# Problem 5.1. Suggested Solution

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1 Problem 5.1 A 12µF capacitor has an accumulated charge of 480µC. Determine the voltage across the capacitor after 4 s. t 2 1 ( 2) ( 1) C it () 100 µ ( ) 1 vt vt = dt C= F it = ma= I t1 vt vt t t t t I ( 1) = 0. so, ( 2) = C ( 2 1) where 2 1= 4sec vt ( 2) = 40 V

2 Problem 5.2 A 12µF capacitor has an accumulated charge of 480µC. Determine the voltage across the capacitor. Q C = Q= 480µ C C = 12µ F V = 40V C

3 Problem 5.3 A capacitor has an accumulated charge of 600µC with 5V across it. What is the value of capacitance? Q C = Q= 600µ C V = 5V C = 120µ F V

4 Problem 5.4 A 25- µ F capacitor initially charged to 10V is charged by a constant current of 2.5 µ A. Find the voltage across the capacitor after 2 ½ minutes. t 1 C 0 V = idt+ v(0) x10 0 = dt x10 dt 10 = 10 = t 10 V = = 5V 150 0

5 Problem 5.5 The energy that is stored in a 25-µF capacitor is w(t)=12sin 2 377t J. Find the current in the capacitor. () = ( ) = 12sin wt 2 Cv t t () ( ) C = 25 µ F v ( t) = sin 377t = sin 377t vt ( ) =± 979.8sin 377 tv dv t it C t t dt 6 ( ) = =± (979.8)(377) cos 377 =± 9.23cos 377 A

6 Problem 5.6 An uncharged 10-uF capacitor is charged by the current I(t)=10cos377t ma. Find (a) the expression for the voltage across the capacitor and (b) the expression for the power. it ( ) = 10 cos(377 tma ) C= 10 µ F v(0) = 0V a) t C C vt ( ) = itdt ( ) = sin(377 t) vt ( ) = 2.65sin(377 t) V t 0 b) Pt ( ) = vtit ( ) ( ) = (2.65)(0.01)sin(377 t)cos(377 t) but cos( x)sin( x) = sin(2 x) 1 2 pt ( ) = 13.3sin(754 t) mw so,

7 Problem 5.7 The voltage across a 100-uF capacitor is given by the expression v(t)=120sin(377t) V. Find (a)the current in the capacitor and (b) the expression for the energy stored in the element. vt ( ) = 120sin(377 t) V, C= 100 µ F dv a i t C t t dt 1 1 () 2 2 b) w t = Cv () t = ( 10 4 )( 120) sin 2 ( 377t) = 720sin 377 t mj 6 ) ( ) = = ( )(120)(377) cos(377 ) = 4.52 cos(377 ) A ( ) = cos 754 t mj

8 Problem 5.8 A capacitor is charged by a constant current of 2mA and results in a voltage increase of 12V in a 10-s interval. What is the value of the capacitance? t 2 1 C t1 vt (2) vt (1) = itdt () it () = 2mA= I I vt ( ) vt ( ) = 12 = ( t t) = (10) I 2 1 C 2 1 C = (2 m)(10) /12 = 1.67mF C = 1.67mF C

9 Problem 5.9 The current in a 100µF capacitor is shown in Figure P5.9. Determine the waveform for the voltage across the capacitor if it is initially uncharged. i(t)(ma) Figure P5.9 t(ms) C = F v t = i t dt µ () C () Time(ms) i(t)(ma) v(t)(v) 0 t t t > v(2 m) = 0.2V i(t)(ma) t(ms)

10 Problem 5.10 The voltage across a 100-µF capacitor is shown in Figure P5.10. Compute the waveform for the current in the capacitor v(t)(v) Figure P5.10 t(ms) C = 100 µ F i( t) = Cc Time(ms) dv(v/ms) i(t)(ma) dt 0 t t t v(t)(v) Figure P5.10 t(ms)

11 Problem 5.11 The voltage across a 6µF capacitor is shown in Figure P5.11. Compute the waveform for the current in the capacitor v(t)(v) Figure P5.11 t(ms) C = 6 µ F i( t) = C dv() t dt Time(ms) dv(v/ms) i(t)(ma) dt 0 t t v(t)(v) t(ms)

12 Problem 5.12 The voltage across a 50-µF capacitor is shown in Figure P5.12. Determine the current waveform. v(t)(v) t(ms) Figure P5.12 C = 50 µ F i( t) = C Time(ms) dv(v/ms) i(t)(ma) dt 0 t t t t t dv dt t > v(t)(v) t(ms)

