EE 221
Final Exam
Name: ____________________________________
12/09/2002
_____ / 100 pts
You need to show all of your steps in detail to get full credit!
1. Determine the current I using current division. Find the
voltage V and the power P supplied by the source.
(8 pts)
+
IS 12A
V
I
6Ω
18Ω
18Ω
3Ω
−
GND
I = (1/18) / (1/18 + 1/18 + 1/9) * (−12A) = 1/4 (−12A) = −3 A
V = I R = −3A 18Ω = −54 V
Psupplied = −V I = −(−54V) (12A) = 648 W
I = _−3 A__
V = _−54 V__
P = _648 W__
Example 1 of 10
2. Given the following circuit, determine the voltage V and
current I.
(8 pts)
R1=20Ω
R3=45Ω
VS=24V
R4=90Ω
+
−
R2=50Ω
−
V
I
+
R’ = R3 || R4 = R3 R4 / (R3 + R4) = 45 90 / 135 = 30 Ω
V’ = VS R’ / (R1 + R’ + R2) = 24V 30/(20 + 30 + 50) = 24V 3/10 = 7.2 V
I = V’ / R4 = 7.2V / 90Ω = 80 mA
V = VS R2 / (R1 + R’ + R2) = 24V 50/100 = 12 V
I = __80 mA__
V = _12 V__
Example 2 of 10
3. For the circuit shown below, use nodal analysis to find a consistent
set of equations using the assigned node voltages V1, V2, V3, and V4.
Do not substitute, simplify, or solve.
V2
V1
3IR3
R2
V3
+
−
R1
R3
IR3
−
+
Vs
(12 pts)
Is
V4
R4
GND
A) (KVL)
V2 = −VS
B) (Supernode KCL) (V1 − V2) / R1 + V4 / R4 + (V4 − V3) / R3 = 0
C) (Supernode KVL) V1 − V4 = 3 IR3 = 3 (V3 − V4) / R3
D) (KCL)
(V3 − V2) / R2 + (V3 − V4) / R3 = IS
Example 3 of 10
4. Use mesh analysis in the phasor domain to find a consistent set
of equations using the assigned mesh currents I1, I2, I3, and I4. Do
not substitute, simplify, or solve. Use ω = 100 rad/s.
(12 pts)
+
−
5I
I4
20Ω
4Ω
20mH
−
+
I2
I1
5mF
I3
I
GND
5 cos(ω t+50º)A
4 cos(ω t+25º)V
Circuit into phasor-domain:
L: 20mH ⇒ jωL = j 100 0.02 = j2 Ω
C: 5mF ⇒ 1/jωC = −j 1/(100 0.005) = −j2 Ω
Signals into phasor-domain:
I: 5 cos(ωt + 50°)A ⇒ 5∠50° A
V: 4 cos(ωt + 25°)V ⇒ 4∠25° V
A) (KCL) I1 = −5∠50° A
B) (KVL) −4∠25° V + 20Ω (I2 − I1) + j2Ω (I2 − I4) = 0
C) (KVL) 4∠25° V − j2Ω I3 = 0
D) (KVL) 5 I + 4Ω I4 + j2Ω (I4 − I2) = 0 where I = I3
Example 4 of 10
5. Find the output voltage VOUT of the following ideal OpAmp-circuit.
(10 pts)
R3=3kΩ
R1=6kΩ
VOUT
Vs=8V
+
−
R2=5kΩ
R4=5kΩ
V + = V- = V
A) KCL at (-) node: (V − VS) / 6k + (V − VOUT) / 3k = 0
B) KCL at (+) node: V / 5k + (V-VOUT) / 5k = 0
------------
A) V/2 − VS/2 + V − VOUT = 0
B) 2V − VOUT = 0
-----------B) ⇒ A)
3/2 VOUT/2
= VOUT + VS/2
VOUT (3/4 - 1) = VS/2
VOUT = −2 VS = −16 V
VOUT = ___−16 V___
Example 5 of 10
6. The current through an inductor (4H) is shown below. Sketch the
voltage across the inductor’s terminals and determine the energy
stored in the inductor at t = 2.5s.
(10 pts)
i(t) in A
1.5
1
1
3
5
t (s)
1
3
5
t (s)
v(t) in V
4
2
-2
-4
uL = L di / dt = 4H ∆I / ∆t
time (s)
0-1
1-2
2-3
3-4
4-5
∆I / ∆t (A/s)
1
½
0
-1/2
-1
uL (V)
4
2
0
-2
-4
w(2.5s) = ½ L I(2.5)2 = ½ 4H (1.5A)2 = 2 9/4 J = 4.5 J
w(2.5s) = __4.5 J__
Example 6 of 10
7. For the circuit below, find the equivalent input impedance of the
network at ω = 20 rad/s.
