EE 221
Exam #3
11/21/2002
Name: ____________________________________
_____ / 100 pts
You need to show all of your steps in detail to get full credit!
1. Determine the voltage and current for each of the passive
elements in the circuit. Find the values of the energy stored in
the capacitor and the inductor.
R1=20Ω
Vs=15V
(15 pts)
L=8mH
C=5µF
+
−
R2=10Ω
DC-voltage source ⇒ IC = 0 A (open loop for DC) and VL = 0 V
(short circuit for DC). The resulting circuit consists only of the
voltage source in series with the two resistors:
I = VS / (R1 + R2) = 15V /(20Ω+10Ω) = 0.5 A
VR1 = VS R1 / (R1 + R2) = 15V 2/3 = 10 V
VR2 = VS R2 / (R1 + R2) = 15V 1/3 = 5 V
The inductor’s current is IL = I = 500 mA and the capacitor’s
voltage VC = VR2 = 5 V. Therefore, the values for energy stored
are:
WL = ½ L IL2 = 0.5 * 0.008 * ¼ Ws = 1 mWs = 1mJ
WC = ½ C VC2 = 0.5 * 5e-6 * 25 Ws = 62.5 µWs = 62.5 µJ
VR1 = _10 V__
IR1 = _0.5 A__
VL = _0 V __
IL = _0.5 A__
VC = _5 V___
IC = _0 A___
VR2 = _5 V___
IR2 = _0.5 A__
WC = _62.5 µJ_
WL = __1 mJ__
Example 1 of 6
2. The voltage across the terminals of a 1mF
capacitor is shown below. Sketch the current
through the capacitor and the energy stored in the
capacitor as function of time.
v(t) in V
(15 pts)
iC(t) = C dv/dt :
from 0 to 1s:
10
1
3
5
t (s)
voltage rises linearly,
iC = 1mF *10V/1s = 10mA
from 1 to 2s:
voltage is constant,
iC = 0A
-10
from 2 to 3s:
i(t) in mA
voltage drops linearly,
iC = -1mF *20V/1s = -20mA
20
10
from 3 to 4s:
1
3
5
t (s)
voltage is constant,
iC = 0A
-10
from 4 to 5s:
-20
voltage rises linearly,
iC = 1mF *10V/1s = 10mA
w(t) in mJ
w c(t)=½CVC2:
from t = 1-2s:
50
wc = ½ 1mF (10V)2 = 50mWs
1
0
3
5
t (s)
from t = 3-4s:
wc = ½ 1mF (-10V)2 = 50mWs
between 0-1s, 2-3s, and 4-5s:
increases and decreases
between the two extreme values
of zero and the maximum of
50mWs with v2
Example 2 of 6
3. For the circuit shown below, use nodal analysis in the phasordomain to find a consistent set of equations. Do not substitute,
simplify, or solve.
(15 pts)
V2
-j4Ω
2Ω
V1
V3
j7Ω
2∠75ºA
-j8Ω
10Ω
5∠-45ºA
GND
3 equations in 3 unknowns:
A) V1 / 10Ω + (V1 − V2) / (−j4Ω) + (V1 − V3) / (j7Ω) = 2A∠75º
B) (V2 − V1)/ (−j4Ω) + (V2 − V3) / 2Ω = 0
C) V3 / (−j8Ω) + (V3 − V1) / (j7Ω) + (V3 − V2) / 2Ω = −5A∠-45º
Example 3 of 6
4.
a.) Solve for IC and VC in the phasor-domain and draw a
phasor diagram showing both.
b.) Add the source current to the phasor diagram; finally,
indicate (approximately = without computations) the
current IL
c.) What is the capacitor’s current in the time domain iC(t)
assuming ω = 100 rad/s?
