 0 (20 ns is default). The circuit is simulated by selecting Analysis/ Simulate. In the Probe menu, select Trace/Add and display V(R2:2) as shown in Fig. (b). P.P.7.17



(a)



(b)



The schematic is shown in Fig. (a). While constructing the circuit, rotate L1 counterclockwise through 270 ° so that current i(t) enters pin 1 of L1 and set IC = 10 for L1. After saving the schematic, select Analysis/Setup/Transient to change the Final Time to 1 s. Set the Print Step slightly greater than 0 (20 ns is default). The circuit is simulated by selecting Analysis/ Simulate. After simulating the circuit, select Trace/Add in the Probe menu and display I(L1) as shown in Fig. (b). P.P.7.18



(a)



(b)



P.P.7.19



v(0) = 0 . When the switch is closed, we have the circuit shown below. 10 k



R



a



+ 9V



80 F



b



We find the Thevenin equivalent at terminals a-b. 10 ( R + 4) R th = ( R + 4) || 10 = R + 14 v th = v(∞) =



R + 4 (9) R + 14



v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ ,



τ = R th C



v( t ) = v(∞) ( 1 − e - t τ ) Since v(0) = 0 , v( t ) 9 ( 1 − e -t τ ) mA i( t ) = = R + 4 R + 4 Assuming R is in k Ω, 9 ( 1 − e -t 0 τ ) × 10 -3 R + 14 R + 14 = 1 − e -t 0 τ (0.12) 9 0.12R + 1.68 7.32 − 0.12R  = e -t 0 τ = 1 − 9 9



120 × 10 -6 =



or



⎛   ⎞ 9 ⎟ t 0 = τ ln ⎜ ⎝ 7.32 − 0.12R  ⎠ 10 ( R + 4) ⎛   ⎞ 9 ⎟ t0 = × 80 × 10 -6 × ln ⎜ ⎝ 7.32 − 0.12R  ⎠ R + 14 When R = 0,



⎛  9  ⎞ 40 × 80 × 10 -6 ⎟ = 0.04723 s t0 = × ln ⎜ ⎝ 7.32 ⎠ 14



4k



When R = 6 k Ω, 100 ⎛  9  ⎞ × 80 × 10 - 6 × ln ⎜ t0 = ⎟ = 0.124 s 20 ⎝ 6.6 ⎠ The time delay is between 47.23 ms and 124 ms. P.P.7.20 



(a) (b) (c) (d) (e)



q = CV = (2 × 10 -3 )(80) = 0.16 C 1 1 CV 2 = ( 2 × 10 -3 )(6400) = 6.4 J 2 2 Δ q  0.16 ΔI= = = 200 A Δ t 0.8 × 10 -3 Δw 6.4 = = 8 kW  p = Δ t 0.8 × 10 -3 Δ q  0.16 Δt = = = 32 s Δ I 5 × 10 -3 W=



L 500 × 10 -3 τ= = = 2.5 ms P.P.7.21 R  200 110 i (0 ) = 0 , = 550 mA i(∞) = 200 i( t ) = 550 ( 1 − e - t τ ) mA 350 mA = i( t 0 ) = 550 ( 1 − e - t 0 τ ) mA 35 20 = 1 − e - t 0 τ  ⎯   ⎯→ e - t 0 τ = 55 55 e t0 τ =



55 20



⎛ 55 ⎞ ⎛ 55 ⎞ t 0 = τ ln ⎜ ⎟ = 2.5 ln ⎜ ⎟ ms ⎝ 20 ⎠ ⎝ 20 ⎠ t 0 = 2.529 ms P.P.7.22 



(a) (b) (c)



5L 5 × 20 × 10 -3 = = 20 ms t = 5τ = R  5 2 ⎛   ⎞ 1 2 1 12 W = LI = ( 20 × 10 -3 ) ⎜ ⎟ = 57.6 mJ ⎝  5  ⎠ 2 2 ⎛  12 5  ⎞ di ⎟ = 24 kV V = L = 20 × 10 -3 ⎜ ⎝ 2 × 10 -6  ⎠ dt



February 5, 2006



CHAPTER 8



P.P.8.1



(a)



At t = 0-, we have the equivalent circuit shown in Figure (a). 10



i



+ 2



v



vL



a +



24V



+ −



2



vC



(a)



+ i



50mF



+ −



(b)



i(0-) = 24/(2 + 10) = 2 A, v(0-) = 2i(0-) = 4 V hence, v(0+) = v(0-) = 4V. (b)



At t = 0+, the switch is closed. L(di/dt) = vL, leads to (di/dt) = vL/L But,



vC(0+) + vL(0+) = 24 = 4 + vL(0+), or vL(0+) = 20 (di(0+)/dt) = 20/0.4 = 50 A/s C(dv/dt) = iC  leading to (dv/dt) = iC/C



But at node a, KCL gives i(0+) = iC(0+) + v(0+)/2 = 2 = iC(0+) + 4/2 or iC(0+) = 0, hence (dv(0+)/dt) = 0 V/s (c) As t approaches infinity, the capacitor is replaced by an open circuit and the inductor is replaced by a short circuit. v(∞) = 24 V, and i(∞) = 12 A.



24V



P.P.8.2



(a)



At t = 0-, we have the equivalent circuit shown in (a).



5



5



a iR



2A



iL 3A



+ vC



+



b



vR



−



10 F



+ vL



2H 3A



(a)



(b)



iL(0-) = -3A, vL(0-) = 0, vR (0-) = 0 At t = 0+, we have the equivalent circuit in Figure (b). At node b, iR (0+) = iL(0+) + 3, since iL(0+) = iL(0-) = -3A, iR (0+) = 0, and vR (0+) = 5iR (0+) = 0. Thus, iL(0) = –3 A, vC(0) = 0, and vR (0+) = 0. (b)



dvC(0+)/dt = iC(0+)/C = 2/0.2 = 10 V/s. To get (dvR/dt), we apply KCL to node b, iR = iL + 3, thus diR /dt = diL/dt. Since vR = 5iR , dvR/dt = 5diR/dt = 5diL/dt. But LdiL/dt = vL, diL/dt = vL/L. Hence, dvR(0+)/dt = 5vL(0+)/L. Applying KVL to the middle mesh in Figure (b), -vC(0+) + vR(0+) + vL(0+) = 0 = 0 + 0 + vR (0+), or vR (0+) = 0 Hence, dvR (0+)/dt = 0 = diL(0+)/dt; diL(0+)/dt = 0, dvC(0+)/dt = 10 V/s, dvR (0+)/dt = 0.



(c)



As t approaches infinity, we have the equivalent circuit shown below.



5 2A



iL 3A



2



= 3 + iL(∞) leads to iL(∞) = -1A vC(∞) = vR (∞) = 2x5 = 10V



Thus, iL(∞) = –1 A, vC(∞) = vR (∞) = 10 V



P.P.8.3



(a)



α 



= R/(2L) = 10/(2x5) = 1, s1,2 =



(b)



Since α
0, we have a source-free RLC circuit. ωo = 1



α 



LC



=1



1x



1 9



=



3



= R/(2L) = 5/(2x1) = 2.5



Since α < ωo, we have an underdamped case. s1,2 =



−α±



α



2



i(t) = e



2



− ωo = −2.5 ±



6.25 − 9 = -2.5 ± j1.6583



-2.5t



[A1cos1.6583t + A2sin1.6583t]



We now determine A1 and A2 . i(0) = 5 = A1 -2.5t



di/dt = -2.5{e [A1cos1.6583t + A2 sin1.6583t]} -2.5t + 1.6583e [-A1sin1.6583t + A2cos1.6583t] di(0)/dt = -(1/L)[Ri(0) + v(0)] = -2.5A1 + 1.6583A2 = -1[25] = -2.5(5) + 1.6583A2 A2 = -7.5378 -2.5t



Thus, i(t) = e



P.P.8.5



α 



-3



= 1/(2RC) = 1/(2x2x25x10 ) = 10 ω0 = 1



since α =



ωo,



[5cos1.6583t – 7.538sin1.6583t] A



LC



=1



0.4x 25x10 −3



=  10



we have a critically damped response. Therefore, -10t



v(t) = [(A1 + A2t)e



] -10t



v(0) = 0 = A1 + A2x0 = A1, which leads to v(t) = [A2 te



].



-3



dv(0)/dt = -(v(0) + Ri(0))/(RC) = -2x3/(2x25x10 ) = -120 dv/dt = [(A2 - 10A2t)e At t = 0,



-10t



] –10t



-120 = A2  therefore, v(t) = (–120t)e



 volts



P.P.8.6 For t < 0, the switch is closed. The inductor acts like a short circuit while the capacitor acts like an open circuit. Hence,



i(0) = 2 and v(0) = 0. α



-3



= 1/(2RC) = 1/(2x20x4x10 ) = 6.25 ωo = 1



Since α >



ωo,



LC



10 x 4 x10 −3 = 5



=1



this is an overdamped response.



s1,2 =



−α±



α



2



− ωo = −6.25 ±



(6.25) 2 -2.5t



Thus, v(t) = A1e



−



25 = -2.5 and –10 -10t



+ A2 e



v(0) = 0 = A1 + A2 , which leads to A2 = -A1 dv(0)/dt = -(v(0) + Ri(0))/(RC) = -12.5(2x20) = -500 But, dv/dt = -2.5A1e



-2.5t



-10t



-10A2e



At t = 0, -500 = -2.5A1 -10A2 = 7.5A1 since A1 = -A2 A1 = -66.67,



–10t



v(t) = 66.67(e



Thus,



P.P.8.7



A2 = 66.67 –e



–2.5t



) V



The initial capacitor voltage is obtained when the switch is in position a. v(0) = [2/(2 + 1)]12 = 8V



The initial inductor current is i(0) = 0. When the switch is in position b, we have the RLC circuit with the voltage source. α



= R/(2L) = 10/(2x2.5) = 2



ωo = 1



LC



=1



(5 / 2) x (1 / 40) = 4



Since



α




0, α=



0, ωo



=1



LC



=1



0.2x5 = 1



i(t) = is + A1cost + A2sint = 20 + A1 cost + A2 sint i(0) = 0 = 20 + A1 , therefore A1  = -20 Ldi(0)/dt = v(0) = 0



But di/dt = -A1sint + A2cost At t = 0, di(0)/dt = 0 = 0 + A2 leading to i(t) = 20(1 – cost) A v(t) = Ldi/dt = 5x20sint = 100sint V



P.P.8.9



At t = 0, the switch is open so that v(0) = 0, i(0-) = 0



(1)



For t > 0, the switch is closed. We have the equivalent circuit as in Figure (a).



iC



i



10



iC 4



2A (1/20)F



i



10 +



2H



4 2A



v



(a)



(b)



v(0+) = 0, i(0+) = 0



(2)



-2 + iC + i = 0



(3)



From (3), i(0+) = 0 means that iC(0+) = 2, but iC (0+) = Cdv(0+)/dt which leads to dv(0+)/dt = iC (0+)/C = 2/(1/20) = 40 V/s As t approaches infinity, we have the equivalent circuit in (b). i(∞) = 2 A, v(∞) = 4i(∞) = 8V



(5)



 Next we find the network response by turning off the current source as shown in Figure (c).



iC



i



10



4 i



+ v



(1/20)F



2H (c)



Applying KVL gives



-v – 10iC + 4i + 2di/dt = 0



Applying KCL to the top node,



(6)



i – iC = 0



 Namely,



i = iC = -Cdv/dt = -(1/20)dv/dt



(7)



Combining (6) and (7),



-v – (10/20)dv/dt – (4/20)dv/dt – (2/20)d v/dt   = 0.



2



or



2



2



2



(d  v/dt ) + 7(dv/dt) + 10 = 0 2



The characteristic equation is s + 7s + 10 = 0 = (s + 2)(s + 5) -2t



This means that vn = (Ae Thus, v = vf  + vn



-2t



v = 8 + (Ae



-5t



+ Be ) and vf  = v(∞) = 8. -5t



+ Be )



(8)



v(0) = 0 = 8 + A +B, or A + B = -8 -2t



(9)



-5t



dv/dt = (-2Ae   -5Be ) dv(0)/dt = 40 = -2A – 5B 2A + 5B = -40



(10)



From (9) and (10), A = 0 and B = -8. -5t



Thus, v(t) = 8(1 – e ) V for all t > 0.



But, from (3), i = 2 – iC  = 2 –(1/20)dv/dt = 2 –(1/20)(40)e -5t



i(t) = 2(1 – e ) A for all t > 0.



-5t



P.P.8.10



For t 0, the circuit is as shown in Figure (a).