13 Problem 5.13 The voltage across a 2-µF capacitor is given by the waveform in Figure P5.13. Compute the current waveform. v(t)(v) t(ms) -12 Figure P5.13 C = 2 µ F i( t) = C Time(ms) dv(v/ms) i(t)(ma) dt 0 t t t t > dv dt v(t)(v) t(ms) -12

14 Problem 5.14 The voltage across a 0.1-F capacitor is given by the waveform in Figure P5.14. Find the waveform for the current in the capacitor v c (t)(v) t(s) Figure P5.14 C = 0.1 F i( t) = C Time(s) dv(v/ms i(t)(a) dt 0 t t t dv dt t > v c (t)(v) t(s)

15 Problem 5.15 The waveform for the current in a 200-µF capacitor is shown in Figure P5.15. Determine the waveform for the capacitor voltage. i(t)(ma) t(ms) Figure P5.15 C = F v t = i t dt µ ( ) C ( ) 0 t 4ms it ( ) = 1.25t A vt t v 2 ( ) = (0) assuming v(0) = 0 V, vt ( ) = 3125t t > 4ms it () = 0A vt () = 0 + v(4 m) 2 2 vt ( ) = (4 m) = 50mV tV 0 t 4ms vt () = 50mV t > 4ms

16 Problem 5.16 Draw the waveform for the current in a 12-µF capacitor when the capacitor voltage is as described in Figure 5.16 v(t)(v) t(µs) Figure P5.16 C = 12 µ F i( t) = C Time( µ s) dv(v/ µ s) i(t)(a) dt 0 t 2 dv dt 6 t t t >

17 v(t)(v) t(µs) i(t)(ma) t(µs) v(t)(v) t(µs) i(t)(ma) t(µs)

18 Problem 5.17 Draw the waveform for the current in a 3-µF capacitor when the voltage across the capacitor is given in Figure P5.17 v(t)(v) t(ms) Figure P5.17 C = 3 µ F i( t) = C Time(ms) dv(v/ µ s) i(t)(a) dt 0 t t t t t t > 0 dv dt 0 0

19 v(t)(v) t(µs) i(t)(ma) t(µs)

20 Problem 5.18 The waveform for the current in a 100-µF initially uncharged capacitor is shown in Figure P5.18. Determine the waveform for the capacitor s voltage. v(t)(ma) t(ms) -5 Figure P5.18 v(t)(ma) t(ms) -5

21 C = F v t = i t dt v(0) = 0V µ ( ) C ( ) 0 t 1 ms i( t) = 5mA vt () = 50t V 1ms t 2 ms i( t) = 5mA vt ( ) = 50( t 1 m) + v(1 m) vt ( ) = 50t+ 0.1 V 2ms t 3 ms i( t) = 5mA vt ( ) = 50( t+ 2 m) + v(2 m) vt ( ) = 50t 0.1 V 3ms t 4 ms i( t) = 5mA vt ( ) = 50( t 3 m) + v(3 m) vt ( ) = 50t+ 0.2 V 4ms t 5 ms i( t) = 5mA vt ( ) = 50( t 4 m) + v(4 m) v( t) = 50t 0.2 V t > 5 ms i( t) = 0 vt () = 50mV 50t V 0 t 1ms 50t+ 0.1 V 1ms t 2ms 50t 0.1 V 2ms t 3ms vt ( ) = 50t+ 0.2 V 3ms t 4ms 50t 0.2 V 4ms t 5ms 50 mv t > 5ms

22 Problem 5.19 The voltage across a 6-µF capacitor is given by the waveform in Figure P5.19. Plot the waveform for the capacitor current. v(t)(ma) t(ms) -5 Figure P5.19 dv C = 6 µ F i( t) = C dt if vt ( ) = 10sin( ωtv ) ω = 2π f = 2 π / T where T = 1m sec, vt ( ) = 10sin(2000 πtv ) it ( ) = (6 µ )(10)(2000 π)cos(2000 π f) it ( ) = 377 cos(2000 πtma ) v(t)(v) t(ms) i(t)(v) t(µs)

23 Problem 5.20 The current in an inductor changes from 0 to 200mA in 4ms in 4 ms and induces a voltage of 100mV. What is the value of the inductor? vt () = L di dt i = 200mA t = 4ms v = 100mV Assuming the current change is linear, i i v = L L= v ( ) = 2mH ind t L= 2mH ind t ind