VC
IS 2∠0ºA
IC
(15 pts)
j2Ω
IL
−j2Ω
R=2Ω
GND
Current division to find IC:
IC = I (2 + j2Ω) / (2 + j2 − j2Ω) = 2A∠0º (2 + j2Ω) / (2Ω) =
= 2 + j2 A = √8∠45º A ≅ 2.83∠45º A
Im
IC
IL
Ohm’s law in phasor-domain:
VC = ZC IC = −j2Ω ( 2 + j2 A) = −j4 + 4 V =
= 2√8∠−45º ≅ 5.66∠−45º
(note: VC lags the current IC by 90 degrees)
Re
IS
IL
VC
The current in the time-domain:
IC = √8∠45º A ⇒ iC(t) = √8 cos(100t + 45º)
VC = __2√8∠−45º V__
IC = ___√8∠45º A____
iC(t) = ___√8 cos(100t + 45º)___________
Example 4 of 6
5. Find the Thévenin-equivalent (VTH and ZTH) of the circuit and the
load ZL that maximizes the average power delivered to the load.
What is the maximum average power P?
1Ω
Vs=15∠0ºV
j1.3Ω
1.1Ω
(20 pts)
a
-j3Ω
+
−
b
The Thevenin-impedance as seen from the two terminals a and b (voltage source
replaced by short-circuit):
ZTH = (1Ω || −j3Ω ) + (1.1Ω + j1.3Ω) = −j3Ω / (1 − j3Ω) + 1.1Ω + j1.3Ω =
= (−j3 + 9Ω) / (1 + 9) + 1.1Ω + j1.3Ω = 2 + j Ω = √5∠26.6º Ω
VTH = 15V (−j3Ω) / (1Ω − j3Ω) = 15V (0.9Ω − j0.3Ω) =
= 13.5 − j4.5 V = 14.23∠−18.43º V
ZL = Z*TH (the conjugate complex) = 2 − j Ω = √5∠ −26.6º Ω
Connecting the load results in an overall impedance of
Z = ZL + ZTH = 2 − j Ω + 2 + j Ω = 4Ω
Therefore, the current through the load equals
I = V / Z, also: |I| = |V| / |Z| and Im = Vm / |Z|.
Using the last relationship of Im = Vm / |Z| = 14.23 / 4Ω = 3.558 A, allows to
compute the consumed power P = ½ Im2 RL = ½ 3.5582 2 = 12.66 W.
Note: You could have used the voltage magnitude directly: P = ½ (Vm/2)2 / RL
ZTH = __2 + j Ω_, VTH = __14.23∠−18.43º V__
ZL = __2 − j Ω__, P = ___12.66 W_______
Example 5 of 6
6. The circuit shown has reached steady-state conditions. Determine
the average power generated by each source and the average power
delivered to each impedance.
(20 pts)
Z1 = 1.41∠45º Ω = 1+j Ω, Z2 = 3.61∠-33.7º Ω = 3 − j2 Ω
1∠90ºA
2∠0ºA
Z1
Z2
I = I1 + I2 = 1∠90º + 2∠0º = 2 + j A = √5∠26.6º A
I1 = I Z2 / (Z1 + Z2) = √5∠26.6º A (3.61∠−33.7º) / (4 −j) = 1.96∠6.9º
I2 = I Z1 / (Z1 + Z2) = √5∠26.6º A (1.41∠45º) / (4 −j) = 0.77∠85.6º
P1 = ½ Im12 R1 = ½ (1.96A)2 1Ω = 1.92 W
P2 = ½ Im22 R2 = ½ (0.77A)2 3Ω = 0.89 W
The voltage across the elements equals
V = I1 Z1 = 1.96∠6.9ºA 1.41∠45ºΩ = 2.77∠51.9º V
PS1 = ½ Vm Im,S1 cos(ϕV - ϕI,S1) = ½ 2.77 cos(51.9 - 90) = 1.09 W
PS2 = ½ Vm Im,S2 cos(ϕV - ϕI,S2) = ½ 2.77 2 cos(51.9 - 0) = 1.71 W
PS1 = ___1.09 W_
PS2 = ___1.71 W_
PZ1 = __1.92 W_
PZ2 = __0.89 W_
ΣP = _≈0 W_(0.01W)
Example 6 of 6