1



1



v1



1



v2



1 +



5V



+ −



0.5F



(1/3)F



5V



+



+



v1



−



(a)



v2



(b)



i1  = C1dv1/dt, or dv1 /dt = i1 /C1; likewise dv2/dt = i2/C2 i2(0+) = (v1(0+) – v2(0+))/1 = (0 – 0)/1 = 0 (5 – v1(0+))/1 = i1(0+) + i2(0+), or 5 = i1(0+) Hence,



dv1 (0+)/dt = 5/(1/2) = 10 V/s



(2a)



dv2(0+)/dt = 0



(2b)



As t approaches infinity, the capacitors can be replaced by open circuits as shown in Figure (b). Thus, v1(∞) = v2(∞) = 5V



(3)



 Next we obtain the network response by considering the circuit in Figure (c).



1



1



v1



0.5F



(c)



v2 (1/3)F



Applying KCL at node 1 gives (v1/1) + (1/2)(dv1/dt) + (v1 – v2)/1 = 0 or



v2 = 2v1  + (1/2)dv1/dt



(4)



Applying KCL at node 2 gives (v1 – v2)/1 = (1/3)dv2/dt or v1 = v2 + (1/3)dv2/dt



(5)



Substituting (5) into (4) yields, 2



v2 = 2v2 +(2/3)(dv2/dt) + (1/2)(dv2/dt) + (1/6)d  v2/dt or,



2



2



2



(d  v2/dt ) + (7dv2/dt) + 6v2 = 0 2



 Now we have, s + 7s + 6 = 0 = (s + 1)(s + 6) -t



-6t



Thus, v2n  = (Ae  + Be ) and v2f  = v2(∞) = 5V. -t



-6t



v2 = v2n + v2f  = 5 + (Ae  + Be ) v2(0) = 0 which implies that A + B = -5 -t



(6)



-6t



dv2/dt = (-Ae  - 6Be ) dv2(0) = 0 = -A – 6B



(7)



From (6) and (7), A = -6 and B = 1. -t



-6t



Thus, v2(t) = (5 – 6e  + e ) V From (5),



v1 = v2 + (1/3)dv2/dt -t



-6t



Thus, v1(t) = (5 – 4e  – e ) V



 Now we can find,



–t



–6t



vo = v1 – v2 = (2e  – 2e )V, t > 0



Let v1 equal the voltage at non-inverting terminal of the op amp. P.P.8.11 Then vo is equal to the output of the op amp. At the non-inverting terminal, (vs – vo)/R 1 = C1dv1/dt



(1)



At the output terminal of the op amp, (v1 – vo)/R 2 = C2dvo/dt



(2)



We now eliminate v1 from (2), v1 = vo + R 2C2dvo/dt



(3)



From (1)



(4)



vs = v1 + R 1C1dv1/dt



Substituting (3) into (4) gives 2



2



vs = vo + R 2C2dvo/dt + R 1C1dvo/dt + R 1C1R 2C2d  vo/dt or



2



2



d  vo/dt  + [(1/(R 1C1)) + (1/(R 2C2))]dvo/dt + vo/(R 1R 2C1C2) = vs/(R 1R 2C1C2)



With the given parameters, 4



4



-6



-6



-2



(R 1R 2C1C2) = 10 x10 x20x10 x100x10   = 2x10 1/(R 1R 2C1C2) = 5 -4



-6



-6



[(1/(R 1C1)) + (1/(R 2C2))] = 10 [(1/20x10 ) + (1/200x10 )] = 6 2



Hence, we now have s + 6s + 5 = 0 = (s +1)(s + 5) -t



-5t



-t



-5t



Therefore von = Ae  + Be , and vof   = 4V Thus,



vo = 4 + Ae  + Be  



(5)



For t < 0, vs  = 0, v1(0-) = 0 = vo(0-) For t > 0, vs  = 4, but v1(0+) - vo(0+) =0 From (2),



dvo(0+)/dt = [v1(0+) – vo(0+)]/R 2C2 = 0



(6) (7)



Imposing these conditions on vo(t), 0 = 4+A+B



(8)



0 = -A – 5B or A = -5B



(9)



From (8) and (9), A = -5 and B = 1 -t



-5t



vo(t) = (4 – 5e  + e ) V, t > 0



P.P.8.12 We follow the same procedure as in Example 8.12. The schematic is shown in Figure (a). (a). The current marker is inserted inserted to display the inductor current. current. After simulating the circuit, the required inductor current is plotted in Figure (b).



P.P.8.13 When the switch is at position a, the schematic is as shown in Figure (a). We carry out dc analysis on this to obtain initial conditions. It is evident that vC(0) (0) = 8 volt volts. s.



(a)



With the switch in position b, the schematic is as shown in Figure (b). A voltage marker is inserted to display the capacitor voltage. When the schematic is saved and run, the output is as shown in Figure (c).



(b)



P.P.8.14 The dual circuit is obtained from the original circuit as shown in Figure (a). It is redrawn redrawn as shown in Figure (b). (b). 3H 50mA



3F 4H 10



50mV



+ −



0.1



(a)



4 F



3H 0.1 50mV



+



4F



−



(b)



P.P.8.15



The dual circuit is obtained in Figure (a) and redrawn in Figure (b).



5



0.2F



4H



0.2 4F



0.2 H



2A



1/3



2V



+



+



3



−



20A



−



(a)



20 V



1/3



4F



0.2 H



2V



0.2



+



20A



−



(b)



P.P.8.16



Since 12 = 4i + vL + vC



or vC = 12 12 – 4i - vL



-250t



-(vC – 12) = 4i + vL = e



-250t



vC(t) = [12 – 12e



(12cosωdt + 0.2684sinωdt – 268sinωd t) -250t



cos11.18t + 267.7e



sin11.18t] V



We follow the same procedure as in Example 8.17. The schematic P.P.8.17 is as shown in Figure (a) with two voltage markers to display both input and output voltages. When the schematic is saved saved and run, the result is as displayed in Figure (b). (b).



(a)



(a)



(b)



February 5, 2006



CHAPTER 9



P.P.9.1



amplitude = 5  phase = -60 angular frequency ( ω) = 4π = 12.57 rad/s 2π  period (T) =  = 0.5 s



ω



frequency (f) =



1  = 2 Hz T



i1 = -4 sin(ωt + 25°) = 4 cos(ωt + 25° + 90°) i1 = 4 cos(ωt + 115°) , ω = 377  rad/s Compare this with i 2 = 5 cos(ωt − 40°) indicates that the phase angle between i1 and i 2 is 115° + 40°  = 155° i1 leads i2 by 155 Thus, P.P.9.2



P.P.9.3



P.P.9.4



(a)



(5 + j2)(-1 + j4) = -5 + j20 – j2 – 8 = -13 + j18 5∠60° = 2.5 + j4.33 (5 + j2)(-1 + j4) – 5 ∠60° = -15.5 + j13.67 [ (5 + j2)(-1 + j4) – 5 ∠60 ]* = -15.5 – j13.67 = 20.67 221.41



(b)



3∠40° = 2.298 + j1.928 10 + j5 + 3∠40°  = 12.298 + j6.928 = 14.115 ∠29.39 ° -3 + j4 = 5∠126.87 ° 10 +  j5 + 3∠40° 14.115∠29.39° = = 2.823∠ - 97.48° - 3 +  j4 5∠126.87° 2.823∠-97.48 ° = -0.3675 – j2.8 10∠30° = 8.66 + j5 10 +  j5 + 3∠40° + 10∠30° = 8.293 + j2.2 - 3 +  j4



(a)



-cos(A) = cos(A – 180°) = cos(A + 180°) Hence, v = -7 cos(2t + 40°) = 7 cos(2t + 40° + 180°) v = 7 cos(2t + 220°) The phasor form is V = 7 220  V



(b)



Since sin(A) = cos(A – 90°),



i = 4 sin(10t + 10°) = 4 cos(10t+10° – 90°) i = 4 cos(10t – 80°) The phasor form is I = 4 -80  A P.P.9.5



(a)



Since -1 = 1∠-180 ° = 1∠180° V  = -10∠30° = 10∠(30°+180 °) = 10∠210° The sinusoid is v(t) = 10 cos( t + 210 ) V



(b)



I  = j (5 – j12) = 12 + j5 = 13 ∠22.62°



The sinusoid is i(t) = 13 cos( t + 22.62 ) A P.P.9.6 



Let V = -10 sin( ωt + 30°) + 20 cos( ωt − 45°) Then, V = 10 cos( ωt + 30° + 90°) + 20 cos(ωt − 45°) Taking the phasor of each term V = 10∠120° + 20∠-45° V = -5 + j8.66 + 14.14 – j14.14 V  = 9.14 – j5.48 = 10.66 ∠-30.95 ° Converting V to the time domain v(t) = 10.66 cos( t – 30.95 )V



P.P.9.7



Given that dv 2 + 5v + 10 v dt = 20 cos(5t − 30°) dt we take the phasor of each term to get 10 ω = 5 2jω V +5 V + V = 20∠-30°,  jω V [j10 + 5 – j(10/5)] = V (5 + j8) = 20∠-30° 20∠ - 30° 20 ∠ - 30° = V = 5 +  j8 9.434∠58° V  = 2.12∠-88° Converting V to the time domain v(t) = 2.12 cos(5t – 88 )V



∫



P.P.9.8



For the capacitor, V = I / (jωC),where V = 6∠-30°, ω  = 100 -6 I = jωC V  = (j100)(50x10 )(6∠-30°) I = 30∠60° mA i(t) = 30 cos(100t + 60 ) mA



P.P.9.9



Vs = 5∠0°,



ω  = 10



Z = 4 + jωL = 4 + j2



5∠0° 5 (4 −  j2) =   = 1 – j0.5 = 1.118∠-26.57 ° 4 +  j2 16 + 4 V = jωL I  = j2 I = (2∠90°)(1.118∠-26.57°) = 2.236∠63.43 ° I = Vs / Z =



Therefore,



v(t) = 2.236 sin(10t + 63.43 ) V i(t) = 1.118 sin(10t – 26.57 ) A



P.P.9.10 Z1 = impedance of the 2-mF capacitor in series with the 20-Ω resistor Let Z2 = impedance of the 4-mF capacitor Z3 = impedance of the 2-H inductor in series with the 50- Ω resistor



1 1 = 20 + = 20 −  j50  jωC  j (10)(2 × 10 -3 ) 1 1 Z2 = = = -j25  jωC  j (10)(4 × 10 -3 ) Z3 = 50 +  jωL = 50 +  j (10)(2) = 50 +  j 20 Z1 = 20 +



Zin = Z1 + Z2 || Z3 = Z1 + Z2 Z3 / (Z2 + Z3)



- j25x (50 +  j20) - j25 + 50 +  j20 = 20 – j50 + 12.38 – j23.76 = 32.38 – j73.76



Zin = 20 −  j50 + Zin Zin



In the frequency domain, the voltage source is Vs = 10∠75° the 0.5-H inductor is j ωL = j (10)(0.5) = j5 1 1 1 the -F capacitor is = = - j2 20  jωC  j (10)(1 20)



P.P.9.11



Let and



Z1 = impedance of the 0.5-H inductor in parallel with the 10- Ω resistor Z2 = impedance of the (1/20)-F capacitor



(10)( j5) Z2  = -j2 = 2 + j4 and 10 +  j5 Vo = Z2 / (Z1 + Z2) Vs −  j (10∠75°) 10∠(75° − 90°) −  j2 = Vo = (10∠75°) = 2 +  j4 −  j2 1 +  j 2 ∠45° Vo  = 7.071∠-60° vo(t) = 7.071 cos(10t – 60 ) V Z1  = 10 || j5 =



We need to find the equivalent impedance via a delta-to-wye transformation as shown below. P.P.9.12



c Zcn



Zan 30 0  V



n



Zbn



+



− b



a 5 10 -2



4 (-5 +  j8)  j4 (8 +  j5)  = = 0.32 + j3.76 8 +  j6  j4 + 8 +  j5 −  j3 3 (5 −  j8)(8 −  j6) - j3 (8 +  j5) =  =  = -0.24 – j2.82 100 8 +  j6  j4 (- j3) 12 (8 −  j6) =  =  = 0.96 – j0.72 100 8 +  j6



Zan = Z bn Zcn



The total impedance from the source terminals is Z = Zcn + (Zan + 5 – j2) || ( Z bn + 10) Z = Zcn + (5.32 + j1.76) || (9.76 – j2.82) (5.32 +  j1.76) (9.76 −  j2.82) Z = Zcn + (5.32 +  j1.76) + (9.76 −  j2.82) Z = 0.96 – j0.72 + 3.744 + j0.4074 Z  = 4.704 – j0.3126 = 4.714 ∠-3.802 ° Therefore, I = V / Z =



30 ∠0° 4.714 ∠ − 3.802°



I = 6.364 3.802  A



Let us now check this using PSpice. The solution produces the magnitude of I = 6.364E+00, and the phase angle = 3.803E+00, which agrees with the above answer.



