Circuits
Name _______________________ ECSE 2010
Spring 2012
Section _________
Circuits
Quiz 1
Spring 2012
1.
/20
2.
/20
3.
/20
4.
/20
5.
/20
Total
/100
Name __________________
Notes:
1)
2)
3)
4)
The positive side of a voltage source has a ‘longer’ line.
One crib sheet.
Sit with at least one chair between you and your neighbor.
Calculators are okay as long as they don’t have a wireless.
J. Braunstein
Rensselaer Polytechnic Institute
Revised: 2/10/2016
Troy, New York, USA
1
Circuits
Name _______________________ ECSE 2010
Spring 2012
Section _________
1) Short Answers (20 points)
Question 1 (3 points)
1
1
R
1
R
R
R
R
R
2
2
R
2
R
A
B
C
When considering the total resistance between node 1 and node 2 in the above circuits (circle the
correct answer),
a) Circuit A has a larger total resistance than Circuit B and C
b) Circuit B has a larger total resistance than Circuit A and C
c) Circuit C has a larger total resistance than Circuit A and B
d) None of the above are true
Question 2 (3 points)
IR1
R1
+
VR1
+
IV1
+
VR2
-
V1
-
R2
IR2
Based on the above schematic with the indicated voltage and current polarities, indicate whether
the following are True or False (circle the correct answer).
T / F a) IR1*(R1) + IR2*(R2) + IV1*(V1) = 0
T / F b) IV1*(V1) < 0
T / F c) IR1 = IR2 = -IV1
Question 3 (3 points)
An inverting amplifier with a gain of -2 and voltage supply of ±9 V
T / F a) will always output negative voltage (relative to ground)
T / F b) outputs -6 V (DC) when the input is 3 V (DC)
T / F c) outputs -12 V (DC) when the input is 6 V (DC)
J. Braunstein
Rensselaer Polytechnic Institute
Revised: 2/10/2016
Troy, New York, USA
2
R
R
Circuits
Name _______________________ ECSE 2010
Spring 2012
Section _________
Question 4 (6 points)
9V
9V
Vin=0.5V
U27 V+
+ OS2
0
OUT
2
- 4 OS1
uA741 V3
7
U1 V+
+ OS2
OUT
2
- 4 OS1
uA741 V3
-9v
5
6
V1
R3
2k
1
5
Vout
6
1
-9v
R1
6k
R2
2k
0
In the above circuit determine the voltage at V1 (output of the first amplifier) and Vout (output
of the second amplifier). The amplifiers have 9/-9V supply voltages. Pay attention to the input
connections.
V1:
_____2.0V______________
Vout: _____-9.0V__________
Question 5 (3 points)
In the following schematic voltages at some of the nodes are provided (1V, 3V and 5V as
indicated). In the dashed lines, determine the resistance of R4, the voltage of V3 and the current
of I1. Pay attention to polarity.
3V
R4
5V
V3
I2 1E-3
1V
R1
2k
I1
2k
R2
R3
0
J. Braunstein
Rensselaer Polytechnic Institute
Revised: 2/10/2016
Troy, New York, USA
3
Circuits
Name _______________________ ECSE 2010
Spring 2012
Section _________
2) Circuit analysis I (20 points)
R3
R2
6k
2k
R4
2k
R1
R6
4k
R7
4k
R5
6k
I2
2E-3
8k
V1
4
I1
1E-3
0
Use superposition to find the voltage across R5 (indicated by the box). For each source, draw
the circuit you are analyzing, before doing any circuit reduction. In your analysis, you can
perform circuit reduction and use any circuit analysis method. For each source circuit, indicate
the voltage contribution at the bottom of the next page. (There is an additional question on the
next page.)
V1 circuit:
VR 5V 1 
R5
6k
V1 
4  1.5V
R5  R 4  R 2  R3  R 4 || R1
16k
J. Braunstein
Rensselaer Polytechnic Institute
Revised: 2/10/2016
Troy, New York, USA
4
Circuits
Name _______________________ ECSE 2010
Spring 2012
Section _________
I1 circuit:
V R 5 I 1 
R3  R4 || R1
4k
1E  36k   1.5V
I1R5V 1 
R5  R 4  R 2  R3  R 4 || R1
16k
I2 circuit:
VR 5 I 2 
R3  R4 || R1  R2
6k
2E  36k   4.5V
I1R5V 1 
R5  R 4  R 2  R3  R 4 || R1
16k
Based on your above expressions, change V1 such that the current through R6 is 1.5mA.
IR6 = IR5 = 1.5mA, when VR5 = 9V
The voltage across R5 is the sum of the individual contributions
from each independent source. Keeping I1 and I2 the same
results in 9 – 1.5 – 4.5 = 3V contribution from the V1 voltage
source. When, V1 is 4 V it contributes 1.5 V across R5. Voltage
scales linearly as V1 changes, therefore V1 must be 8 V to
contribute a 3 V drop across R5, giving a total drop of 9 V across
R5.