24 Problem 5.21 The current in a 100-mH inductor is i(t)=2sin(377t)a. Find (a) the voltage across the inductor and (b) the expression for the energy stored in the element L= 100 mh i( t) = 2sin(377 t) A a) L di vt ( ) = L = (0.1)(2)(377) cos(377 t) = 75.4 cos(377 tv ) b) dt wt ( ) = i ( t) = (2)2sin (377 t) = 0.2sin (377 t) but sin x = cos(2 x) wt ( ) = cos(754 t) J so

25 Problem 5.22 A 10-mH inductor has a sudden current change from 200mA to 100mA in 1ms. Find the induced voltage. L = 10mH i = 100mA 200mA = 100mA t = 1ms vt () = L v v IND IND di dt i = (0.01) = 1V = 1V t assuming the current changed linearly

26 Problem 5.23 The induced voltage across a 10-mH inductor is v(t)=120cos(377t) V. Find (a) the expression for the inductor current and (b) the expression for the power L L= 10 mh v( t) = 120 cos(377 t) V Assume i(0) = 0. a) it ( ) = vtdt ( ) = sin(377 t) = 31.83sin(377 t) A b) pt ( ) = vtit ( ) ( ) = (120)(31.83)cos(377 t)sin(377 t) but sin( x)cos( x) = sin( x) pt ( ) = 1910sin(754 tw ) 1 2 so,

27 Problem 5.24 The current in a 25-mH inductor is given by the expressions i(t)=0 t<0 i(t)=10(1-te -4t )ma t>0 Find (a) the voltage across the inductor, (b) the expression for the energy stored in it. L= 25mH 0 t < 0 a) i( t) = t 10(1 e ) ma t > 0 vt () = L di dt t < 0: v( t) = 0 t t > 0 : v( t) = (0.025)(0.01) e = 250e 0 t < 0 vt () = t 250e µ V t > 0 t µv b) wt = i t = e + e tµ J L 2 t 2 ( ) 2 ( ) 1.25[1 2 ]

28 Problem 5.25 Given the data in the previous problem, find the voltage across the inductor and the energy stored in it after 1 s. t t 2 vt ( ) = 250 e µ V w( t) = 1.25[1 2 e + e t ] µ J at t = 1sec. v(1) = µ V w(1) = 0.5µ J

29 Problem 5.26 The current in a 50-mH inductor is given by the expressions i(t)=0 t<0 i(t)=2te -4t A t>0 Find (a) the voltage across the inductor, (b) the time at which the current is a maximum, and (c) the time at which the voltage is a minimum 0 t < 0 L= 50 mh i( t) = t 2te A t > 0 di() t 4 t 4t ) ( ) = dt = (0.05)(2 8 ) for > 0 a v t L e te t 0 t < 0 vt () = > b) 4t 0.1 e (1 4 t) V t 0 The current will be at its maximum when di/dt =0, or, v(t)= 0 vt ( ) = 0 = 0.1 e (1 4 t ) t = 0.25sec c) 4t mx max max max The voltage will be at its max. or min. when dv/dt =0 = [0.1 e ( 4) e (4)(4t 1)] = 0 di 4t 4t dt t= tmin t= min yields : t = 0.5 min sec.

30 Problem 5.27 The current i (t) = 0 t<0 i (t) = 100te -t/10 A t>0 flows through a 150-mH inductor. Find both the voltage across the inductor and the energy stored in it after 5 seconds. vt () = L 3 d t / 10 ( ) = dt (100 ) t /10 () = 1000 ( 10 )( e ) vt ( ) = 1.5e t /10 1/2 vs ( ) = 1.5e = 0.91V wt () = 1 2 di dt vt x e vt Li 2 2t () = 2 (5010 )(10 ) wt x e 1 t 5 ( ) = 2 (50)(10) e ) wt wt ( ) = 250e t 5 w(5) = 91.97J

31 Problem 5.28 The current in a 10-mH inductor is shown in Figure P5.28. Find the voltage across the inductor. i(t)(ma) t(ms) -10 Figure P5.28 L= 10 mh v( t) = C Time(ms) di(a/ms) v(t)(mv) dt 0 t t t t t > 7 0 di dt 0

32 Problem 5.29 The current in a 50-mH inductor is given in Figure P5.29. Sketch the inductor voltage. i(t)(ma) t(ms) -100 Figure P5.29 L= 50 mh v( t) = L Time(ms) di(a/s) v(t)(v) dt 0 t t t t di dt t >