To show that the circuit in Fig. (a) meets the requirement, consider the equivalent circuit in Fig. (b). P.P.9.13



Z = -j10 || (10 – j10) =



10



V1



10



+ Vi



- j10 (10 −  j10) - j (10 −  j10)  = = 2 – j6 Ω 10 −  j20 1 −  j2 10 +



+ - 10



-j10



(a)



Vo



Vi = 10 V



+



V1



− (b)



Z = 2  j6



V1 =



2 −  j6 10 (10) = (1 −  j) 10 + 2 −  j6 3



⎛  - j  ⎞⎛ 10 ⎞ 10 - j10 ⎟⎜ ⎟ (1 −  j)  = - j V1 = ⎜ 3 10 −  j10 ⎝ 1 −  j ⎠⎝  3  ⎠ 10 Vo = ∠ - 90° 3 This implies that the RC circuit provides a 90 ° lagging phase shift. 10 The output voltage is  = 3.333 V 3 Vo =



P.P.9.14



the 1-mH inductor is j ωL =  j (2π)(5 × 10 3 )(1 × 10 -3 ) = j31.42 the 2-mH inductor is j ωL =  j (2π)(5 × 10 3 )(2 × 10 -3 ) = j62.83 Consider the circuit shown below. 31.42



V1



 j62.83



+ Vi



+ 10



50



Vo



(10)(50 +  j62.83) 60 +  j62.83 Z  = 9.205 + j0.833 = 9.243 ∠5.17° Z = 10 || (50 + j62.83) =



9.243∠5.17° (1)   = 0.276∠-68.9° 9.205 +  j32.253 50 (0.276 ∠ - 68.9°) 50 Vo = V1 =  = 0.172∠-120.4 ° 80.297 ∠51.49° 50 +  j62.83 V1 = Z / (Z + j31.42) Vi =



Therefore, magnitude = 0.172  phase = 120.4  phase shift is lagging



P.P.9.15



Zx = (Z3 / Z1) Z2 Z3  = 12 k Ω Z1  = 4.8 k Ω Z2  = 10 + j ωL = 10 +  j ( 2π )(6 × 10 6 ) (0.25 × 10 -6 ) = 10 + j9.425 Zx =



12k  (10 + j9.425) = 25+ j23.5625 Ω 4.8k 



R x = 25, Xx  = 23.5625 = ωLx Xx 23.5625 = = 0.625 μH Lx = 2πf  2π (6 × 10 6 ) i.e. a 25-



resistor in series with a 0.625- H inductor.



February 5, 2006



CHAPTER 10



P.P.10.1



 ⎯→ 10 ∠0°, ω = 2 10 sin( 2t )  ⎯ 



 ⎯→  jωL =  j4 2 H  ⎯  1  ⎯→ = - j2.5 0.2 F  ⎯   jωC Hence, the circuit in the frequency domain is as shown below. -j2.5



V1



4



V2



+ 10 0  A



At node 1,



At node 2,



10 =



2



V1



+



V1



Vx



4



3Vx − V2   where Vx = V1  j4 -  j2.5 4 - j2.5V2 =  j4 (V1 − V2 ) + 2.5 (3V1 − V2 ) 0 = - (7.5 +  j4) V1 + (2.5 +  j1.5) V2  



=



V1



3Vx



− V2



2 - j2.5 100 = (5 +  j4) V1 −  j4V2   V2



+



− V2



(1)



+



(2)



Put (1) and (2) in matrix form. ⎡ 5 +  j4 -  j4 ⎤⎡ V1 ⎤ ⎡100 ⎤ ⎢⎣ - (7.5 +  j4) 2.5 +  j1.5⎥⎦⎢⎣ V ⎥⎦ = ⎢⎣ 0 ⎥⎦ 2 where Δ = (5 +  j4)( 2.5 +  j.15) − (-j4)(-(7.5 +  j4)) = 22.5 −  j12.5 = 25.74 ∠ - 29.05 °



⎡2.5 +  j1.5  j4 ⎤ ⎡ V1 ⎤ ⎢⎣ 7.5 +  j4 5 +  j4⎥⎦ ⎡100⎤ ⎢V ⎥ = ⎢ 0 ⎥ 22.5 −  j12.5 ⎣ 2⎦ ⎣ ⎦ 2.5 +  j1.5 2.915∠30.96° V1 = (100) = (100) = 11.32∠60.01° 22.5 −  j12.5 25.74∠ - 29.05 ° 7.5 +  j4 8.5∠28.07° V2 = (100) = (100) = 33.02∠57.12° 22.5 −  j12.5 25.74∠ - 29.05°



In the time domain, v1 (t ) = 11.32 sin(2t + 60.01 ) V v 2 ( t ) = 33.02 sin(2t + 57.12 ) V P.P.10.2



The only non-reference node is a supernode. 15 − V1 V1 V2 V2 = + + 4  j4 -  j 2 15 − V1 = - j V1 +  j4V2 + 2V2 15 = (1 −  j) V1 + (2 +  j4) V2  



(1)



The supernode gives the constraint of V1 = V2 + 20∠60°



(2)



Substituting (2) into (1) gives 15 = (1 − j)(20∠60°) + (3 +  j3) V2 15 − (1 −  j)(20 ∠60°) 14.327 ∠210.72° = = 3.376∠165.7° V2 = 3 +  j3 4.243∠45° V1 = V2 + 20∠60° = (-3.272 +  j0.8327) + (10 +  j17.32) V1 = 6.728 +  j18.154



=



Therefore,



V1



19.36 69.67  V,



P.P.10.3



Consider the circuit below.



V2



=



3.376 165.7  V



2 0 A



I3 -2



8



For mesh 1,



6



I1



(8 −  j2 +  j4) I1 −  j4 I 2 = 0 (8 +  j2) I1 =  j4 I 2



4



I2



+



−



10 30  V



(1)



For mesh 2,



(6 +  j4) I 2 −  j4 I 1 − 6 I 3 + 10∠30° = 0



For mesh 3,



I3



= -2



Thus, the equation for mesh 2 becomes (6 + j4) I 2 − j4 I 1 = -12 − 10∠30°   From (1),



I2



=



(2)



8 +  j2 I = (0.5 −  j2) I1    j4 1



(3)



Substituting (3) into (2), (6 +  j4) (0.5 −  j2) I 1 −  j4 I1 = -12 − 10∠30° (11 −  j14) I 1 = -(20.66 +  j5) - (20.66 +  j5) I1 = 11 −  j14 Hence,



P.P.10.4



Io



= - I1 =



Io



=



20.66 +  j5 21.256∠13.6° = 11 −  j14 17.8∠ - 51.84°



1.194 65.44  A



Meshes 2 and 3 form a supermesh as shown in the circuit below. 10 -j4



 j8 I2



50 0  V



+



−



I1 I3 5



-j6



− 50 + (15 −  j4) I 1 − (− j4) I 2 − 5 I 3 = 0 (15 −  j4) I 1 +  j4 I 2 − 5 I 3 = 50  



(1)



For the supermesh,



( j8 −  j4) I 2 + (5 −  j6) I 3 − (5 −  j4) I 1 = 0  



(2)



Also,



I3



For mesh 1,



= I2 + 2  



(3)



Eliminating I 3 from (1) and (2) (15 −  j4) I 1 + (-5 + j4) I 2 = 60   (-5 +  j4) I 1 + (5 −  j2) I 2 = -10 +  j12  



(4) (5)



From (4) and (5), ⎡15 −  j4 - 5 +  j4⎤⎡ I 1 ⎤ ⎡ 60 ⎤ ⎢⎣ - 5 +  j4 5 - j2 ⎥⎦⎢⎣ I ⎥⎦ = ⎢⎣ - 10 +  j12 ⎥⎦ 2 15 −  j4 - 5 +  j4 = 58 −  j10 = 58.86∠ - 9.78° - 5 +  j4 5 -  j2



Δ= Δ1 =



60



- 5 +  j4



- 10 +  j12



5 -  j2



= I1 =



Δ1 = Δ



= 298 −  j20 = 298.67 ∠ - 3.84°



Thus,



Io



P.P.10.5



Let I o = I 'o + I "o , where I 'o and I "o are due to the voltage source and



5.074 5.94  A



current source respectively. For I 'o consider the circuit in Fig. (a). -2 Io 8



6



'



I1



4



I2



+



−



10 30  V



(a)



For mesh 1,



For mesh 2,



(8 + j2) I1 −  j4 I 2 = 0 I 2 = (0.5 −  j2) I1



(1)



(6 +  j4) I 2 − j4 I 1 − 10∠30° = 0  



(2)



Substituting (1) into (2), (6 +  j4)(0.5 −  j2) I 1 −  j4 I 1 = 10∠30° 10∠30° = 0.08 +  j0.556 I 'o = I 1 = 11 −  j14



For I "o consider the circuit in Fig. (b). 2 0 A



-2



6



"



Io 8



4



(b)



Let



 j24 = 1.846 +  j2.769 Ω 6 +  j4 Z2 (2)(1.846 +  j2.769) = 0.4164 +  j0.53 I "o = (2) = Z1 + Z 2 9.846 +  j0.77 Z1



= 8 −  j2 Ω ,



Z2



= 6 ||  j4 =



= I 'o + I "o = 0.4961 +  j1.086 I o = 1.1939 65.45  A



Therefore,



Io



P.P.10.6



Let v o = v 'o + v "o , where v 'o is due to the voltage source and v "o is due to



the current source. For v 'o , we remove the current source.



 ⎯→ 30 ∠0°, ω = 5 30 sin(5t )  ⎯  1 1 0.2 F  ⎯   ⎯→ = = - j  jωC  j (5)(0.2)  ⎯→  jωL =  j (5)(1) =  j5 1 H  ⎯  The circuit in the frequency domain is shown in Fig. (a). 8 + 30 0  V



+



−



Vo



'



-



(a)



5



-  j ||  j5 = -j1.25



 Note that



By voltage division, - j1.25 (30) = 4.631∠ - 81.12 ° 8 −  j1.25 v 'o = 4.631sin(5t − 81.12°)



Vo'



Thus,



=



For v "o , we remove the voltage source.



 ⎯→ 2 ∠0°, ω = 10 2 cos(10 t )  ⎯  1 1 0.2 F  ⎯   ⎯→ = = - j0.5  jωC  j (10)(0.2)  ⎯→  jωL =  j (10)(1) =  j10 1 H  ⎯  The corresponding circuit in the frequency domain is shown in Fig (b). I



+ 8



 j10



Vo



"



-j0.5



2 0



(b)



Let



Z1  = - j0.5 ,



Z2



= 8 ||  j10 =



 j80 = 4.878 +  j3.9 8 +  j10



By current division, I= Vo"



Z2 Z1 + Z 2



(2)



= I (-j0.5) =



Z2 Z1 + Z 2



(2)(-j0.5) =



- j (4.877 +  j3.9) 4.878 +  j3.4



Thus,



6.245∠ - 51.36° = 1.051∠ - 86.24° 5.94 ∠34.88° v "o = 1.051 cos(10t − 86.24°)



Therefore,



v o = v 'o + v "o



Vo"



=



v o = 4.631 sin(5t – 81.12 ) + 1.051 cos(10t – 86.24 ) V



If we transform the current source to a voltage source, we obtain the circuit shown in Fig. (a). P.P.10.7



4



-3



2 Io



VS



1



+



 j5



−



-2



(a) Vs



= I s Z s = ( j4)(4 −  j3) = 12 +  j16



We transform the voltage source to a current source as shown in Fig. (b). V 12 +  j16 Z = 4 −  j3 + 2 +  j = 6 −  j2 . Then, Is = s = = 1.5 +  j3 . Let Z 6 −  j2 Io 6



1



IS



 j5 -j2



-j2



(b)



 Note that



Z ||  j5 =



(6 −  j2)( j5) 10 = (1 +  j) . 6 +  j3 3



By current division, Io



=



Io



=



10 (1 +  j) 3



(1.5 +  j3) 10 (1 +  j) + (1 −  j2) 3 − 20 +  j40 44.72∠116.56° = Io = 13 +  j4 13.602∠17.1° 3.288 99.46  A



When the voltage source is set equal to zero, Z th = 10 + (- j4) || (6 +  j2) (-j4)(6 +  j2) Z th = 10 + 6 - j2 Z th = 10 + 2.4 −  j3.2



P.P.10.8



Z th



=



12.4 – j3.2



By voltage division,



- j4 (- j4)(30∠20°) (30∠20°) = 6 +  j2 −  j4 6 −  j2 ( 4 ∠ - 90°)(30 ∠20°) Vth = 6.324 ∠ - 18.43° Vth



=



Vth



=



18.97 -51.57  V



To find Vth , consider the circuit in Fig. (a).