(10 pts)
25mF 2Ω
50mF
0.2H
Circuit into phasor-domain:
L: 200mH ⇒ jωL = j 20 0.2 = j4 Ω
C: 25mF ⇒ 1/jωC = −j 1/(20 0.025) = −j2 Ω
C: 50mF ⇒ 1/jωC = −j 1/(20 0.05) = −j Ω
Z1 = j4Ω || -jΩ = j4 (−j) / (j4 − j) = −j4/3 Ω
ZEQU = 2Ω − j2Ω − j4/3Ω = 2Ω − j10/3Ω = 2Ω − j3.33Ω
ZEQU = ___2Ω − j10/3Ω___
Example 7 of 10
8. For the circuit shown below, find the Thevenin- and Norton
equivalent circuit (VTH, IN, and ZTH). Find the load impedance ZL
that maximizes it’s own power consumption. Compute the value of
the maximum power transfer (P).
(10 pts)
IS
-j20Ω
20Ω
+
−
Vs=100∠0ºV
j10Ω
IS
10Ω
VTH: (open terminals)
(KCL, 1 equation in 1 unknown) 2 (100∠0 − VTH) / (20 − j20Ω) = VTH / (10 + j10Ω)
⇒ 2 100∠0 / (20 - j20Ω) = VTH ( 1/(10 − j10Ω) + 2/(20 − j20Ω) )
⇒ VTH = (20 + j20Ω)/40 100∠0 = (1/2 + j/2Ω) 100∠0 = 100/√2∠45
ISC: (terminals shortened)
(KCL, 1 equation in 1 unknown)
IN = 2 IS = 100∠0 / (20 - j20Ω) = 100∠0 / (√200∠-45) = 100/√200 ∠45 A
use ZTH = VTH / ISC = (100/√2∠45) / (100/√200 ∠45 A) = 10 Ω
ZL = ZTH* = 10 Ω
P = ½ (VTH/2)2 / RL= 1/8 1002/2 1/10 = 62.5 W
VTH = ___100/√2∠45__, IN = __100/√200 ∠45 A__, ZTH = _____10 Ω____
ZL = ____10 Ω___,
P = _____62.5 W___
Example 8 of 10
9. From the figure below, determine
a) the admittance Y
b) the functions v(t) and i(t) in the time domain (ω = 120π rad/s)
c) the value of the power factor
d) is the PF leading or lagging?
e) is the load more inductive or capacitive?
f) the average and reactive power absorbed by the impedance.
I
+
(10 pts)
Im
I = 45ARMS
V Z
V = 60VRMS
110°
60°
−
Re
ω = 2πf = 377 rad/s
V: V = 60∠60 VRMS ⇒ v(t) = 60 √2 cos(377t + 60º) V
I:
I = 45∠110 ARMS ⇒ i(t) = 45 √2 cos(377t + 110º) A
Y = I / V = 45∠110º / 60∠60º = 0.75∠50º S
ϕLOAD = −50º (voltage lags current by 50º ⇒ capacitive load)
PF = cos(ϕLOAD) = cos(−50º) = 0.643 leading
S = V I* = 60∠60º 45∠-110º = 2700∠-50º = 1735.5 W − j2068.3 VAR
Y = __0.75∠50º S_____
v(t) = _60 √2 cos(377t + 60º) V__
PF = _0.643 leading__
......capacitive.... load
P = ____1735.5 W__
i(t) = _45 √2 cos(377t + 110º) A_
Q = _−2068.3 VAR__
Example 9 of 10
10. Find the total average and reactive power consumed by the two
loads (impedances). How much reactive power must a capacitor
absorb to achieve a PF = 1. Determine the capacitor’s value
assuming ω = 120π rad/s.
Z1: absorbs 2kW and 1kVAR
(10 pts)
Z2: absorbs 1kVA at 0.8 lagging
+
100VRMS Z1
Z2
C
−
S1 = 2 kW + j1kVAR
S2 = 1000 * 0.8 W + j 1000 sin(cos-1(0.8)) VAR = 800 W + j 600 VAR
S = S1 + S2 = 2800 W + j 1600 VAR
capacitor must absorb -1600VAR to achieve PF = 1
SC = V I* = V V* / Z* = V2 / (-j/ωC)* = -jωCV2
-1600 = -120π C 1002 ⇒ C = 1600 / (120π 1002) = 424.4 µF
P = ___2800 W__
Q = _1600 VAR__
QC = _-1600 VAR__
C = ___424.4 µF__
Example 10 of 10