EE 221
Comprehensive
Final Exam
December 2002
Analyzing and solving basic electrical circuits in DC and sinusoidal steady-state:
The final exam will require you to understand and apply the following topics and principles
1. Understand the definitions of voltage, current, and power
2. Be aware that quantities consist of three parts: sign + number + unit. Know how to
assign voltages and currents and how this affects the resulting sign.
3. Know the Passive Sign Convention
4. Calculate power of an electrical component in a circuit and be able to know if the
component delivers or absorbs power
5. Be able to apply Kirchhoff’s Voltage Law around a loop
6. Be able to apply Kirchhoff’s Current Law at a node
7. Be able to apply Ohm’s Law
8. Equivalent resistances of series and parallel connections of resistors
9. Voltage divider circuit for resistors in series
10. Current divider circuit for resistors in parallel
11. Solve a basic electrical circuit with dependent sources in DC
12. Know node-voltage method
13. Know mesh-current method
14. Be able to apply source transformation to circuits
15. Be able to find Thevenin and Norton equivalents for circuits containing independent and
dependent sources.
16. Be able to solve circuits using superposition.
17. Know inverting and non-inverting amplifiers.
18. Be able to solve circuits containing operational amplifiers
19. Understand the V-I-relationships of capacitors and inductors in both time and phasor
domain
20. Be able to determine the energy stored in inductors and capacitors
21. Know the behavior of inductors and capacitors in DC circuits
22. Be able to determine the impedance and admittance of circuits containing resistors,
inductors, and capacitors
23. Be able to transform circuits and signals from time-domain to frequency (phasor)
domain and back.
24. Be able to solve for complex, apparent, average (real) and reactive powers of a circuit
element
25. Be able to determine the power factor
26. Know condition under maximum power transfer to a load.
The following provides examples taken from two previous final exams.
Page 1 of 1
1. Given the circuit below, use current-division to find current I. Also find
the voltage (V) and the power (P) supplied by the source. Be sure to
include the proper unit with your answer.
(7 pts)
+
IS 22A
2Ω
3Ω
V
I = _−3.67 A__
6Ω
V = _−22 V__
GND
P = _484 W__
−
I = (1/6) / (1/2 + 1/3 + 1/6) * (−22A) = −3.67 A
V = I R = −22 V
Psupplied = −V I = −(−22V) (22A) = 484 W
+
IS 16A
2Ω
V
I = _−4 A__
6Ω
V = _−24 V__
P = _384 W__
GND
−
I = 2 / (2 + 6) * (−16A) = −4 A
V = I R = −24 V
Psupplied = −V I = −(−24V) (16A) = 384 W
Page 2 of 2
2. Given the following circuit, determine the voltage V and current I.
1Ω
IS 22A
6Ω
+
V
(7 pts)
I = _10 A_
4Ω
V = __−48 V_
−
I
GND
I = 5 / (5 + 6) * (22A) = 10 A
I1 = 22 - 10 = 12 A
V = −R I1 = −(12A) (4Ω) = −48 V
1.6Ω
VS=8V + V −
+
−
I = _1.2 A_
6Ω
4Ω
V = __3.2 V_
I
V = 1.6 / (1.6 + 4 || 6) * (8V) = 3.2 V
VR = 8 - 3.2 = 4.8 V
I = V / R = 4.8V / 4Ω = 1.2 A
Page 3 of 3
3. Fill in the boxes in the equations below that result from a nodevoltage analysis. Be certain to clear all fractions and include
proper signs, so that all boxes contain signed integers.
(14 pts)
2IX
+
−
2Ω
V1
IX
2Ω
−
+
V2
3Ω
2Ω
2Ω
14
V1 +
-5
V2 =
-6
-8
V1 +
11
V2 =
12
+
−
2Ω
2V
4V
GND
1.) assign VX between the 2Ω resistor and the dependent voltage source, and
2.) enclose the dependent voltage source within a supernode
(V1 + 2)/2 + V1/2 + (V1-V2)/3 + (V1-VX)/2 = 0 ⇒ 11 V1 - 2 V2 - 3 VX = -6 ⇒ 14 V1 − 5 V2 = -6
A) (KCL)
B) (Supernode) (VX - V1)/2 + (V2-V1)/3 + V2/2 + (V2-4)/2 = 0 ⇒ −5 V1 + 8 V2 + 3 VX = 12 ⇒ −8 V1 + 11 V2 =12
C) (Loop)
⇒ VX = V2 - V1
-VX - 2IX + V2 = 0
IX
2Ω
4IX
V1
V2
+
−
2Ω
2Ω
2Ω
2Ω
1
V1 +
2
V2 =
8
-1
V1 +
1
V2 =
0
+
−
2A
4V
GND
A) (KCL at Supernode) -2 + V1/2 + (V1-V2)/2 + (V2-V1)/2 + V2/2 + (V2-4)/2 = 0 ⇒ V1 + 2 V2 = 8
B) (KVL around Supernode) -V1 - 4IX + V2 = 0 where IX = (V1 - V2) / 2
⇒ -V1 + V2 = 0
Page 4 of 4
4. Fill in the boxes in the equations below that result
from a mesh-current analysis. Be certain to clear all
fractions and include proper signs, so that all boxes
contain signed integers.