P.P.10.9



8 + j4 +



8 + j4 +



Vo



Vo



5 0 V2



V1 4 – j2



At node 2,



a



0.2Vo



(a)



At node 1,



VS a



4 – j2



b



5 + 0.2Vo +



V1 − V2



8 +  j4



= 0,



Hence, the equation for node 2 becomes



+



0.2Vo



(b)



V − V2 0 − V1 = 5+ 1 4 −  j2 8 +  j4 - (2 +  j)V1 = 50 + (1 −  j0.5)(V1 − V2 ) 50 = (1 −  j0.5)V2 − (3 +  j0.5)V1  



Is



−



b



(1) where Vo = V1 − V2 .



1 0



5 + 0.2 (V1 − V2 ) +



= V2 −



V1



− V2 =0 8 +  j4



V1



50   3 +  j0.5



(2)



Substituting (2) into (1), 50 = (1 −  j0.5)V2 − (3 +  j0.5)V2 + (50)



3 +  j0.5 3 −  j0.5



50 (35 +  j12) 37 - 2.702 +  j16.22 = 7.35∠72.9° V2 = 2 +  j



0 = -50 − ( 2 +  j) V2 +



Vth



= V2 =



7.35 72.9  V



To find Z th , we remove the independent source and insert a 1-V voltage source between terminals a-b, as shown in Fig. (b). At node a,



But, So, and



Is



= -0.2Vo +



Vs



8 +  j4 + 4 −  j2



8 +  j4 V 8 +  j4 + 4 −  j2 s 1 8 +  j4 2.6 +  j0.8 I s = (0.2) + = 12 +  j2 12 +  j2 12 +  j2 V 1 12 +  j2 12.166∠9.46° = = Z th = s = Is Is 2.6 +  j0.8 2.72∠17.10° Vs



=1



Z th



=



and



4.473 –7.64



– Vo =



P.P.10.10



To find Z N , consider the circuit in Fig. (a).



4



 j2



4



 j2



I3 8



1



-3



8



1



-3



a



a ZN



20 0



+



I1



−



b



(a)



(b)



Z N



= (4 +  j2) || (9 −  j3) =



Z N



=



IN



I2



-j4



b



(4 +  j2)(9 −  j3) 13 −  j



3.176 + j0.706



To find I N , short-circuit terminals a-b as shown in Fig. (b). Notice that meshes 1 and 2 form a supermesh. For the supermesh,



- 20 + 8 I 1 + (1 −  j3) I 2 − (9 −  j3) I 3 = 0  



Also,



I1



For mesh 3,



(13 −  j) I 3 − 8 I 1 − (1 −  j3) I 2 = 0  



= I 2 +  j4  



Solving for I 2 , we obtain I N



= I2 =



I N



=



(2)



50 −  j62 79.65∠ - 51.11° = 9 −  j3 9.487 ∠ - 18.43°



8.396 -32.68  A



Using the Norton equivalent, we can find I o as in Fig. (c). Io IN



ZN



(c)



(1)



10 – j5



(3)



By current division,



3.176 +  j0.706 (8.396∠ - 32.68°) Z N + 10 −  j5 13.176 −  j4.294 (3.254 ∠12.53°)(8.396 ∠ - 32.68°) Io = 13.858∠ - 18.05° Io



=



Io



=



Z N



I N



=



1.971 -2.10  A



P.P.10.11



1 1 = = -j20 k Ω  jωC1  j (5 × 10 3 )(10 × 10 -9 ) 1 1  ⎯→ = = -j10 k Ω 20 nF  ⎯   jωC 2  j (5 × 10 3 )(20 × 10 -9 )



 ⎯→ 10 nF  ⎯ 



Consider the circuit in the frequency domain as shown below.



- 20 k 10 k



20 k V1



2 0  V



V2



+



Io Vo



−



+



- 10 k



−



As a voltage follower, V2 = Vo At node 1,



At node 2,



2 − V1 V1 − Vo V1 − Vo = + 10 -  j20 20 4 = (3 +  j)V1 − (1 +  j)Vo   V1



− Vo



=



Vo



(1)



−0



20 -  j10 V1 = (1 +  j2) Vo



(2)



Substituting (2) into (1) gives 4 =  j6Vo



or



Vo



2 3



= ∠ - 90°



Hence,



v o ( t ) = 0.667 cos(5000 t − 90°) V v o ( t ) = 0.667 sin(5000t) V



 Now,



Io



=



But from (2)



Vo



− V1 = - j2Vo =



Io



=



Vo



− V1



-  j20k  -4 3



-4 3 = - j66.66 μA - j20k 



Hence,



i o ( t ) = 66.67 cos(5000 t − 90°) μA i o ( t ) = 66.67 sin(5000t) A



P.P.10.12 



Let Z = R ||



1 R  =  jωC 1 +  jωRC



Vs R  = Vo R + Z The loop gain is 1/ G =



Vs R  = = Vo R + Z



R  1 +  jωRC = R  2 +  jωRC R + 1 +  jωRC



where ωRC = (1000)(10 × 10 3 )(1 × 10 -6 ) = 10 1/ G =



1 +  j10 10.05∠84.29° = 2 +  j10 10.2∠78.69°



G = 1.0147 –5.6



P.P.10.13



The schematic is shown below.



 ⎯→ f  = 477.465 Hz . Setup/Analysis/AC Sweep as Since ω = 2πf  = 3000 rad / s  ⎯  Linear for 1 point starting and ending at a frequency of 447.465 Hz. When the schematic is saved and run, the output file includes Frequency 4.775E+02



IM(V_PRINT1) 5.440E-04



IP(V_PRINT1) -5.512E+01



Frequency 4.775E+02



VM($N_0005) 2.683E-01



VP($N_0005) -1.546E+02



From the output file, we obtain Vo = 0.2682 ∠-154.6 ° V



and



Io



Therefore, v o ( t ) = 0.2682 cos(3000t – 154.6 ) V i o ( t ) = 0.544 cos(3000t – 55.12 ) mA



=  0.544∠-55.12 ° mA



P.P.10.14



The schematic is shown below.



We select ω = 1 rad/s and f = 0.15915 Hz. We use this to obtain the values of capacitances, where C = 1 ωX c , and inductances, where L = X L ω . Note that IAC does not allow for an AC PHASE component; thus, we have used VAC in conjunction with G to create an AC current source with a magnitude and a phase. To obtain the desired output use Setup/Analysis/AC Sweep as Linear for 1 point starting and ending at a frequency of 0.15915 Hz. When the schematic is saved and run, the output file includes Frequency 1.592E-01



IM(V_PRINT1) 2.584E+00



IP(V_PRINT1) 1.580E+02



Frequency 1.592E-01



VM($N_0004) 9.842E+00



VP($N_0004) 4.478E+01



From the output file, we obtain Vx = 9.842 44.78  V  P.P.10.15



P.P.10.16



and



Ix



=



2.584 158  A



⎛  R 2 ⎞ ⎛  10 × 10 6 ⎞ ⎟ C = ⎜1 + ⎟( × -9 ) = C eq  = ⎜1 + 3 10 10 ⎝  10 × 10  ⎠ ⎝  R 1  ⎠



10 F



C = C1 = C 2 = 1 nF If R  = R 1 = R 2 = 2.5 k Ω   and 1 1 = = 63.66 kHz f o = 2πRC (2π)(2.5 × 10 3 )(1 × 10 -9 )



February 5, 2006



CHAPTER 11



P.P.11.1



i( t ) = 15 sin(10 t + 60°) = 15 cos(10 t − 30°) v( t ) = 80 cos(10 t + 20°)  p ( t ) = v( t ) i( t ) = (80)(15) cos(10 t + 20°) cos(10 t − 30°) 1  p( t ) = ⋅ 80 ⋅ 15 [cos(20 t + 20° − 30°) + cos( 20 − -30°)] 2  p ( t ) = 600 cos( 20t 10 ) 385.7 W



P= P.P.11.2



1 V I cos(θ v − θi ) = 385.7 W 2 m m



V = I Z = 200 ∠8°



1 V I cos(θ v − θi ) 2 m m 1 P = ( 200)(10) cos(8° − 30°) = 927.2 W 2 P=



P.P.11.3 3



8 45  V



I=



+ −



I



 j1



8∠45° = 2.53∠26.57° 3 +  j



For the resistor, I R  = I = 2.53∠26.57° VR  = 3 I = 7.59∠26.57° 1 1 PR  = Vm I m = ( 2.53)(7.59) = 9.6 W 2 2



For the inductor, I L = 2.53∠26.57° VL =  j I L = 2.53∠(26.57° + 90°) = 2.53∠116.57° 1 PL = ( 2.53) 2 cos(90°) = 0 W 2 The average power supplied is 1 P = (8)(2.53) cos( 45° − 26.57°) = 9.6 W 2 P.P.11.4



Consider the circuit below. 8



40 V



+ −



 j4



I1



I2



-j2



+ −



j20 V



For mesh 1,



- 40 + (8 −  j2) I1 + (- j2) I 2 = 0 (4 −  j) I1 −  j I 2 = 20  



(1)



- j20 + ( j4 −  j2) I 2 + (- j2) I 1 = 0 - j I 1 +  j I 2 =  j10  



(2)



For mesh 2,



In matrix form, ⎡ 4 −  j -  j⎤⎡ I 1 ⎤ ⎡ 20 ⎤ ⎢ -  j  j ⎥⎢ I ⎥ = ⎢ j10 ⎥ ⎣ ⎦⎣ 2 ⎦ ⎣ ⎦



Δ = 2 +  j4 , I1 =



Δ 1 = -10 +  j20 ,



Δ1 = 5∠53.14°   and Δ



I2 =



Δ 2 = 10 +  j60 Δ2 = 13.6∠17.11° Δ



For the 40-V voltage source, Vs = 40∠0° I 1 = 5∠53.14° -1 Ps = ( 40)(5) cos(-53.14°) = - 60 W 2



For the j20-V voltage source, Vs = 20∠90° I 2 = 13.6∠17.11° -1 Ps = (20)(13.6) cos(90° − 17.11°) = - 40 W 2



For the resistor, I = I1 = 5 V = 8 I 1 = 40 1 P = (40)(5) = 100 W 2 The average power absorbed by the inductor and capacitor is zero watts. P.P.11.5 We first obtain the Thevenin equivalent circuit across Z L . Z Th is obtained from the circuit in Fig. (a). -j4



10



8



Zth



5



(a) Z Th = 5 || (8 −  j4 +  j10) =



(5)(8 +  j6) = 3.415 +  j0.7317 13 +  j6



VTh  is obtained from the circuit in Fig. (b). -4



 j10 I



8



2A



5



(b)



By current division, I=



8 −  j4 (2) 8 −  j4 +  j10 + 5



+ Vth



VTh = 5 I =



(10)(8 −  j4) = 6.25∠ - 51.34° 13 +  j6



Z L = Z *Th = 3.415  j0.7317



Pmax = P.P.11.6



Let



VTh



2



8 R L



(6.25) 2 = = 1.429 W (8)(3.415)



We first find Z Th and VTh across R L . Z1 = 80 +  j60



(90)(- j30) = 9 (1 −  j3) 90 −  j30 (80 +  j60)(9 −  j27) = Z1 || Z 2 = = 17.181 −  j24.57 Ω 80 +  j60 + 9 −  j27



Z 2 = 90 || (- j30) = Z Th



VTh =



Z2 (9)(1 −  j3) (120∠60°) = (120∠60°) Z1 + Z 2 89 +  j33



VTh = 35.98∠ - 31.91°



R L = Z Th = 30 The current through the load is VTh 35.98∠ - 31.91° = = 0.6764 ∠ - 4.4° I= Z Th + R L 47.181 −  j24.57 The maximum average power absorbed by R L is 1 2 1 Pmax = I R L = (0.6764) 2 (30) = 6.863 W 2 2 P.P.11.7



⎧ 4t 0 < t P1 For a Y-connected load, VL 208 I p = I L , V p = = 3 3 P p



=



V p I p cos θ



Z p



=



Z p



=



V p I p



=



120 6.575



 ⎯   ⎯→



I p



=



= 120



V



80 (120)(0.1014)



=



6.575 A



= 18.25



Z p ∠θ = 18.25 84.18



The impedance is inductive . P.P.12.15



ZΔ



30 −  j40 = 50∠ - 53.13°



=



The equivalent Y-connected load is ZΔ ZY = = 16.67 ∠ - 53.13° 3 440 V p = = 254 V 3 V p 254 IL = = = 15.24 16.67 ZY



P1



=



VL I L cos(θ + 30°)



P1



=



(440)(15.24) cos(-53.13° + 30°) = 6.167 kW



P2



=



VL I L cos(θ − 30°)