33 120 i(t)(ma) v(t)(v) time (ms) time (ms) 10 10

34 Problem 5.30 The current in a 16-mH inductor is given by the waveform in Figure P5.30. Find the waveform of the voltage across the inductor. i(t)(a) t(ms) Figure P5.30 L= 16 mh v( t) = L Time(ms) di(a/ms) v(t)(v) dt 0 t t t t di dt 11 t t > i(t)(a) time (ms)

35 Problem 5.31 Draw the waveform for the voltage across a 10-mH inductor when the inductor current is given by the waveform shown in Figure P5.31 i(t)(a) t(s) Figure P5.31 L= 10 mh v( t) = L Time(s) di(a/s) v(t)(mv) dt 0 t 3 2/ di dt 3 t t t t >

36 4 2 i(t)(a) time(s) 30 v(t)(mv) time(s)

37 Problem 5.32 The voltage across a 10-mH inductor is shown in Figure P5.32. Determine the waveform for the inductor current v(t)(mv) Figure P5.32 t(ms) L= mh i t = v t dt 1 10 () L () 0 t 1 ms v( t) = 10t mv assuming 2 i(0) = 0 A, i( t) = 500t A ( ) = 10 10( 10 ) t ms v t t V it ( ) = ( tdt ) + K 2 it ( ) = 2t 500t + K where K = integration constant A Both equations for i(t) must be equal at t =1ms (10 ) = 2(10 ) 500(10 ) + = 1 So 2 3 it ( ) = 2t 500t 10 A K K ma 2 500t A 0 t 1ms 1 ma t > 2ms 2 3 it ( ) = 2t 500t 10 A 1 t 2ms

38 Problem 5.33 The waveform for the voltage across a 20-mH inductor is shown in Figure P5.33. Compute the waveform for the inductor current. v(t)(mv) t(ms) -20 Figure P5.33 L= mh i t = v t dt 1 20 () L () 0 t 2 ms v( t) = 10mV if i(0) = 0 A, i( t) = 0.5t A 2 t 3 ms v( t) = 20mV it () = ( t 2 m) + i(2 m) = t+ 2m+ 1m 3 it ( ) = (3x10 t) A t > 3 ms v( t) = 0V it () = 0 + i(3 m) = 0A 0.5t A 0 t 2ms 3 it () = 310 x t A 2 t 3ms 0 A t > 3ms

39 10 v(t)(mv) time (ms)

40 Problem 5.34 The voltage across a 2-H inductor is given by the waveform shown in Figure P5.34. Find the waveform for the current in the inductor. v(t)(mv) Figure P5.34 t(ms)

41 L= H i t = v t dt 1 2 ( ) L ( ) 0 t 1 ms assume i(0) = 0A vt () = 1mV it ( ) = 500t µ A 1 t 2 ma v( t) = 0V it ( ) = i(1 ms) = 0.5µ A 2ms t 3 ms v( t) = 1mV it ( ) = 500( t 2 m) + Kµ A K1 is an integrator constant, i(2 m) = 0.5µ A= 500(2m 2 m) + K K 1 = 0.5µ A so, it ( ) = 500t 0.5µ A 3ms t 4 ms v( t) = 0 it () = 0 + i(3 m) = 1µ A 4ms t 5 ms v( t) = 1mV it ( ) = 500( t 4 m) + Kµ A i(4 m) = 1µ A= 500(4m 4 m) + K K2= 1µ A it ( ) = 500t 1µ A t > 5 ms i( t) = 1.5µ A 500t µ A 0 t 1ms 0.5 µ A 1 t 2ms 500t 0.5 µ A 2 t 3ms it () = 1.0 µ A 3 t 4ms 500t 1 µ A 4 t 5ms 1.5 µ A t > 5ms

42 Problem 5.35 Find the possible capacitance range of the following capacitors. a) µf with a tolerance of 10% b) 120pF with a tolerance of 20% c) 39µF with a tolerance of 20% a) Minimum capacitor value = 0.9C = 61.2nF Maximum capacitor value = 1.1C = 74.8nF b) Minimum capacitor value = 0.8C = 96pF Maximum capacitor value = 1.2C = 144pF c) Minimum capacitor value = 0.8C = 31.2µF Maximum capacitor value = 1.2C = 46.8µF

