EE 202-Winter 2013-2014 (132)
HW5 Solution
Prepared By: Dr. Mohammad S. Sharawi
Due 13/4/2014
Q1
A voltage of 60 cos 4πt V appears across the terminals of
a 3-mF capacitor. Calculate the current through the capacitor and
the energy stored in it from t = 0 to t = 0.125 s.
i=C
dv
= 3x10 −3 x 60(4π)(− sin 4π t)
dt
= –0.72 π sin (4πt) A
P = vi = 60(-0.72)π cos 4π t sin (4π t) = -21.6π sin (8π t) W
W=
∫
t
o
1
pdt = − ∫ 8 21.6π sin(8πt ) dt
o
=
21.6π
cos 8π
8π
1/ 8
o
= –5.4J
Q2 Find the equivalent capacitance between terminals a and b in the
circuit of Fig. 1. All capacitances are in µF.
We combine 10-, 20-, and 30- µ F capacitors in parallel to get 60 µ F. The 60 - µ F capacitor in
series with another 60- µ F capacitor gives 30 µ F.
30 + 50 = 80 µ F, 80 + 40 = 120 µ F
120- µ F capacitor in series with 80 µ F gives (80x120)/200 = 48
48 + 12 = 60
60- µ F capacitor in series with 12 µ F gives (60x12)/72 = 10 µ F
In the circuit in Fig. 2, let i s = 30e-2t mA and v 1 (0) = 50 V, v 2 (0) = 20 V. Determine:
(a) v 1 (t) and v 2 (t), (b) the energy in each capacitor at t = 0.5 s.
Q3
(a)
v1 =
C eq = (12x60)/72 = 10 µ F
10 − 3
t
∫ 30e
12 x10 − 6
− 2t
dt + v1 (0) = − 1250e − 2 t 0 + 50 = − 1250e − 2 t + 1300V
t
0
v2 =
10 − 3
t
∫ 30e
60 x10 − 6
− 2t
dt + v 2 (0) = 250e − 2 t 0 + 20 = − 250e − 2 t + 270V
t
0
(b)
At t=0.5s,
v1 = −1250e −1 + 1300 = 840.2,
w12 µF =
v 2 = −250e −1 + 270 = 178.03
1
x12 x10 −6 x(840.15) 2 = 4.235 J
2
w 20µF =
1
x 20 x10 − 6 x (178.03) 2 = 0.3169 J
2
w 40µF =
1
x 40 x10 − 6 x (178.03) 2 = 0.6339 J
2
Q4 Determine L eq at terminals a-b of the circuit in Fig. 3.
1
1
1
1
1
=
+
+
=
L 60 20 30 10
L eq = 10 (25 + 10 ) =
= 7.778 mH
10 x 35
45
L = 10 mH
Q5 Determine L eq that may be used to represent the inductive network of
Fig. 4 at the terminals.
Let v = L eq
di
dt
(1)
v = v1 + v 2 = 4
di
+ v2
dt
(2)
i = i1 + i2
v2 = 3
i2 = i – i1
(3)
di1 di1 v 2
or
=
dt
dt
3
(4)
and
− v2 + 2
v2 = 2
di
di
+5 2 = 0
dt
dt
di
di
+5 2
dt
dt
Incorporating (3) and (4) into (5),
v2 = 2
v
di
di
di
di
+5 −5 1 = 7 −5 2
3
dt
dt
dt
dt
di
 5
v 2 1 +  = 7
dt
 3
v2 =
21 di
8 dt
Substituting this into (2) gives
v=4
di 21 di 53 di
+
=
dt 8 dt
8 dt
Comparing this with (1),
L eq =
53
= 6.625 H
8
(5)
Q6 Find the time constant of each of the circuits in Fig. 5.
(a) RTh = 10 // 10 = 5kΩ,
τ = RTh C = 5 x10 3 x 2 x10 −6 = 10 ms
(b) RTh = 20 //(5 + 25) + 8 = 20Ω,
τ = RTh C = 20 x0.3 = 6s
Q7 In the circuit of Fig. 6, v(0) = 20 V. Find v(t) for t > 0.
v( t ) = v(0) e- t τ ,
τ = R eq C
Req = 2 + (8 || 8) + (6 || 3) = 2 + 4 + 2 = 8 Ω
τ = R eq C = (0.25)(8) = 2
v( t ) = 20 e -t 2 V
Q8 Consider the circuit of Fig. 7. Find v 0 (t) if i(0) = 2 A and v(t) = 0.
i ( t ) = i ( 0) e - t τ ,
τ=
14 1
L
=
=
R eq
4 16
i( t ) = 2 e -16t
v o ( t ) = 3i + L
di
= 6 e-16t + (1 4)(-16) 2 e-16t
dt
v o ( t ) = - 2 e -16t V