EE240 – Circuits I
Mid Examination (Fall 2020) - Solutions
November 07, 2020 02:00 pm–04:30 pm
INSTRUCTIONS:
We require you to solve the exam in a single time-slot of two hours without any
external or electronic assistance.
We encourage you to solve the exam on A4 paper, use new sheet for each question
and write sheet number on every sheet.
Clearly outline all your steps in order to obtain any partial credit.
The exam is closed book and notes. You are allowed to have one A4 sheet with you
with hand-written notes on both sides. Calculators can be used.
For the sake of completeness, we require you to write the following statement on your
first page of submission: I commit myself to uphold the highest standards of (academic)
integrity.
If you are ready, please proceed to the next page.
Mapping between exam parts and course learning outcomes (CLOs)
Part 1: R, L, C Basics, Sources and I-V Characteristics (CLO1)
Part 2: Network Topology, Network Equations and Kirchhoff’s Laws (CLO2)
Part 3: Additional Analysis Techniques (CLO3)
Part 1: R, L, C Basics, Sources and I-V Characteristics
Problem 1. (10 pts)
(a) (5 pts) For a circuit given below, assume that the switch is initially open and is
closed at t = 0 and the inductor is not carrying any current before the switch is
closed. Label the voltages across resistor and inductor as vR (t) and vL (t) respec-
tively and plot the waveforms of the voltages.
Solutions:
The combination of sources is equivalent to 10V voltage source. Plots given below:
(b) (5 pts) For a circuit given below, assume that the switch is initially open and is
closed at t = 0 and the inductor is not carrying any current before the switch is
closed. Determine the voltage across the inductor and the total energy supplied
to the inductor.
1
Solutions:
di 1
v(t) = L = 2e−t = e−t
dt 2
1
(Energy) w(t) = Li2 (t) = 1 − 2e−t + e−2t
2
(TotalEnergy) w(∞) = 1 J
2
Problem 2. (5 pts) For the circuit given below, determine the equivalent resistance Req and
the current io indicated in the circuit. Req is the equivalent resistance across terminals
A and B.
Solutions:
Req is (80k20 + 12k6)k60k15 = (20)k12 = 7.5 Ω.
The current io = 40/12.5 = 3.2A
3
Problem 3. (3 pts) For the circuit given below, determine Vo , Io and I1 indicated in the
circuit. Provide brief justification or working to support your answer.
Io = 2A, Vo = 18V, I1 = 3A
Problem 4. (2 pts) For the circuit given below, determine Vo . Provide brief justification or
working for your answer.
Solutions:
Vo = −2 − 12 + 6 − 4 = −12V
4
Problem 5. (5 pts) The figure below shows the dots marked for three windings L1 , L2 and
L3 on a magnetic flux-conducting core. If the dots are marked using the dot convention,
draw the windings on the core with directions consistent with the dots.
Solutions:
5
Part 2: Network Topology, Network Equations and
Kirchhoff ’s Laws
Problem 6. (2 pts)
i) What do we mean by electrically equivalent circuits?
Solutions:
Same i − v characteristics.
ii) Do you agree with the following statement (support your answer with the justifica-
tion)?
The two electrically equivalent circuits may not be be topologically equivalent but two
‘topologically equivalent’ circuits are electrically equivalent.
Solutions:
No! We cannot relate topological equivalence (related to graphical representation) and
electrical equivalence (related to i − v characteristics).
6
Problem 7. (18 pts)
Consider the circuit given below.
(a) (4 pts) Draw the graph and one tree of the circuit. Determine the number of
nodes and number of branches in a circuit.
Solutions:
Number of nodes = n = 5
Number of branches = b = 10
(b) (2 pts) Determine the number of network equations required for carrying out i)
nodal analysis and ii) loop analysis.
Solutions:
Nodal analysis: number of equations = n − 1 = 4
Loop analysis: number of equations = b − n + 1 = 6
7
(c) (9 pts) Use Kirchhoff current law to determine the nodal voltages indicated on
the circuit.
Solutions:
Using KCL to write nodal equations.
Node 1 and Node 4 (Super node):
V1 V1 − V2 V4 − V2 V4 − V3
+ + + − 10 = 0
4 8 2 5
Node 2:
V2 − V1 V2 − V4 V2 − V3
+ + − 2Vx = 0
8 2 10
Node 3:
V3 = 20V
Super-node equation:
V1 − V4 = 4Ix
Dependent voltage source:
V3 − V2
Ix =
10
Dependent current source:
Vx = V4 − V3
We manipulate these equations to obtain
1 1
− 18 − 21 1 1
8 + 4 2 + 5 v1 14
−1 1 1 1 1
8 2 + 8 + 10 −2 − 2 v2 = −38
5 2 −5 v4 40
v1 21.145
v2 = 9.074 V
v4 16.774
(d) (3 pts) Determine the power supplied by the voltage source.
Solutions:
We use I to denote the current supplied by 20V voltage source. We can apply
KCL at node 3 to determine I as
V3 − V2 V3 − V4
I= + − 6 = 1.0926 + 0.6452 − 6 = −4.2622A
10 5
Power is given by
P = 20 × I = −85.2445 W
8
Part 3: Additional Analysis Techniques
Problem 8. (8 pts) Determine the voltage vx in the following circuit using the source
transformation technique.
Solutions:
Applying source transformation:
9
Problem 9. (12 pts) For the circuit given below, determine the current io using superpo-
sition principle. You can use any of the techniques (nodal analysis, loop analysis and
source transformation) to carry out analysis when you keep one independent source in
the circuit.
Solutions:
10