43 Problem 5.36 The capacitor in Figure P5.36a is 51 nf with a tolerance of 10%. Given the voltage waveform in Figure P5.36b graph the current i(t) for the minimum and maximum capacitor values. i(t) -60 v(t) C v(t)(v) 0 (a) -60 Figure P time (ms) (b) Maximum capacitor value = 1.1C = 56.1 nf Minimum capacitor value = 0.9C = 45.9 nf The capacitor voltage and current are related by the equation it () = C dv() t dt

44 Problem 5.37 Find the possible inductance range of the following inductors a) 10 mh with a tolerance of 10% b) 2.0 nh with a tolerance of 5% c) 68 µh with a tolerance of 10% a) Minimum inductor value = 0.9L = 0.9 mh Maximum inductor value = 1.1L = 1.1 mh b) Minimum inductor value = 0.95 = 1.9 nh Maximum inductor value = 1.05L = 2.1 nh c) Minimum inductor value = 0.9L = 61.2 nh Maximum inductor value = 1.1L = 74.8 nh

45 Problem 5.38 The inductor in Figure P5.38a is 330µH with a tolerance of 5%. Given the current waveform in Figure P5.38b, graph the voltage v(t) for the minimum and maximum inductor values. i(t) v(t) L (a) Figure P5.38 Maximum inductor value = 1.05L = µh Minimum inductor value = 0.95L = µh The inductor voltage and current are related by the equation vt () = L di() t dt

46 Problem 5.39 The inductor in Figure P5.39a is 4.7 µh with a tolerance of 20%. Given the current waveform in Figure P5.39b, graph the voltage v(t) for the minimum and maximum inductor values. i(t) v(t) L (a) Figure P5.39 Maximum inductor value = 1.2L = 5.64 µh Minimum inductor value = 0.8L = 3.76 µh The inductor voltage and current are related by the equation vt () = L di() t dt

47 Problem 5.40 What values of capacitance can be obtained by interconnecting a 4-µF capacitor, a 6-µF capacitor, and a 12-µF capacitor? Combo A C1 Combo B Combo C C2 C1 C3 C1 C2 C3 C2 C3 C A = 1 / ( 1/C 1 + 1/C 2 + 1/C 3 ) = 2µF C B = C 1 + C 2 + C 3 = 22µF C C = [ C 1 (C 2 + C 3 ) ] / [C 1 + (C 2 + C 3 )] C C = 3.27µF Combo D Combo E Combo F C2 C3 C1 C3 C1 C3 C2 C1 C2 C D = [ C 2 (C 1 + C 3 ) ] / [C 2 + (C 1 + C 3 )] C D = 4.36µF C C = [ C 3 (C 2 + C 1 ) ] / [C 3 + (C 2 + C 1 )] C C = 5.45µF C F = C 3 + C 1 C 2 / (C 1 + C 2 ) = 14.4µF Combo G C1 C2 C3 C G = C 2 + C 1 C 3 / (C 1 + C 3 ) = 9µF Combo H C2 C1 C3 C H = C 1 + C 2 C 3 / (C 2 + C 3 ) = 8µF Possibilities: CA = 2µF CB = 22µF CC = 3.27µF CD = 4.36 µf CE = 4.45µF CF = 14.4µF CG = 9 µf CH = 8 µf

48 Problem 5.41 Given a 1, 3, and 4-µF capacitor, can they be interconnected to obtain an equivalent 2-µF capacitor? C1 = 1µF C2 = 3µF C3 = 4µF C 1 C 2 C 3 C eq ( C1+ C2) C3 (1µ + 3 µ )4µ = = = 2µ F C1+ C2+ C3 1µ + 3µ + 4µ

49 Problem 5.42 Given four 2-µF capacitors, find the maximum value and minimum value that can be obtained by interconnecting the capacitors in series/parallel combinations. Minimum combo Maximum combo C 1 C 2 C 3 C 4 C 1 C 2 C 3 C 4 1 C = = 0.5µ F C = C + C + C + C 8µ F min max = C + 1 C + 2 C + 3 C4 C C min max = 0.5µ F = 8µ F

50 Problem 5.43 The two capacitors in Figure P5.42 were charged and then connected as shown. Determine the equivalent capacitance, the initial voltage at the terminals, and the total energy stored in the network. - 4V + + 1V - 6µF 3µF Figure 5.43 Ceq + - V V 1 V C 1 C 2 C 1 = 6µF V 1 =- 4V C 2 = 3µF V 2 = 1V Ceq = C1C2/( C1+ C2) = 2µF V = V1+ V 2= 3V W = CV + CV W = 49.5µ J C eq = 2µ F V = 3V W = 49.5µ J