P2



=



(440)(15.24) cos(-53.13° − 30°) = 0.8021 kW



PT



=



P1 + P2



=



6.969 kW



3 ( P2



−



P1 ) =



QT



=



QT



=



- 9.292   kVAR



3 (802.1 − 6167 )



February 5, 2006



CHAPTER 13 P.P. 13.1



For mesh 1,  j6 = 4(1 + j2)I1 + jI2 



(1)



For mesh 2,



0 = jI1 + (10 + j5)I2 



(2)



For the matrix form



 j ⎤ ⎡ I 1 ⎤ ⎡ j6⎤ ⎡4 +  j8 = ⎢ 0 ⎥ ⎢  j 10 +  j5⎥⎦ ⎢⎣ I 2 ⎥⎦ ⎣ ⎦ ⎣



Δ = j100, Δ2 = 6 I2 = Δ2/Δ = 6/j100 Vo = 10I2 = 60/j100 = 0.6 -90 V Since I1 enters the coil with reactance 2Ω and I2 enters the coil with P.P. 13.2 reactance 6Ω, the mutual voltage is positive. Hence, for mesh 1, 12∠60o = (5 + j2 + j6 – j 3x2)I1 – j6I2 + j3I2 12∠60o = (5 + j2)I 1 – j3I2 



or For mesh 2,



0 = (j6 – j4)I2 – j6I1 + j3I1



or



I2 = 1.5I1 



Substituting this into (1),



(1)



(2)



12∠60o = (5 – j2.5)I1



I1 = (12∠60o)/(5.59∠ –26.57o) = 2.147 86.57o A I2 = 1.5I1 = 3.22 86.57o A



P.P. 13.3



The coupling coefficient is, k = m/ L1 L 2 = 1 / 2x1  = 0.7071



To obtain the energy stored, we first obtain the frequency-domain circuit shown below. 20cos(ωt) becomes 20∠0o, ω = 2



1H becomes jω1 = j2 2H becomes jω2 = j4 (1/8) F becomes 1/jωC = -j4



-j4



4



VS



+ –



I1



 j4



 j2



For mesh 1,



20 = (4 – j4 + j4)I1 – j2I2



or



10 = 2I1 – jI2 



For mesh 2,



–j2I1 + (2 + j2)I2 = 0



or



I1 = (1 – j)I 2 



Substituting (2) into (1),



I2



2



(1)



(2)



(2 – j3)I2 = 10 I2 = 10/((2 – j3) = 2.78∠56.31o I1 = 3.93∠11.31o



In the time domain,



At t = 1.5,



i1 = 3.93cos(2t + 11.31o) i2 = 2.78cos(2t + 56.31o)



2t = 3 rad = 171.9o i1 = 3.93cos(171.9o + 11.31o) = –3.924 A i2 = 2.78cos(171.9o + 56.31o) = -1.85 A



The total energy stored in the coupled inductors is given by, W = 0.5L1(i1)2 + 0.5L2(i2)2 – 0.5M(i1i2) = 0.5(2) (-3.924)2 + 0.5(1)(-1.85)2 – (1)(-3.924)(-1.85) = 9.85 J



P.P. 13.4



Zin = 4 + j8 + [32/(j10 – j6 + 6 + j4)] = 4 + j8 + 9/(6 + j8) = 8.58 58.05o ohms



The current from the voltage is, I = V/Z = 10 ∠0o/8.58∠58.05o = 1.165 –58.05o A



P.P. 13.5



L1  = 10, L2  = 4, M = 2 L1L2 – M2  = 40 – 4 = 36 LA = (L1L2 – M2)/(L2 – M) = 36/(4 – 2) = 18 LB = (L1L2 – M2)/(L1 – M) = 36/(10 – 2) = 4.5 LC = (L1L2 – M2)/M = 36/2 = 18



Hence, we get the π equivalent circuit as shown below.



18 H



18 H



4.5 H



If we reverse the direction of i 2 so that we replace I2 by –i2, we P.P. 13.6 have the circuit shown in Figure (a).  j3



-j4



+ –



 j3



i1



 j6



i2



12



o



12 0



(a)



We now replace the coupled coil by the T-equivalent circuit and assume ω  = 1. La  = 5 – 3 = 2 H L b  = 6 – 3 = 3 H Lc  = 3 H Hence the equivalent circuit is shown in Figure (b). We apply mesh analysis.



-j4



 j2



 j3  j3



o



12 0



+ –



I1



I2



(b)



12



12 = i1(-j4 + j2 + j3) + j3i 2 or 12 = ji1 + j3i2  Loop 2 produces,



(1)



0 = j3i1 + (j3 + j3 + 12)i2 or i1 = (-2 + j4)i2 



Substituting (2) into (1),



(2)



12 = (-4 + j)i2, which leads to i2 = 12/(-4 + j) I2 = -i2  = 12/(4 – j) = 2.91 14.04o A



I1 = i1 = (-2 + j4)i2 = 12(2 – j4)/(4 – j) = 13 -49.4o A



P.P. 13.7



(a)



n = V2/V1  = 110/3300 = 1/30 (a step-down transformer)



(b)



S = V1I1  = 3300x3 = 9.9 kVA



(c)



I2 = I1/n = 3/(1/30) = 90 A



P.P. 13.8 resulting in



The 16 – j24-ohm impedance can be reflected to the primary



Zin = 2 + (16 – j24)/16 = 3 – j1.5 I1  = 100/(3 – j1.5) = 29.82∠26.57o I2 = –I1/n = –7.454∠26.57o Vo  = -j24i2  = (24∠ –90o)(–7.454∠26.57o) = 178.92 116.57oV S1 = V1I1  = (100)( 29.82∠26.57o) = 2.982 -26.57okVA



P.P. 13.9



8



+ v0 –  4



i1



1



2



1:2



2



i2 + – 60 0o



+



+



v1



v2



 – 



 – 



+ v3



10



 – 



Consider the circuit shown above. At node 1,



(60 – v1)/4 = i1 + (v1 – v3)/8



(1)



At node 2,



[(v1 – v3)/8] + [(v 2 – v3)/2] = (v3)/8



(2)



At the transformer terminals, v2  = -2v1 and i2 = -i1/2



(3)



But i2 = (v2 – v3)/2 = -i1/2 which leads to i1 = (v3 – v2)/1 = v3 + 2v1. Substituting all of this into (1) and (2) leads to,



(60 – v1)/4 = v3 + 2v1 + (v1 – v3)/8 which leads 120 = 19v1 + 7v3 



(4)



[(v1 – v3)/8] + [(-2v1 – v3)/2] = v3/8 which leads to v3  = -7v1/6



(5)



From (4) and (5), 120 = 10.833v1 or v1  = 11.077 volts v3  = -7v1/6 = -12.923 vo = v1 – v3 = 24 volts



We should note that the current and voltage of each winding of the P.P. 13.10 autotransformer in Figure (b) are the same for the two-winding transformer in Figure (a). 6A 0.5A



+



6A



10V



 –  +



6.5A



+



+



120V



10V



 – 



 – 



+



120V



120V



 – 



0.5A



(a)



 – 



+ 130V



 – 



(b)



For the two-winding transformer, s1  = 120/2 = 60 VA s2  = 6(10) = 60 VA For the autotransformer, s1  = 120(6.5) = 780 VA s2  = 130(6) = 780 VA



P.P. 13.11



i2 = s2/v2  = 16,000/800 = 20 A



Since s1 = v1i1 = v2i2 = s2, v2/v1 = i1/i2, 800/1250 = i1/20, or i1  = 800x20/1250 = 12.8 A. At the top, KCL produces i1 + io = i2, or io = i2 – i1  = 20 – 12.8 = 7.2 A.



P.P. 13.12



(a)



sT = (√3)vLiL, but sT = pT/cosθ  = 40x106/0.85 = 47.0588 MVA iLS = sT/(√3)vLS  = 47.0588x106/[(√3)12.5x103] = 2.174 kA



(b)



vLS  = 12.5 kV, vLP  = 625 kV, n = vLS/vLP  = 12.5/625 = 0.02



(c)



iLP = niLS  = 0.02x2173.6 = 43.47 A or iLP = sT/[(√3)vLP] = 47.0588x106/[(√3)625x103] = 43.47 A



(d)



The load carried by each transformer is (1/3)sT = 15.69 MVA



The process is essentially the same as in Example 13.13. We are P.P. 13.13 given the coupling coefficient, k = 0.4, and can determine the operating frequency from the value of ω = 4 which implies that f = 4/(2π) = 0.6366 Hz.



Saving and then simulating produces, io = 100.6cos(4t + 68.52o) mA



Following the same basic steps in Example 13.14, we first assume P.P. 13.14 ω = 1. This then leads to following determination of values for the inductor and the capacitor.  j15 = jωL leads to L = 15 H -j16 = 1/(ωC) leads to C = 62.5 mF



The schematic is shown below.



FREQ



VM($N_0005,0)



VP($N_0005,0)



1.592E-01



7.652E+01



2.185E+00



FREQ



VM($N_0001,0)



VP($N_0001,0)



1.592E-01



1.151E+02



2.091E+00



Thus, V1 = 76.52 2.18 V V2 = 115.1 2.09 V Note, if we divide V2 by V1 we get 1.5042 –.09  which is in good agreement that the transformer is ideal with a voltage ratio of 1.5:1! ˚



P.P. 13.15



V2/V1 = 120/13,200 = 1/110 = 1/n



P.P. 13.16



VS



+ –



Z1



+ v1



ZL /n



2



 – 



As in Example 13.16, n2 = ZL/Z1  = 100/(2.5x103) = 1/25, n = 1/5 = 0.2 By voltage division, v1 = vs/2 (since Z1 = ZL/n2), therefore v1  = 30/2 = 15 volts, and v2 = nv1  = (1/5)(15) = 3 volts



P.P. 13.17



(a)



s = 12x60 + 350 + 4,500 = 5.57 kW



(b)



iP  = s/vP  5570/2400 = 2.321 A



February 5, 2006



CHAPTER 14



P.P.14.1



H(ω) =



Vo Vs



=



 j L R  j L



 jω ω0  jωL R  = 1 +  jωL R  1 +  jω ω0 R  where ω0 = . L H(ω) =



H = H(ω) = At ω = 0 , As ω → ∞ , At ω = ω0 ,



H = 0 , H = 1 , 1 H = , 2



ω ω0 1 + (ω ω0 ) 2



φ = ∠H(ω) =



⎛  ω  ⎞ − tan -1 ⎜ ⎟ 2 ⎝ ω0 ⎠



π



φ = 90° φ = 0° φ = 90° − 45° = 45°



Thus, the sketches of H and φ are shown below. H 1 0.7071



0



ω0 = R/L



ω



φ 90° 45° 0



ω0 = R/L



ω



P.P.14.2



The desired transfer function is the input impedance. Vo (s) ⎛  1  ⎞ ⎟ || (3 + 2s) = ⎜5 + Z i (s) = I o (s) ⎝  s 10 ⎠ Z i (s) =



(5 + 10 s)(3 + 2s) 5 (s 2)(s 1.5) = s 2 4s 5 5 + 10 s + 3 + 2s



The poles are at  p1, 2 =



- 4 ± 16 − 20 = - 2  j 2



The zeros are at z1 = - 2 , P.P.14.3



H (ω) =



z 2 = - 1.5



1 +  jω 2 ( jω)(1 +  jω 10)



H db = 20 log10 1 +  jω 2 − 20 log10  jω − 20 log10 1 +  jω 10



φ = -90° + tan -1 (ω 2) − tan -1 (ω 10) The magnitude and the phase plots are shown in Fig. 14.14. P.P.14.4



H (ω) =



50 400 jω (1 +  jω 4)(1 +  jω 10) 2



H db = -20 log10 8 + 20 log10  jω − 20 log10 1 +  jω 4 − 40 log10 1 +  jω 10



φ = 90° − tan -1 (ω 4) − 2 tan -1 (ω 10) The magnitude and the phase plots are shown in Fig. 14.16. P.P.14.5



H (ω) =



10 400



⎛   jω8 ⎛  jω ⎞2 ⎞ ( jω)⎜⎜1 + + ⎜ ⎟ ⎟⎟ ⎝ 20 ⎠  ⎠ 40 ⎝ 



H db = -20 log10 40 − 20 log10  jω − 20 log10 1 +  jω 5 − ω2 400



⎛  0.2 ω  ⎞ ⎟ φ = -90° − tan -1 ⎜ ⎝ 1 − ω2 400 ⎠



The magnitude and the phase plots are shown in Fig. 14.18. P.P.14.6



1 +  jω 0.5 1 A pole at ω = 1 , 1 +  jω 1 1 Two poles at ω = 10 , (1 +  jω 10) 2 A zero at ω = 0.5 ,



Hence, H (ω) =



1 +  jω 0.5 (1 0.5)(0.5 +  jω) 2 = (1 +  jω 1)(1 +  jω 10) (1 100)(1 +  jω)(10 +  jω) 2 200 (s