(12 pts)
3IX
1Ω
+
−
2Ω
I2
2Ω
IX
1As
I1
3Ω
−
+
I3
2V
1
I1 +
0
I2 +
0
I3
=
1
-5
I1 +
5
I2 +
1
I3
=
0
-3
I1 +
-2
I2 +
5
I3
=
2
1
I1 +
0
I2 +
0
I3
=
2
-8
I1 +
14
I2 +
0
I3
=
0
-5
I1 +
-3
I2 +
5
I3
=
4
GND
A) I1 = 1 A
B) -3IX + I2 + 2(I2-I3) +2(I2-I1) = 0 where IX = I1 - I3 = 1A - I3
C) 3(I3-I1) + 2(I3-I2) - 2 = 0
3Ω
8Ω
IX
I2
3IX
+
−
2As
I1
5Ω
I3
−
+
4V
GND
A) I1 = 2 A
B) 3I2 + 3IX + 8(I2-I1) = 0 where IX = I2
C) -4 + 5(I3-I1) - 3IX = 0
Page 5 of 5
5. Find the output voltage VOUT of the following ideal OpAmp-circuit.
(10 pts)
20kΩ
10V
+
−
10kΩ
20V
10kΩ
-20V
VOUT
VOUT = _−20 V_
(not −30V)
10kΩ
V+ = V- = V
KCL at (+) node: (Vx-10) / 10k + (VX-VOUT) / 10k = 0 ⇒ 2VX - VOUT = 10V
KCL at (-) node: VX / 10k + (VX-VOUT) / 20k = 0 ⇒ VOUT = 3 VX
⇒ 2/3 VOUT - VOUT = 10 V
⇒ VOUT = −30 V, supply limits to −20 V
20V
1mA
10kΩ
VOUT
VOUT = _15 V_
-20V
20kΩ
10kΩ
V+ = V- = V
KCL at (-) node: -1mA + V / 10k + 0 = 0 ⇒ V = 10V
KCL at (+) node: V / 20k + (V-VOUT) / 10k + 0 = 0 ⇒ VOUT = 3/2 V = 15V
Page 6 of 6
6. The voltage across an inductor (2H) is shown below. Sketch the
current through the inductor. What is the current at t = 2s. What
amount of energy has been stored in the inductor at t=1s?
(Assume all initial conditions are zero.)
(10 pts)
v(t) in V
v(t) = L di/dt
8
i = 1/L ∫ v dt = ½ 8 t2/2
1
2
t (s)
(from t = 0 to 1s)
i(t=1s) = 2A
(constant afterwards)
w = ½ L i(t)2
i(t) in A
2
1
2 t (s)
i(t=2s) = __2 A__
w(t=1s) = __4 J__
The current through an inductor (10mH) is shown below. Sketch
the voltage across the inductor. What is v at t = 1.5s. What amount
of energy has been stored in the inductor at t=1s?
Page 7 of 7
i(t) in A
v(t) = L di/dt
5
di/dt = 5A/1s
1
2
t (s)
v = 10mH * 5A/s = 50 mV
w = ½ L i(t)2 = ½ 10mH 52
v(t) in mV
50
t (s)
1
v(t=1.5s) = _50 mV_
2
w(t=1s) = _125 mJ_
-50
7. For the circuit below, find the equivalent input impedance of the
network at ω = 50 rad/s. Express in polar form.
20mF
(10 pts)
20mF
0.04H
ZEQU = 0.71∠−45º Ω
1Ω
1/(jωC) = -j1Ω, jωL = j2Ω
Z1 = 1Ω || (-j1Ω + j2Ω) = 1/√2 ∠45º Ω
ZEQU = -j1Ω + Z1 = -j1Ω + 1/√2 ∠45º Ω = ½ - j/2 = 0.71∠-45º Ω
For the circuit shown below: If v(t) = 10 cos(ωt) find ω so that v(t)
and i(t) are in phase.
Page 8 of 8
i(t)
20mF
+
ω = 50 rad/s
0.02H
v(t)
1Ω
−
v(t) and i(t) are in phase when the impedance of the capacitor cancels the inductor’s impedance:
1/(jωC) + jωL = 0
⇒ -j / C + j ω2 L = 0 ⇒ 1 / 0.02 - ω2 0.02 = 0 ⇒ 50 - 0.02ω2 = 0
⇒ ω = √2500 = 50 rad/s
8. For the circuit shown below, find the Thevenin- equivalent circuit
(VTH and ZTH). Find the load impedance ZL that maximizes it’s own
power consumption. Compute the value of the maximum power
transfer (P).