51 Problem 5.44 Two capacitors are connected in series as shown in Figure P5.44. Find Vo i(t) V o V - C 1 C 2 C 1 = 12µF C 2 = 6µF i(t) V o V - C 1 C 2 C 1 = 12µF C 2 = 6µF vc = CV 1 C 1 o 2 i() t dt (24) V = 24( C / C ) o Since same current charged both caps, = C 2 1 V o = 12V

52 Problem 5.45 Three capacitors are connected as shown in Figure P5.45. Find V 1 and V 2. V 1 V µF 4µF V Figure 5.45 V 1 V C 1 =8µF C 2 =4µF V V1+ V 2= V CV V V V V Assuming C 1 and C 2 are charged by the same surrent, = CV C 2 = 12 = 4V C1+ C2 C1 = 12 = 8V C1+ C2 = 4V = 8V (same as Q 1= Q 2)

53 Problem 5.46 Select the value of C to produce the desired total capacitance of C T =2µF in the circuit in Figure P5.46 C C T C 1 =2µF C 2 =4µF C C T C 1 =2µF C 2 =4µF C T CC ( 1+ C2) 6C 2 C+ C1+ C2 6 + C = = C = 6C 4C = 12 C = 3µ F

54 Problem 5.47 Select the value of C to produce the desired total capacitance of C T =1µF in the circuit in Figure P5.47. C C 1µF C T 1µF 2µF 1µF Figure P C (1 + C)(1) CT = 1 = C (1 + C ) + 1 3C 1+ C 1 = C 2 + C (3 + C)(2 + C) = 3 C(2 + C) + (3 + C)(1 + C) 2 0= C C 5± 25 4(3)( 3) C1, C2 = 2(3) C = 0.47µ F

55 Problem 5.48 Find the equivalent capacitance at terminals A-B in Figure P5.48 5µF 3µF A 2µF 6µF 2µF 6µF B 6µF 12µF Figure 5.48 C 1 C 5 A C 2 C 6 C 3 C 7 B C 4 C 8 C 1 =5µF C 4 =7µF C 6 =6µF C 2 =2µF C 5 =3µF C 7 =6µF C 3 =2µF C 8 =12µF

56 CC 5 6 Ceq1 = = C5+ C6 C eq2=c 2+C eq1=4µf CC 2µF 7 8 Ceq3 = = C7+ C8 C eq4=c 3+C eq3=6µf 4µF = + Ceq5 = 2.4µF Ceq5 Ceq2 Ceq = + + CAB = 1.32µF CAB C1 C4 Ceq5

57 Problem 5.49 Determine the total capacitance of the network in Figure P5.49 4µF 1µF 2µF C T 4µF 2µF 12µF 4µF Figure P5.49 C 1 C 4 C 6 C 2 C 3 C 5 C 7 C 1 =4µF C 4 =1µF C 6 =2µF C 2 =4µF C 5 =2µF C 7 =4µF C 3 =12µF C C AB AB = ( C4+ C6)( C5+ C7) ( C4+ C6) + ( C5+ C7) = 2µ F = + + = + + CT C2+ CAB C1 C CT = 2µ F

58 Problem 5.50 Find C T in the network in Figure P5.50 if (a) the switch is open and (b) the switch is closed. 3µF 6µF C T 6µF 12µF Figure P5.50 The networks can be reduced as follows: a) 3 6 2µF µF C T C T =(4 2)=6µF b) 3 6 9µF 18µF 6 12 C T CT =9(18)/(9+18)=6µF

59 Problem 5.51 Find the total capacitance C T of the network in Figure P µF C T 6µF 1µF 8µF 3µF Figure P5.51 C 1 = 4µF C 1 C T C 4 C 2 = 1µF C 3 = 8µF C 2 C 4 = 6µF C 3 C 5 C 5 = 3µF C eq1 = C 2 + C 3 + C 5 = 12µF C eq2 = C 1 C eq1 /(C 1 + C eq1 ) = 3µF C T = C 4 + C eq2 = 9µF C T = 9µF

60 Problem 5.52 Compute the equivalent capacitance of the network in Figure P5.52 if all the capacitors are 5µF. Figure P C eq = 4(8)/ = 32/3 µf

61 Problem 5.53 If all the capacitors in Figure P5.53 are 6 µf, find C eq C eq Figure P5.53 C eq C eq C eq /15 = 15.6 Ceq = 4(15.6)/(19.6)=3.18µF