H (ω) =



(s



1)(s



0. 5 ) 10 ) 2



P.P.14.7



(a)



Q=



ω0 L R 



ω0 =



 ⎯   ⎯→ ω0 =



QR  (50)(4) = = 8 × 10 3 rad / s -3 L 25 × 10



1 1 1  ⎯   ⎯→ C = 2 = 6 ω0 L (64 × 10 )(25 × 10 -3 ) LC



C = 0.625 F (b)



ω0



8 × 10 3 = = 160 rad / s B= Q 50



Since Q > 10 , B = 8000 − 80 = 7920 rad / s 2 B ω2 = ω0 + = 8000 + 80 = 8080 rad / s 2



ω1 = ω0 −



(c)



At ω = ω0 ,



Vin2 100 2 = = 1.25 kW P= 2R  8



At ω = ω1 ,



Vin2 = 0.625 kW P = 0.5 ⋅ 2R 



At ω = ω2 ,



Vin2 = 0.625 kW P = 0.5 ⋅ 2R 



P.P.14.8



ω0 =



1 LC



=



1 (20 × 10 )(5 × 10 ) -3



-9



= 10 5 = 100 krad / s



R  100 × 10 3 = = 50 Q= ω0 L (10 5 )(20 × 10 -3 )



ω0



10 5 B= = = 2 krad / s Q 50



Since Q > 10 , B = 100,000 − 1,000 = 99 krad / s 2 B ω2 = ω0 + = 100,000 + 1,000 = 101 krad / s 2



ω1 = ω0 −



P.P.14.9



1 50 =  jω +  jω0.2 5 +  j10ω 10 (1 −  j2ω) Z =  jω + 1 + 4ω2 Z =  jω1 + 10 ||



Im(Z) = 0  ⎯   ⎯→ ω −



20ω =0 1 + 4ω2



20ω  ⎯   ⎯→ 1 + 4ω2 = 20 2 1 + 4ω 19 ω= = 2.179 rad / s 2



ω=



R 2 || sL s =  jω , Vi R 1 + R 2 || sL sR 2 L H(s) = R 1R 2 + sR 1L + sR 2 L  jωR 2 L H(ω) = R 1R 2 +  jωL (R 1 + R 2 ) H (0) = 0  jR 2 L R 2 = H(ω) = lim ω→∞ R  R  ω +  jL ( R  + R  ) R 1 + R 2 1 2 1 2 i.e. a highpass filter. P.P.14.10



H(s) =



Vo



=



The corner frequency occurs when H (ωc ) =



1 ⋅ H(∞) . 2



⎛  R 2  ⎞⎛   ⎞  jωL ⎜ ⎟⎜ ⎟ ⎝ R 1 + R 2 ⎠⎝  jωL + R 1R 2 (R 1 + R 2 ) ⎠ ⎛  R 2  ⎞⎛   jω  ⎞ R 1 R 2 ⎟⎜ ⎟, where k = H (ω) = ⎜ (R 1 + R 2 ) L ⎝ R 1 + R 2 ⎠⎝  jω + k  ⎠ H (ω) =



At the corner frequency, R 2 R 2  jωc 1 ⋅ = ⋅ 2 R 1 + R 2 R 1 + R 2  jωc + k 



ω 1 = 2c 2 2 ωc + k  Hence,



H (ω) =



 ⎯   ⎯→ ωc = k =



R2 R1



R 2  j



R1R 2 (R 1



R2 )L



 j c



and the corner frequency is (100)(100) ωc = = 25 krad / s (100 + 100)(2 × 10 -3 ) P.P.14.11



B = 2π (20.3 − 20.1) × 10 3 = 400π



Assuming high Q,



ω0 =



ω1 + ω2 2



(2π)(40.4 × 10 3 ) = = 40.4π × 103 rad / s 2



ω0



40.4π × 10 3 = = 101 Q= B 400π R  R  20 × 10 3  ⎯   ⎯→ L = = = 15.916 H B= L B 400π 1 1  ⎯   ⎯→ C = ω0 CR  ω0 QR  1 = 3.9 pF C= (40.4π × 10 3 )(101)(20 × 10 3 ) Q=



P.P.14.12



Given H (∞) = 5 and f c = 2 kHz 1 ωc = 2πf c = R i C i



1 1 = 2πf c C i ( 2π)(2 × 10 3 )(0.1 × 10 -3 ) R i = 795.8 ≅ 800 R i =



H(∞) = P.P.14.13



- R f  = -5  ⎯   ⎯→ R f  = 5R i = 3,978 ≅ 4 k R i



ω0 = 20 krad / s



Q = 10 ,



B=



ω0 Q



= 2 krad / s



B = 19 krad / s 2 B ω2 = ω0 + = 21 krad / s 2



ω1 = ω0 −



Since ω1  =



1 , C 2 R  1 1 = = 5.263 nF 3 ω1R  (19 × 10 )(10 × 10 3 ) 1 1 = = 4.762 nF C1 = ω2 R  (21 × 10 3 )(10 × 10 3 ) R f  = 5  ⎯   ⎯→ R f  = 5R i = 50 k K = R i C2 =



P.P.14.14



ω′c 2π × 10 4 = = 2π × 10 4 K f  = ωc 1 C C 1 10 4  ⎯   ⎯→ K m = = = C′ = K m K f  C′ K f  (15 × 10 -9 )(2π × 10 4 ) 3π 10 4 R ′ = K m R = (1) = 1.061 k Ω 3π K m 10 4 2 ⋅ = 33.77 mH L′ = L= K f  3π 2π × 10 4



Therefore,



R 1′ = R ′2 = 1.061 k C1′ = C′2 = 15 nF L ′ = 33.77 mH



P.P.14.15



The schematic is shown in Fig. (a).



(a)



Use the AC Sweep option of the Analysis Setup. Choose a Linear  sweep  sweep type with the following Sweep Parameters : Total Pts  = 100, Start Freq  = 1, and End Freq  = 1K. After saving and simulating the circuit, we obtain the magnitude and phase plots are shown in Figs. (b) and (c) .



(b)



(c) P.P.14.16



The schematic is shown in Fig. (a).



(a)



Use the AC Sweep option of the Analysis Setup. Choose a Decade sweep type with these Sweep Parameters : Pts/Decade  = 20, Start Freq  = 1K, and End Freq  = 100K. Save and simulate the circuit. For the magnitude magnitude plot, choose choose DB( ) from the the Analog Operators and Functions list. Then, select the voltage V(R1:1) and OK. Another option option would be to type DB(V(R1:1)) as the Trace Expression. For the the phase plot, choose choose P( ) from from the Analog Operators and Functions list. Then, select the voltage voltage V(R1:1) and OK. Another option option would would be to type VP(R1:1) as the Trace Expression. The resulting magnitude and phase plots are shown in Figs. (b) and (c).



(b)



(c)



P.P.14.17



or



ω0 = 2πf 0 = C=



1 LC



1 4π f 02 L 2



For the high end of the band, f 0 = 108 MHz 1 = 0.543 pF C1 = 2 2 4π (108 × 1012 )(4 × 10 -6 ) For the low end of the band, f 0 = 88 MHz 1 C2 = = 0.818 pF 2 2 4π (88 × 1012 )(4 × 10 -6 ) Therefore, C must be adjustable and be in the range 0.543 pF to 0.818 pF . P.P.14.18



For BP6 , f 0 = 1336 Hz and it passes frequencies in the range 1209 Hz < f  < 1477 Hz . B = 2π (1477 − 1209 ) = 1683 .9



P.P.14.19 



L=



R  600 = = 0.356 H B 1683.9



C=



1 1 = 2 = 39.83 nF 2 4π f 0 L 4π (1336) 2 (0.356) 2



C = 10 μF



2πf c =



L=



and



R 1 = R 2 = 8 Ω



1 1 1  ⎯   ⎯→ f c = = = 1.989 kHz R 1C 2πR 1C ( 2π )(8)(10 × 10 -6 )



R 2 8 = = 0.64 mH 2πf c (2π)(1.989 × 10 3 )



February 5, 2006



CHAPTER 15



[ t u ( t )] = ∫0



∞



L



P.P.15.1



t e -st dt



Using integration by parts, u dv = uv − v du



Let



 ⎯→ u = t  ⎯ 



du = dt .



 ⎯→ e -st dt = dv  ⎯ 



[ t u ( t )] =



L



-t s



e



-st ∞ 0



-1



v=



+∫



s



∞1



e



s



0



e -st



-st



dt = 0 +



e -st s



∞ 0



2



1



=



s2



Also,



[ e at u ( t )] = ∫0 e at e -st ∞



L



P.P.15.2



-1



dt =



s−a



e -( s− a ) t



∞



1 [ cos(ωt )] = ∫-∞ ( e jωt + e - jωt ) e -st



s



a



∞



L



dt



2



[ cos(ωt )] =



L



1



=



0



1



∫ 2



∞



0



e -(s - jω) t dt +



1



∫ 2



∞



0



e -(s+ jω) t dt



⎛  1 s 1  ⎞ ⎟= 2 + 2 ⎝ s −  jω s +  jω ⎠ s



1 [ cos(ωt )] = ⎜



L



P.P.15.3



If



2



f ( t ) = cos(2 t ) + e -3t , F(s) = F(s) =



s s2 + 4 2s 2 (s



+



1 s+3



3s



3)(s



2



=



s 2 + 3s + s 2 + 4 (s 2 + 4)(s + 3)



4 4)



Given f ( t ) = t 2 cos(3t ) s From P.P.15.2, L[ cos(3t )] = 2 s +9 P.P.15.4



Using Eq. 15.34,



F(s) =



[t



L



2



cos(3t )] = ( - 1)



2



d 2 ⎛  s  ⎞



⎜



⎟



ds 2 ⎝ s 2 + 9 ⎠



d 2



F(s) =



2



[ s(s



+ 9)



2



-1



]=



d 2 2



[ (1) (s



2



-1 -2 + 9) − (s)( 2s) ( s 2 + 9) ]



ds ds -2 -2 -3 F(s) = ( - 2s) ( s 2 + 9 ) − ( 4s) ( s 2 + 9 ) + ( 4s 2 ) ( 2s) ( s 2 + 9 ) F(s) = ( - 6s) ( s + 9 ) 2



2s s 2



F(s) =



P.P.15.5



-2



+ ( 8s )( s + 9) = 3



2



-3



2s 3 − 54s



( s 2 + 9) 3



27



s2



9



3



h ( t ) = 10 u ( t ) − u ( t − 2) + 5 u ( t − 2) − u ( t − 4)



⎛ 1 e -2s  ⎞ ⎛ e -2s e -4s  ⎞ ⎟ + 5⎜ ⎟ H (s) = 10 ⎜ − − s  ⎠ s  ⎠ ⎝ s ⎝  s H (s) =



P.P.15.6



e - 2s



2



s



e -4s



T = 5 f 1 ( t ) = u ( t ) − u ( t − 2) 1 F1 (s) = (1 − e - 2s ) s F(s) =



P.P.15.7



5



F1 (s)



=



1 − e -Ts



1 e -2s s (1 e -5s )



g (0) = lim sF(s) = lim s →∞



s →∞



1+



2



s 3 + 2s + 6 (s 2 + 2s + 1)(s + 3)



+



6



s2 s3 =1 s →∞ ⎛  2 1  ⎞⎛  3 ⎞ ⎜1 + + 2 ⎟⎜1 + ⎟ ⎝  s s  ⎠⎝  s ⎠



g (0) = lim



Since all poles s = 0, - 1, - 1, - 3 lie in the left-hand s-plane, we can apply the final-value theorem. s 3 + 2s + 6 g (∞) = lim sF(s) = lim 2 s→0 s → 0 (s + 1) (s + 3) 6 g (∞) = lim 2 =2 s → 0 (1) (3) P.P.15.8



F(s) = 1 +



4 s+3



−



5s s 2 + 16



f ( t ) =



-1



L



⎡ 4 ⎤ -1⎡ 5s ⎤ −L ⎢ 2 ⎣ s + 3 ⎥⎦ ⎣ s + 16 ⎥⎦



[ 1 ] + L-1⎢



-3t f ( t ) = δ( t ) + (4e − 5 cos(4 t ))u ( t ),



P.P.15.9



F(s) =



A s +1



B s+3



+



A = F(s) (s + 1)



s = -1



=



B = F(s) (s + 3)



s = -3



=



C = F(s) (s + 4)



s = -4



=



F(s) =



1 s +1



f ( t ) = (e



P.P.15.10



+



G (s) =



-t



+



3 s+3



−



t≥0



C s+4



6 (s + 2) (s + 3)(s + 4) 6 (s + 2)



s = -1



=



(s + 1)(s + 4) 6 (s + 2)



s = -3



=



(s + 1)(s + 3)



s = -4



=



6 ( 2)(3) (6)(-1)