j3Ω
-j2Ω
+
−
3∠30ºV
4Ω
(10 pts)
VTH = _3.37∠28.49º V_
3I
ZTH = _4.04∠10.98º Ω_
2Ω
ZL = _4.04∠−10.98º Ω_
I
P = _____0.495 W___
GND
VTH: (open terminals)
A) (KCL) (V-3∠30)/-j2 + V/2 + 3(V/2) = 0 ⇒ V = 3∠30 / (1-j4)= 0.73∠105.96
B) (KVL) -V + j3 (3V/2) + VTH = 0 ⇒ VTH = V (1 - j 9/2) = 3.37∠28.49 V
ISC: (terminals shortened)
A) (KCL) (V1-3∠30)/-j2 + V1/2 + (V1-V2)/j3 = 0
⇒ V1(-1+j3) -2V2 = -9∠30
B) (KCL) (V2-V1)/j3 + 3(V1/2) + V2/4 = 0 ⇒ V1 = V2 (-4-j3)/(-4+j18) =
= V2 (0.27∠114.34)
⇒ V2 = 3.34∠17.51 V and ISC = V2 / 4 = 0.84∠17.51 A
need to use ZTH = VTH / ISC = 4.04∠10.98,
ZL = ZTH* = 3.97 - j0.77Ω
P = ½ (VTH/2)2 / R = 1/8 3.372 / 3.97 = 0.495 W
Page 9 of 9
3I
+
−
10∠45ºV
VTH = _10.91∠−149.04º V_
4Ω
-j2Ω
2Ω
ZTH = _____5∠36.87º Ω__
ZL = _____5∠−36.87º Ω__
j3Ω
I
P = _______3.72 W_____
GND
VTH:
⇒ I = 10∠45 / (2 - j8) = 1.21∠120.96 A
A) (KVL) -10∠45) - j2 (4I) + 2I = 0
B)
VTH = (3I) (j3) = 3 1.21∠120.96 3∠90 = 10.91∠-149.04 V
ISC: (note: KVL as in equation A) is still applicable)
ISC = j3 / (4 + j3) (3I) = 3∠90/(5∠36.87) (3 * 1.21∠120.96 A) = 2.18∠174.09 A
need to use ZTH = VTH / ISC = 5∠36.87,
ZL = ZTH* = 4 - j3 Ω
P = ½ (VTH/2)2 / R = 1/8 10.912 / 4 = 3.72 W
9. From the figure below, determine
a) the impedance Z
b) the functions v(t) and i(t) in the time domain assuming f=60 Hz
c) the value of the power factor
d) is the PF leading or lagging?
e) the average and reactive power absorbed by the impedance.
I
+
(10 pts)
Im
V = 120VRMS
V Z
Z = ____12∠65º Ω__
95° I = 10ARMS
−
30°
Re
v(t) = 120 √2 cos(377t+95º) V
i(t) = 10 √2 cos(377t+30º) A
PF = __0.42 lagging___
Z = V / I = 120∠95º / 10∠30º = 12∠65º Ω
P = ____507.14 W___
ω = 2πf = 377 rad/s
Q = _1087.57 VAR___
PF = cos(θ) = cos(65) = 0.42
S = V I* = 120∠95º 10∠-30º = 1200∠65º = 507.14 W + j1087.57 VAR
Page 10 of 10
From the figure below, determine
a) i(t) in the time domain
b) the value of the power factor
c) is the PF leading or lagging?
d) the average and reactive power absorbed by thecircuit.
i(t)
10mF
+
10 cos(100t+15º)VRM S
30mH
2Ω
i(t) = 6 cos(100t-18.69º) ARMS
PF = __0.83 lagging___
−
P = ____49.92 W___
1/(jωC) = -j1Ω, jωL = j3Ω, Z = -j1Ω + 2Ω || j3Ω = 1.66∠33.69º Ω
Q = __33.28 VAR___
I = V / Z = 10∠15º / 1.66∠33.69º = 6∠-18.69º ARMS
i(t) = 6 cos(100t -18.69º) ARMS
PF = cos(∠ ZEQU) = cos(33.69º) = 0.83 lagging
Complex power supplied by the source equals the power absorbed by the load:
S = V I* = 10∠15º 6∠18.69º = 60∠33.69º = 49.92 W + j 33.28 VAR
10. Find the impedance and the power factor of the equivalent load as
seen from the line’s input terminal. Indicate leading or lagging.
Z1: absorbs 1kW and 2kVAR
Z2: absorbs 2kVA at 0.6 lagging
+
120VRMS Z1
(10 pts)
ZEQU = 3.41∠58.57º Ω
Z2
PF = __0.52 lagging_
−
S1 = 1 kW + j2kVAR
S2 = 2000 * 0.6 W + j 2000 sin(cos-1(0.6)) VAR = 1200W + j 1600 VAR
S = S1 + S2 = 2200W + j 3600VAR = 4219 ∠58.57º VA
ZEQU = |ZEQU| = V2 / S = 1202 / 4219 = 3.41 Ω
PF = cos(θ) = cos(58.57º) = 0.52 lagging
Page 11 of 11