62 Problem 5.54 Given the capacitors in Figure 5.54 are C1=2.0uF with a tolerance of 2% and C2=2.0uF with a tolerance of 20%, find a) the nominal value of CEQ b) the minimum and maximum possible values of CEQ c) the percent errors of the minimum and maximum values C eq C 1 C 2 Figure P5.54 a) The nominal value is = + + = + + = 6 CEQ C1 C2 C3 ( ) x F µ b) The minimum value of CEQ is C = C + C + C = + + x = C EQ,min 1,min 2,min 3,min The maximum value of CEQ is EQ,max 1,max 6 (0.1* *0.8 1*0.9) = C + C2,max + C3,max = (0.1* * *1.1) x10 = 1.606µF µf c) The percent irror of the minimum value is x 100 = 12.3% 1.43 The percent irror of the maximum value is x 100 = 12.3% 1.43

63 Problem 5.55 Given the capacitors in Figure 5.55 are C 1 =0.1µF with a tolerance of 2% and C 2 =0.33µF with a tolerance of 20% and 1 µf with a tolerance of 10%. Find the following. a) the nominal value of C EQ b) the minimum and maximum possible values of C EQ c) the percent errors of the minimum and maximum values C eq C 1 C 2 C 3 Figure P5.55 a) The nominal value is CC 1 2 (2.0)(2.0) 10 6 CEQ = = x = 1.0 µ F C1+ C b) The minimum value of CEQ is C C C C ( 2.0*0.98)(2.0*0.8) 1,min 2,min 6 EQ,min = x = C1,min + C2,min (2.0*0.98) + (2.0*0.8) The maximum value of C C CEQ is ( 2.0*1.02)(2.0*1.2) 1,max 2,max 6 EQ,max = x = C1,max + C2,max (2.0*1.02) + (2.0*1.2) µF µF c) The percent irror of the minimum value is x 100 = 11.9% 1.0 The percent irror of the maximum value is x 100 = 10.3% 1

64 Problem 5.56 Select the value of L that produces a total inductance of L T = 10mH in the circuit in Figure P mH L T = 10mH L 8mH Figure P5.56 L T = (L 1 + L 2 ) L / (L 1 + L 2 + L) L 1 10 = 20 L / (20 + L) L T L L = 20L L = 20mH L 2

65 Problem 5.57 Find the value of L in the network in Figure P5.57 so that the total inductance L T will be 2 mh. 4mH 2mH L L T 6mH Figure P5.57 4mH 2mH L T L All L's in mh 6mH

66 [ + 2]4 = L 6+ L 6L 6+ L 24L 6+ L 6L 6+ L + 8 = L+ 8(6 + L) = (6 L+ (6 + L))2 24L L= 12L L 8L = 24 L= 3mH

67 Problem 5.58 Find the value of L in the network in Figure 5.58 so that the value of L T will be 2 mh. 2mH 1mH L T 6mH L L 4mH Figure P L (4L) / (4+L) [ + 2] L 4+ L 4L 4+ L L L = 2 L L L L = ( L+ 4)( L+ 1) = 0 L= 4mH 6

68 Problem 5.59 Determine the inductance at terminals A-B in the network in Figure P5.59. A 1mH 6mH 2mH 12mH B 1mH 4mH 2mH 2mH Figure 5.59 L 6 A L 1 L 4 L 7 L 2 L 5 L 8 L 3 B L 9 L 1 = L 3 = L 6 = 1mH L 2 = 12mH L 4 = 6mH L 5 = 4mH L 7 = L 8 = L 9 = 2mh

69 L = L + L = 3mH eq1 6 7 L = L L /( L + L ) = 2mH eq2 4 eq1 4 eq1 L = L + L + 4mH eq3 8 9 L = L L /( L + L ) = 2mH L eq4 eq3 5 eq3 5 L ( L + L ) 2 eq2 eq4 eq5 = = L2 + Leq2 + Leq eq5 3mH L = L + L + L = 5mH L AB AB = 5mH

70 Problem 5.60 Compute the equivalent inductance of the network in Figure P5.60 if all inductors are 5 mh L eq Figure 5.60 Redrawing the network Leq = 4(6) / = 24 / / 10 = 44 / 10 = 4.4mA

71 Problem 5.61 Determine the inductance at terminals A-B in the network in Figure P5.61 A 1mH 4mH 12mH 2mH B 2mH 3mH 4mH Figure 5.61 A L 1 L 2 L 3 L 6 L 4 L 7 B L 5 L 1 = 1mH L 2 = L 7 = 4mH L 3 = 12mH L 4 = 3mH L 5 = L 6 = 2mH