=1



(-2)(1) (6)(-2)



=3



(-3)(-1)



= -4



4 s+4



+ 3e -3t − 4e -4t )u ( t ), t ≥ 0



s 3 + 2s + 6 s (s + 1) 2 (s + 3)



=



A s



+



B s +1



+



C (s + 1) 2



+



D s+3



Multiplying both sides by s (s + 1) 2 (s + 3) gives s 3 + 2s + 6 = A (s + 3)(s 2 + 2s + 1) + Bs (s + 1)(s + 3) + Cs (s + 3) + Ds (s + 1) 2



= A (s3 + 5s 2 + 7s + 3) + B (s 3 + 4s 2 + 3s) + C (s 2 + 3s) + D (s3 + 2s 2 + s) Equating coefficients : s0 : 6 = 3A  ⎯   ⎯→



A=2



(1)



s1 :



2 = 7 A + 3B + 3C + D  ⎯   ⎯→



3B + 3C + D = -12



(2)



s2 :



0 = 5A + 4B + C + 2D  ⎯   ⎯→



4B + C + 2D = -10



(3)



s3 :



1 = A + B + D  ⎯   ⎯→



Solving (2), (3), and (4) gives - 13 A  = 2 , B = , 4 G (s) =



2 s



−



13 4 s +1



−



C  =



32 (s + 1)



B + D = -1



2



+



9 4 s+3



-3 2



,



(4)



D =



9 4



g ( t ) = ( 2 3.25 e



1.5 t e



-t



2.25 e



10



G (s) =



P.P.15.11



-t



(s + 1)(s 2 + 4s + 13)



-3t



=



)u( t ),



A s +1



+



t



0



Bs + C s 2 + 4s + 13



Multiplying both sides by (s + 1)(s 2 + 4s + 13) gives 10 = A (s 2 + 4s + 13) + B (s 2 + s) + C (s + 1) Equating coefficients : s2 :



0 = A + B  ⎯   ⎯→



A = -B



s1 :



0 = 4A + B + C  ⎯   ⎯→



s0 :



10 = 13A + C  ⎯   ⎯→ 10 = 10A



Solving (1), (2), and (3) gives A  = 1 , B = -1 , G (s) =



1 s +1



−



s+3 (s + 2) 2 + 9



(1)



C = -3A



(2) (3)



C = -3



=



g ( t ) = (e - t − e - 2t cos(3t ) −



1 s +1



−



s+2 (s + 2) 2 + 9



1 - 2t e sin(3t )), 3



−



1 (s + 2) 2 + 9



t≥0



P.P.15.12



2 For 0 < t < 1 , consider Fig. (a).



x1(t - λ)



x2(λ)



1



 ∫ (1)(1) d λ = t



y( t ) =



t



0



t-1



0



t (a)



1



2



λ



2



For 1 < t < 2 , consider Fig. (b). y( t ) =



∫



1



t −1



(1)(1) d λ +



∫ (1)(2) d λ = λ t



1



t t −1



+ 2λ



1



t 1



y( t ) = 1 − t + 1 + 2 ( t − 1) = t



0



t-1



1



t



2



λ



2



t



λ



(b)



For 2 < t < 3 , consider Fig. (c).



 ∫



y( t ) =



2



t −1



(1)( 2) d λ = 2 λ



2 2 t −1



1



y( t ) = 2 (2 − t + 1) = 6 − 2 t



0



1



t-1



For t  > 3 , there is no overlap so y( t ) = 0 .



(c)



y(t)



Thus, t y( t ) = 6



2t 0



0



t



2



2



t



3



2



otherwise



0



1



2



The result of the convolution is shown in Fig. (d).



3



t



(d)



P.P.15.13



3e



g(t-λ)



t-1



0



t



-λ



1



λ



(a)



∫ (1) 3 e t



0



-λ



t-1 (b)



For 0 < t < 1 , consider Fig. (a). y( t ) =



0



d λ = -3 e -λ



For t  > 1 , consider Fig. (b).



t 0



= 3 (1 − e - t )



1



t



λ



y( t ) =



∫



t



t −1



(1) 3 e -λ d λ = -3 e -λ



t t −1



= 3 e -t (e − 1)



Thus, 3 (1 e-t )



0



y( t ) = 3 e-t (e 1)



t t



0



1 1



elsewhere



The circuit in the s-domain is shown below.



P.P.15.14



1 + Vs



Vo =



2s 1+ 2 s



H(s) =



Vo Vs



+



−



2/s



Vo



Vs



=



2 s+2



 ⎯   ⎯→ h ( t ) = 2 e -2t



v o (t) = h(t) ∗ v s (t) =



∫



t



0



h (λ) v s ( t − λ) d λ



t



= ∫ 2 e- 2λ 10 e-(t − λ ) d λ 0



t



= 20 e- t ∫ e- 2λ eλ d λ = 20 e- t (-e-λ ) 0t 0



= 20 (e -t



e -2t )u( t ) V



Taking the Laplace transform of each term gives 1 [ s 2 V(s) − sv(0) − v′(0) ] + 4 [ sV (s) − v(0) ] + 4 V(s) = s +1 2 1 s + 6s + 6 = (s 2 + 4s + 4) V(s) = s + 5 + s +1 s +1 2 s + 6s + 6 A B C V (s) = = + + (s + 1)(s + 2) 2 s + 1 s + 2 (s + 2) 2



P.P.15.15



s 2 + 6s + 6 = A (s 2 + 4s + 4) + B (s 2 + 3s + 2) + C (s + 1) Equating coefficients :



s2 :



1 = A + B  ⎯   ⎯→



s1 :



 ⎯→ 6 = 4A + 3B + C  ⎯ 



6 = A +3+ C



s0 :



6 = 4A + 2B + C  ⎯   ⎯→



6 = 6− B



Thus,



B = 0 ,



A  = 1 ,



B = 1− A



or  A = 1 − B or  C = 3 − A



or  B = 0



C = 2



and V(s) =



1 s +1



Therefore, v( t ) = (e -t



+



2 (s + 2) 2



2 t e -2t ) u(t )



Note, there were no units give for v(t).



Taking the Laplace transform of each term gives 2 2 sY (s) − y(0) + 3Y (s) + Y (s ) = s s+3 2s [ s 2 + 3s + 2] Y(s) = s+3 2s A B C Y(s) = = + + (s + 1)(s + 2)(s + 3) s + 1 s + 2 s + 3



P.P.15.16



A = Y (s) (s + 1)



s = -1



= -1



B = Y (s) (s + 2)



s = -2



=4



C = Y (s) (s + 3)



s = -3



= -3



Y (s) =



-1 s +1



y( t ) = ( -e -t



+



4 s+2 4 e -2t



−



3 s+3 3 e -3t ) u( t )



February 5, 2006



CHAPTER 16



P.P.16.1



Consider the circuit shown below. s Io 4/s



2/s



Using current division, 4 Io



=



2



s 4 s



⋅ = s



+s+4



Vo (s) = 4 I o 32 s (s + 2) 2



=



32 = A (s 2



= A s



8 s (s 2



+ 4s + 4)



32 s (s + 2) 2



+



B s+2



+



C ( s + 2) 2



+ 4s + 4) + B (s 2 + 2s) + Cs



Equating coefficients : s0 : s1 : s2 :



32 = 4A



 ⎯   ⎯→ A = 8 0 = 4A + 2B + C 0 = A + B  ⎯   ⎯→ B = -A = -8



Hence, 0 = 4A + 2B + C Vo (s) =



8 s



−



8 s+2



v o ( t ) = 8 (1 e -2t



 ⎯   ⎯→ C = -16



−



16 (s + 2) 2 2t e -2t ) u(t ) V



4



+ Vo(s)



P.P.16.2



The circuit in the s-domain is shown below. 1



+



1/(s + 2)



At node o, 1



s+2 1



− Vo =



Vo(s)



2s



−



Vo 2s



+



Vo



+



i(0)



2 s 1  ⎞



i(0)/s



2



where i(0) = 0A



 



= Vo ⎛  ⎜1 + + ⎟ s+2 ⎝  2 2s ⎠ Vo



=



1



2s (s + 2)(3s + 1)



=



2s / 3 (s + 2)(s + 1 3)



=



A s+2



+



B s +1 3



Solving for A and B we get, A = [2(–2)/3]/(–2+1/3) = 4/5, B = [2(–1/3)/3]/[(–1/3)+2] = –2/15 Vo



=



45 s+2



−



2 15 s +1 3



Hence, v o (t) =



⎛ 4 e - 2 t − 2 e - t 3 ⎞ u ( t )V ⎜ ⎟ 15 ⎝ 5  ⎠



v(0) = V0 is incorporated as shown below.



P.P.16.3



V(s) + I0 /s



1/sC



R



V



CV0



We apply KCL to the top node. I0 V 1 ⎞ ⎛  + CV0 = + sCV = ⎜sC + ⎟ V ⎝  s R  R  ⎠ V= V= V= where A =



I0 s (sC + 1 R ) V0 s + 1 RC V0 s + 1 RC



I0 C 1 RC



V(s) =



+



+



CV0 sC + 1 R  I0 C



s (s + 1 RC)



+



A s



+



B s + 1 RC



= I 0 R , V0



s + 1 RC



v( t ) = (( V0



+



B=



I 0 R  s



−



I0 C - 1 RC



= - I 0 R 



I 0 R  s + 1 RC



− I 0 R ) e - t τ + I 0 R ), t > 0, where τ = RC



P.P.16.4



We solve this problem the same as we did in Example 16.4 up to the point where we find V1. Once we have V1, all we need to do is to divide V1 by 5s to and add in the contribution from i(0)/s to find IL. IL = V1/5s – i(0)/s = 7/(s(s+1)) – 6/(s(s+2)) – 1/s = 7/s – 7/(s+1) – 3/s + 3/(s+2) – 1/s = 3/s – 7/(s+1) + 3/(s+2) –t



–2t



Which leads to iL(t) = (3 – 7e  + 3e )u(t)A



P.P.16.5



We can use the same solution as found in Example 16.5 to find iL.



All we need to do is divide each voltage by 5s and then add in the contribution from i(0). Start by letting iL = i1 + i2 + i3. I1 = V1/5s – 0/s = 6/(s(s+1)) – 6/(s(s+2)) = 6/s – 6/(s+1) – 3/s + 3/(s+2) i1 = (3 – 6e –t + 3e –2t)u(t)A



or



I2 = V2/5s – 1/s = 2/(s(s+1)) – 2/(s(s+2)) – 1/s = 2/s – 2/(s+1) – 1/s + 1/(s+2) –1/s  –t



or



 –2t



i2 = (–2e  + e )u(t)A



I3 = V3/5s – 0/s = –1/(s(s+1)) + 2/(s(s+2)) = –1/s + 1/(s+1) + 1/s – 1/(s+2)  –t



or



 –2t



i3 = (e  – e )u(t)A –t



–2t



This leads to iL(t) = i1 + i2 + i3 = (3 – 7e  + 3e )u(t)A



P.P.16.6



Ix



1/s



1 +



+



5/s



Vo



−



+



2



4Ix



(a) Take out the 2 Ω and find the Thevenin equivalent circuit. VTh = Ix



1/s



1 +



5/s



+



−



VTh



+



4Ix



Using mesh analysis we get,  –5/s +1Ix +Ix/s + 4Ix = 0 or (1 + 1/s + 4)Ix = 5/s or Ix = 5/(5s+1) VTh = 5/s – 5/(5s+1) = (25s+5–5s)/(s(5s+1) = 5(4s+1)/(s(5s+1) = 4(s+0.25)/(s(s+0.2)



Ix



5/s



1/s



1



+



+



Isc



−



4Ix



Ix = (5/s)/1 = 5/s Isc = 5/s + 4(5/s)/(1/s) = 5/s + 20 = (20s+5)/s = 20(s+0.25)/s ZTh = VTh/Isc = {4(s+0.25)/(s(s+0.2))}/{20(s+0.25)/s} = 1/(5(s+0.2)) 1 5(s 0.25)



+



s(s



0.2)



−



Vo =



0.25)



s(s



0.2)



1 5(s



(b)



+



4(s



4(s



0.2)



0.2)



Vo



2 = 2



4(s + 0.25) s(s + 0.3)



or 



2



10( 4s + 1) s(10s + 3)



+



Initial value:



vo(0 ) = Lim sVo = 4V s ∞



Final value:



vo(∞) = Lim sVo = 4(0+0.25)/(0+0.3) = 3.333V s 0



(c)



Partial fraction expansion leads to Vo = 3.333/s + 0.6667/(s+0.3) Taking the inverse Laplace transform we get, vo(t) = (3.333 + 0.6667e If x ( t ) = e -3t u ( t ) , then X (s) =



P.P.16.7



Y(s) = H(s) X (s) =



2s (s + 3)(s + 6)



A = Y (s) (s + 3)



s = -3



= -2



B = Y (s) (s + 6)



s = -6



=4



Y (s) =



-2 s+3



+



-3t y( t ) = (-2 e



H(s) =



2s (s + 6)



=



1 s+3



A s+3



+



B s+6



4 s+6



+ 4 e -6t )u ( t ) =



2 (s + 6 − 6) s+6



= 2−



12 s+6



h ( t ) = 2 (t ) 12 e -6t u(t ) P.P.16.8



I1



=



By current division, 2 + 1 2s I s + 4 + 2 + 1 2s 0



H(s) =



I1 I0



=



2 + 1 2s s + 4 + 2 + 1 2s



=



4s 2s 2



1 12s



1



P.P.16.9



2s (a)



Vo Vi



=



1 || 2 s 1 + 1 || 2 s



=



1+ 2 s 2s



1+



1+ 2 s



=



.