J. Braunstein
Rensselaer Polytechnic Institute
VR5V1
[V]
VR5I1
[V]
VR5I2
[V]
New V1
[V]
Revised: 2/10/2016
Troy, New York, USA
5
Circuits
Name _______________________ ECSE 2010
Spring 2012
Section _________
3) Circuit Analysis II – Dependent Source (20 points)
V1
R1
2k
3
R2
4k
I1
2E-3
R3
Vx
+
R4
1k
U1
+
R8
1k
1k
R5
2k
I2
0.004Vx
Vout
OUT
0
-
OPAMP
R7
R6
4k
0
In the above circuit find Vx. You may use any analysis technique. Note, the current source, I2,
is a voltage controlled dependent current source (VCCS). If you do any circuit reduction,
include the redrawn/reduced circuits. Also, determine R7 such that the output of the opamp,
Vout, has a magnitude of 1V. Assume the opamp is ideal.
Vx
[V]
R7
[Ω]
(next page is blank if needed)
J. Braunstein
Rensselaer Polytechnic Institute
Revised: 2/10/2016
Troy, New York, USA
6
Circuits
Name _______________________ ECSE 2010
J. Braunstein
Rensselaer Polytechnic Institute
Spring 2012
Section _________
Revised: 2/10/2016
Troy, New York, USA
7
Circuits
Name _______________________ ECSE 2010
Spring 2012
Section _________
4) Thevenin Circuits I (20 points)
I1
R6
2E-3
2k
R10
5k
R2
5k
R8
1k
R7
8k
RLoad
0
V1
R5
unknown
unknown
For the above circuit, it was found that IN and RTH were -2.5mA (negative) and 8kΩ, as shown
below in the Norton equivalent circuit. In the circuit, R5 and V1 are unknown. V1 is an
independent voltage source.
2.5mA
IN
RTH
8k
RLoad
Using the Norton equivalent circuit, determine the value of the unknown resistor, R5. Include
your answer in the box at the bottom of the next page. Draw the circuit you use in your analysis.
RTH  R5  R8  R6  R10 || R7  8k
R5  8k  4k || 4k  4k
J. Braunstein
Rensselaer Polytechnic Institute
Revised: 2/10/2016
Troy, New York, USA
8
Circuits
Name _______________________ ECSE 2010
Spring 2012
Section _________
Draw the circuit when determining IN (IShortCircuit). Your schematic should not have any circuit
reduction. Use mesh analysis to determine the voltage of the unknown source, V1, such that the
Norton equivalent circuit is correct (IN = -2.5mA). Be careful with your polarity.
Recognizing that i3 = isc
loop 1: i1  1E  3
loop 2: i22k  i21k  i2  isc 8k  i 2  1E  35k  0
loop 3: isc  i 28k  isc 4k  V1  0
From loop 2
J. Braunstein
Rensselaer Polytechnic Institute
Revised: 2/10/2016
Troy, New York, USA
9
V1
[V]
Circuits
Name _______________________ ECSE 2010
Spring 2012
Section _________
R5
J. Braunstein
Rensselaer Polytechnic Institute
[Ω]
Revised: 2/10/2016
Troy, New York, USA
10
Circuits
Name _______________________ ECSE 2010
Spring 2012
Section _________
5) Thevenin Circuits II (20 points)
R3
3k
2
R1
1k
V1
RLoad
R5
Ix
R6
2k
4k
R2
4k
4
V3
0.002Ix
V2
2000Ix
Part a) In the above circuit find the equivalent circuit parameter VTH (VOpenCircuit) using node
analysis. Note, a current controlled dependent voltage source (CCVS) is included in the
schematic. The Thevenin equivalent circuit represents the components to the left of the dashed
line. On the above schematic, label any nodal voltages that are constrained/known by a voltage
source. Circle those values so that the grader may find them quickly.
RTH
VTH
RLoad
0
VTH
J. Braunstein
Rensselaer Polytechnic Institute
[V]
Revised: 2/10/2016
Troy, New York, USA
11
Circuits
Name _______________________ ECSE 2010
Spring 2012
Section _________
Part b) In the circuit on the previous page find the equivalent circuit parameter RTH (RThevenin) by
applying a test voltage at the load terminals. You may use any analysis technique. Redraw the
circuit with the test voltage source.
Part c) Find IN (IShortCircuit)
J. Braunstein
Rensselaer Polytechnic Institute
VTH
[V]
RTH
[Ω]
IN
[A]
Revised: 2/10/2016
Troy, New York, USA
12