Chapter 6, Problem 2. A 20-μF capacitor has energy 2( ) 10cos 377w t t= J. Determine the current through the capacitor.

Chapter 6, Solution 2.

2

2 2 6 26

1 2 20cos 377 10 cos 3772 20 10

W tw Cv v tC x −= ⎯⎯→ = = =

v = ±103cos(377t) V, let us assume the v = +cos(377t) mV, this then leads to,

i = C(dv/dt) = 20x10–6(–377sin(377t)10–3) = –7.54sin(377t) A.

Please note that if we had chosen the negative value for v, then i would have been positive.

Chapter 6, Problem 8. A 4-mF capacitor has the terminal voltage

⎩⎨⎧

≥+≤

=0 V, BeAe

0 V, 50600-100- t

tv tt

If the capacitor has initial current of 2A, find:

(a) the constants A and B, (b) the energy stored in the capacitor at t = 0, (c) the capacitor current for t > 0.

Chapter 6, Solution 8.

(a) tt BCeACedtdvCi 600100 600100 −− −−== (1)

BABCACi 656001002)0( −−=⎯→⎯−−== (2)

BAvv +=⎯→⎯= −+ 50)0()0( (3) Solving (2) and (3) leads to A=61, B=-11

(b) J 5250010421)0(

21Energy 32 === − xxxCv

(c ) From (1),

A 4.264.24104)11(60010461100 60010060031003 tttt eeexxxexxxi −−−−−− +−=−−−= Chapter 6, Problem 9.

The current through a 0.5-F capacitor is 6(1-e-t)A. Determine the voltage and power at t=2 s. Assume v(0) = 0. Chapter 6, Solution 9.

v(t) = ( ) ( )∫ −− +=+−to

t

0tt Vet120dte16

211 = 12(t + e-t) – 12

v(2) = 12(2 + e-2) – 12 = 13.624 V p = iv = [12 (t + e-t) – 12]6(1-e-t) p(2) = [12 (2 + e-2) – 12]6(1-e-2) = 70.66 W Chapter 6, Problem 13.

Find the voltage across the capacitors in the circuit of Fig. 6.49 under dc conditions. Figure 6.49

Chapter 6, Solution 13.

Under dc conditions, the circuit becomes that shown below:

i2 = 0, i1 = 60/(30+10+20) = 1A

30 Ω

+

v1

+ −

+

v2

30 Ohms

10 Ohms 50 Ohms

20 Ohms

60V

v1 = 30i1 = 30V, v2 = 60–20i1 = 40V

Thus, v1 = 30V, v2 = 40V Chapter 6, Problem 37. The current through a 12-mH inductor is 4 sin 100t A. Find the voltage, and also the energy stored in the inductor for 0 < t < π/200 s. Chapter 6, Solution 37.

t100cos)100(4x10x12dtdiLv 3−==

= 4.8 cos 100t V p = vi = 4.8 x 4 sin 100t cos 100t = 9.6 sin 200t

w = t200sin6.9pdtt

o

200/11

o∫ ∫=

Jt200cos200

6.9 200/11o−=

=−π−= mJ)1(cos48 96 mJ Please note that this problem could have also been done by using (½)Li2. Chapter 6, Problem 39.

The voltage across a 200-mH inductor is given by v(t) = 3t2 + 2t + 4 V for t > 0. Determine the current i(t) through the inductor. Assume that i(0) = 1 A.

Chapter 6, Solution 39

)0(iidtL1i

dtdiLv t

0 +∫=⎯→⎯=

1dt)4t2t3(10x200

1i t0

23 +∫ ++= −

1)t4tt(5t

023 +++=

i(t) = 5t3 + 5t2 + 20t + 1 A

Chapter 6, Problem 47. For the circuit in Fig. 6.70, calculate the value of R that will make the energy stored in the capacitor the same as that stored in the inductor under dc conditions.

Figure 6.70

Chapter 6, Solution 47. Under dc conditions, the circuit is equivalent to that shown below:

,2R

10)5(2R

2iL +=

+=

2RR10Riv Lc +

==

2

262

cc )2R(R100x10x80Cv

21w

+== −

232

1L )2R(100x10x2Li

21w

+== −

If wc = wL,

+ vC

5A 2 Ohms

R

2

3

2

26

)2R(100x10x2

)2R(R100x10x80

+=

+

−− 80 x 10-3R2 = 2

R = 5Ω

Chapter 6, Problem 48. Under steady-state dc conditions, find i and v in the circuit in Fig. 6.71.

i 2 mH + 10 mA 30kΩ v 6 μF 20 kΩ - Figure 6.71 For Prob. 6.48. Chapter 6, Solution 48.

Under steady-state, the inductor acts like a short-circuit, while the capacitor acts like an open circuit as shown below. i + 10 mA 30kΩ v 20 kΩ

–

Using current division,

i = [30k/(30k+20k)]10 mA = 6 mA

v = 20ki = 120 V