2 s+4



–0.3t



)u(t)V



H(s) =



Vo



=



Vi



2 s



4



(b)



h ( t ) = 2 e -4t u(t )



(c)



Vo (s) = H (s) Vi (s) =



A



s (s + 4)



=



1



= s Vo (s) s=0 = ,



s



+



B s+4



s = -4



=



-1 2



1 ⎛ 1



1  ⎞ ⎜ − ⎟ 2 ⎝ s s + 4 ⎠



1



v o (t) =



A



B = (s + 4) Vo (s)



2



Vo (s) =



(d)



2



2



(1



e -4t ) u(t ) V



v i ( t ) = 8 cos( 2 t )



 ⎯   ⎯→



Vo (s) = H(s) Vi (s) =



A = (s + 4) Vo (s)



s = -4



Vi (s) =



8s s2



16s (s + 4)(s 2



=



=



+ 4)



+4 A



+



(s + 4)



Bs + C (s 2



+ 4)



- 16 5



Multiplying both sides by (s + 4)(s 2



+ 4) gives 16s = A (s + 4) + B (s 2 + 4s) + C (s + 4)



Equating coefficients : 0= A+B



s1 :



16 = 4B + C



 ⎯   ⎯→



 



(2)



s0 :



0 = 4 A + 4C



 ⎯   ⎯→ C = -A



(3)



Vo (s) =



 ⎯   ⎯→



B = -A =



16



s2 :



C=



16 5



5



 



16 ⎛ 



(1)



−1 s + 1  ⎞ 16 ⎛  − 1 s 1 2  ⎞ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟ + 2 + 2 + ⋅ 2 5 ⎝ s + 4 s + 4 ⎠ 5 ⎝ s + 4 s + 4 2 s + 4 ⎠⎟



v o ( t ) = 3.2



e



- 4t



cos( 2t )



0.5 sin( 2t ) u(t ) V



P.P. 16.10 Consider the circuit below.



iR 



R 1



i



L +



vL



-



+ C



+  _



vs



R 2



v -



i R



= i + C 



vo



= R2i  



But



dv dt 



(1)



−v



=



vs



vs



−v



i R



 R1



Hence,  R1



= i + C 



dv dt 



or •



v=−



v  R1C



i



vs



C



R1C 



− +



 



(2)



Also, v L



v + v L + vo =0



=L



di dt 



= v − vo



But vo = iR2  . Hence •



+ vo



i = v / L − vo / L =



v



−



iR2



   L L Putting (1) to (3) into the standard form



(3)



⎡ 1 −1⎤ ⎡v• ⎤ ⎢ − ⎥ ⎡v ⎤ ⎡ 1 ⎤  R C C  ⎢ ⎥=⎢ 1 ⎥ ⎢ ⎥ + ⎢ R 1C ⎥ vs • ⎥ ⎢ ⎥ ⎢  I R2 ⎥ ⎣ i ⎦ ⎢ i ⎢⎣ 0 ⎥⎦ − ⎥ ⎣ ⎦ ⎢ L⎦ ⎣  L ⎡v ⎤ vo = [ 0 R2 ] ⎢ ⎥ ⎣i ⎦ If we let R 1= 1, R 2 = 2, C = ½, L = 1/5, then ⎡ −2 −2 ⎤ ⎡2⎤ =  A = ⎢ , B ⎥ ⎢0 ⎥ , C = [ 0 2] 5 10 − ⎣ ⎦ ⎣ ⎦ sI



⎡s + 2 − A = ⎢ ⎣ −5



⎤ ⎥ s + 10 ⎦ 2



⎡ s + 10 −2 ⎤ ⎢ 5 ⎥ s + 2⎦ ⎣ −1 ( sI − A) = 2 s + 12s + 30



H(s) = C(sI − A ) −1 B =



=



=



P.P. 16.11



[0



⎡s + 10 − 2 ⎤ ⎡2⎤ 2]⎢ ⎥⎢ ⎥ s + 2 ⎦ ⎣0 ⎦ ⎣ 5 s2



+ 12s + 30



20 s 2 12s + 30



20 s2



+ 12s + 30



Consider the circuit below. i 1



L



vo



2 io



i1



R 1



+ v -



C



R 2



i2



At node 1, i1



=



v  R1



•



or



v=−



•



+ C v+ i 1



v−



 R1C



1 C



i+



i1



  C 



(1)



This is one state equation. At node 2, io = i + i2  



(2)



Applying KVL around the loop containing C, L, and R 2, we get •



−v + L i + io R2 = 0 or •



i=



v



−



R2



io    L L Substituting (2) into (3) gives • v R R i = − 2 i − 2 i2   (4)  L L L vo = v (5) From (1), (3), (4), and (5), we obtain the state model as ⎡ 1 −1⎤ ⎡1 ⎤ 0 ⎥ ⎡v• ⎤ ⎢− ⎥ ⎢  ⎡i1 ⎤  R1C C  ⎡v ⎤ C 



⎢ ⎥=⎢ ⎢•⎥ ⎢ ⎣i ⎦ ⎢ ⎣



1  L



⎥ ⎢ ⎥ + ⎢ ⎥ ⎢ ⎥ R2 ⎥ ⎣i2 ⎦  R2 ⎥ ⎣ i ⎦ ⎢ 0 − − ⎥ ⎢ ⎣  L ⎦⎥ L⎦



⎡vo ⎤ ⎡1 0⎤ ⎡v ⎤ ⎡0 0 ⎤ ⎡ i1 ⎤ ⎢ i ⎥ = ⎢ 0 1 ⎥ ⎢ i ⎥ + ⎢0 1 ⎥ ⎢ i ⎥ ⎦⎣ ⎦ ⎣ ⎦ ⎣ 2 ⎦ ⎣ o⎦ ⎣ Substituting R 1 = 1, R 2 =2, C = ½, L = ¼ yields



⎡v• ⎤ −2 −2 v ⎤ ⎡ ⎤ ⎡2 0 ⎤ ⎡ i1 ⎤ ⎢ ⎥ = ⎡ + ⎢ • ⎥ ⎢⎣ 4 −8⎥⎦ ⎢⎣ i ⎥⎦ ⎢⎣0 −8⎥⎦ ⎣⎢i2 ⎥⎦ ⎣i ⎦ ⎡vo ⎤ ⎡1 ⎢ i ⎥ = ⎢0 ⎣ o⎦ ⎣



0 ⎤ ⎡v ⎤



⎡0 + ⎥⎢ ⎥ ⎢ 1 ⎦ ⎣ i ⎦ ⎣0



0 ⎤ ⎡ i1 ⎤



⎥⎢ ⎥



1 ⎦ ⎣i2 ⎦



(3)



P.P. 16.12



Let so that



x1 = y •



 x1



(1) •



= y  (2)



•



Let Finally, let



•



 x2



= x1 = y  



 x3



=



•



x2



(3)



••



= y 



(4)



then •



•••



 x3



••



•



= y = −6 y− 11 y − 6 y + z   = −6 x3 − 11x2 − 6 x1 + z



(5)



From (1) to (5), we obtain,



⎡ x• ⎤ ⎢ 1 ⎥ ⎡ 0 1 0 ⎤ ⎡ x1 ⎤ ⎡0⎤ ⎢•⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ x2 ⎥ = ⎢ 0 0 1 ⎥ ⎢ x2 ⎥ + ⎢0⎥ z (t ) ⎢ • ⎥ ⎣⎢ −6 −11 −6⎦⎥ ⎣⎢ x3 ⎦⎥ ⎣⎢1⎥⎦ ⎢ x3 ⎥ ⎣ ⎦



 y(t ) = [1 0



P.P.16.13



⎡ x1 ⎤ ⎢ ⎥ 0] x2 ⎢ ⎥ ⎢⎣ x3 ⎥⎦



The circuit in the s-domain is equivalent to the one shown below. Vo



+ Z



Z



Vo



= (βVo ) Z  ⎯   ⎯→



- Vo



where



R 



Z = R || 1 sC =



Thus, - 1 =



- 1 = βZ , 1 + sRC



βR  or 1 + sRC



- (1 + sRC ) = βR 



For stability,



βR  > -1



-1



or



R



From another viewpoint,



Vo



= -(βVo ) Z  ⎯   ⎯→



(1 + βZ) Vo



=0



⎛  βR   ⎞ ⎜1 + ⎟V = 0 ⎝  1 + sRC ⎠ o (sRC + βR + 1) Vo



=0



⎛  βR + 1 ⎞ ⎜s + ⎟V = 0 ⎝  RC  ⎠ o For stability



βR  + 1 RC



must be positive, i.e.



βR + 1 > 0



-1



or



R



P.P.16.14



(a)



Following Example 15.24, the circuit is stable when -10 10 + α > 0 or



(b)



For oscillation, 10 + α = 0



-10



or



P.P.16.15



Vo Vi



=



s⋅



R  R + sL +



1 sC



= s2



+s⋅



R 



L R  L



+



1 LC



Comparing this with the given transfer function,



R  L



= 4 



1



and



LC



If we select R  = 2 , then 2 L = = 0 .5 H   4 P.P.16.16



= 20



C=



and



1 20L



=



1 10



= 0. 1 F



Consider the circuit shown below. Y3 Y4 Y1



Y2



− +



Vin



Clearly,



V1



V2



+



Vo



−



=0



V2



At node 1,



− V1 ) Y1 = (V1 − Vo ) Y3 + (V1 − 0) Y2 Vin Y1 = V1 (Y1 + Y2 + Y3 ) − Vo Y3  



( Vin or



(1)



At node 2, ( V1 − 0) Y2 or



=



V1



- Y4 Y2



= (0 − Vo ) Y4



Vo  



(2)



Substituting (2) into (1), Vin Y1 or



If we select Y1



Vo Vin



=



= 1



R 1



=



- Y4 Y2



Vo ( Y1 + Y2



+ Y3 ) − Vo Y3



- Y1 Y2 Y4 ( Y1 + Y2 , Y2



= sC1 ,



+ Y3 ) + Y2 Y3 Y3



= sC 2 , and



Y4



=



1 R 2



, then



-s⋅



Vo



=



Vin



C1 R 1



⎛  1  ⎞ ⎜ + sC1 + sC 2 ⎟ + s 2 C1C 2 R 2 ⎝ R 1  ⎠ 1



-s⋅



Vo



=



Vin



s2



+s⋅



1 R 1C 2



⎛  1 1  ⎞ 1 ⎜ + ⎟+ R 2 ⎝ C1 C 2  ⎠ R 1 R 2 C1C 2 1



Comparing this with the given transfer function shows that 1 R 1C 2



If R 1



⎛  1 1  ⎞ ⎜ + ⎟ = 6, R 2 ⎝ C1 C 2 ⎠ 1



= 2,



1 R 1 R 2 C1C 2



= 10



= 10 k Ω , then C2



=



1 R 2 C1



1 2 × 10 3



= 0.5 mF 1



= 5  ⎯   ⎯→



R 2



= 5C1



⎛  1 1  ⎞ ⎜ + ⎟ = 6  ⎯   ⎯→ R 2 ⎝ C1 C 2 ⎠ 1



R 2



=



1 5C1



=



1 (5)(0.1 × 10 -3 )



Therefore, C1 = 0.1 mF ,



C2



⎛  ⎝ 



5 ⎜1 +



C1  ⎞



⎟ = 6  ⎯   ⎯→ C 2  ⎠



C1



= 2 k Ω



= 0.5 mF ,



R 2



= 2k



=



C2 5



= 0.1 mF



February 5, 2006



CHAPTER 17 T = 2, ωo = 2π/T = π



P.P.17.1



f(t) = 1,  –1, ao =



1 T



an =



2 T



=



 bn =



=



∫



T



∫



T



0



0



f ( t )dt =



0