72 L = L + L = 6mH eq1 6 7 L = L L /( L + L ) = 3mH eq L = L + L = 6mH eq3 eq2 4 L = L L /( L + L ) L eq4 eq1 eq3 eq1 eq3 eq4 = 3mH L = L + L + L5 L AB AB 1 eq4 = 6mH

73 Problem 5.62 Find the total inductance at the terminals of the network in Figure P5.62 6mH 3mH 6mH 4mH L T 10mH Figure 5.62 L 1 = 6mH L 1 L 2 L 3 L 2 = 6mH L 3 = 3mH L 4 = 4mH L 4 L T L 5 = 10mH L 5 L 1 is shorted out! L = L L /( L + L ) = 2mH eq L = L + L = 10m+ 2m= 12mH eq2 eq1 5 L = L L /( L + L ) T 4 eq2 4 eq2 L T = 3mH

74 Problem 5.63 Given the network shown in Figure P5.63 find (a) the equivalent inductance at terminals A-B with terminals C-D short circuited. And (b) the equivalent inductance at terminals C-D with terminals A-B open circuited. 12H A C 6H 2H B 2H Figure P5.63 D

75 A L 1 = 12H L 1 L 3 L 2 = 2H C D L 3 = 6H L 4 = 2H L 2 L 4 L eq1 = L 1 L 3 /(L 1 + L 3 ) = 4H B L eq2 =L 2 L 4 /(L 2 + L 4 ) = 1H L AB = L eq1 + L eq2 L AB = 5H A L 1 C L 3 C L eq3 = L 1 + L 3 = 18H L eq4 = L 2 + L 4 = 4H 1/L CD = 1/L eq3 + 1/L eq4 L 2 L 4 L CD = 3.27 H B D

76 Problem 5.64 For the network in Figure P5.64 choose C such that v = 10 v dt o C s Rs=10KΩ 70KΩ + + Vs - R L v o - Figure P5.64 C 70KΩ i 1 i 2 0 Rs + Vs R L v o -

77 Rs = 10KΩ v o = 10 v dt s R = Rs + 70K = 80K i eq Using ideal op-amp assumptions, = i 1 2 vs 0 dvo 1 = C vo = vsdt R dt R C eq So, 1 R eq C = C = 1.25µ F 10 eq

78 Problem 5.65 For the network in Figure P5.65, vs(t)=120cos377t V. Find Vo(t) 1KΩ 1µF Vs(t) 0 + vo(t) - Figure P5.65 R Vs(t) C i 1 i vo(t) - C = 1µ F R= 1KΩ v i s = 120cos377t V 1 2 dvs vo dvs C = vo = RC dt R dt v ( t) =+ (1 k)(1 µ )(120)(377)sin(377) t o = i v ( t) = 45.24sin(377 t) V o (ideal op-amp assumptions)

79 Problem 5.66 For the network in Figure P5.66, v s (t)=115sin377t V. Find v o (t) 5µF 5KΩ Vs(t) 0 + vo(t) - Figure P5.66 5µF 5KΩ C Vs(t) R + vo(t) - 1 vo() t = vsdt RC 115 = cos(377 t) (5 K)(5 µ )(377) v ( t) = 12.20cos(377 t) V o

80 Problem 5FE-1 Given three capacitors with values 2µF, 4µF and 6µF, can the capacitors be interconnected so that the combination is an equivalent 3µF? Yes. 2µF 4µF 3µF 6µF

81 Problem 5FE-2 The current pulse shown in Figure 5PFE-2 is applied to a 1µF capacitor. Determine the charge on the capacitor and the energy stored. 6A 0 1 t (µsec) Figure PFE-2 The capacitor voltage is T v() t = i() t dt 6dt 6V C = 10 = Q= CV = = µ C (10 6 )(6) 2 18 W = CV = = µ J 2 2

82 Problem 5FE-3 In the network shown in Figure 5PFE-3, determine the energy stored in the unknown capacitor Cx. + 24V 8V µF Cx - Figure PFE V 8V µF Q = CV = (60x10-6 )(8) = 480µC Vx = 24-8 = 16V Cx = Q/Vx = (480x10-6 ) / 16 = 30µF - Cx W = 1/2 CV 2 = 1/2 (30x10-6 ) (16) 2 W = 3.84mJ

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