The Hong Kong University of Science and Technology

Department of Electronic and Computer Engineering

ELEC2400 ELECTRONIC CIRCUITS SPRING 2024-25

HOMEWORK 1 SOLUTION

Q1. Find the resistance RAB between node A and node B in the circuit shown in Fig. 1.

Fig. 1

Solution:

Finally,
4 × 4.5 36
R AB = 4||(3.5 + 1) = = = 2.12 Ω
4 + 4.5 17

Q2. Determine the electric power in each circuit element (PA, PB, and PC) shown in Fig. 2, and
specify whether each circuit component is supplying power or absorbing power (dissipating
power).
+ VC -

IA IB

IG = 0

Fig. 2

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Solution:
Define the voltages and currents as shown in the diagram. We notice that:
1) IG = 0 (Single ground connection with no external power supplies. See Lecture Slide 2-18.)
2) IA = IB = −I1 = −2 A (by KCL)
3) VC = V1 + V2 = 4 − 3 = 1 V (by KVL)

Then, P = VI, provided V and I follow the reference direction (Lecture Slide 1-40). Hence,
1) PA = V1 IA = 4 × (−2) = −8 W (supplying power)
2) PB = V2 IB = (−3) × (−2) = 6 W (absorbing power)
3) PC = VC I1 = 1 × 2 = 2 W (absorbing power)

The net electric power is zero. Thus, electric power balance is maintained.

Q3. Find V1, V2, and I1 for the circuit shown in Fig. 3.

2A

Fig. 3

Solution:
Apply KCL to node P.
I1 + 1 + 2 = 0
I1 = −3 A

The remaining challenge is that a current source can only tell us its current but not its voltage. To
determine V1 and V2, we need to find some viable paths that can yield useful results.

Going from the ground node to V2 through the 4-Ω resistor,

V2 = 0 − 2 × 4 = −8 V

Apply KVL to the outermost loop, going counterclockwise and treating voltage rises as positive,

V2 + I1 × 1 − 5 + V1 = 0
−8 − 3 × 1 − 5 + V1 = 0
V1 = 16 V

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Q4. Use nodal analysis to find V1 in the circuit shown in Fig. 4.

Fig. 4

Solution:
Apply KCL to node P.
V1 V1 + 4
3= +
4 6
36 = 3V1 + 2V1 + 8
28
V1 = = 5.6 V
5

Q5. Use nodal analysis to find V1 and V2 in the circuit shown in Fig. 5.

Fig. 5

Solution:
We need two equations for two unknowns V1 and V2.

Apply KCL to node V1.

V1 − 2 V1 V1 − V2
+ + =0 (1)
6 3 5
Apply KCL to node V2.
V2 − V1 V2 V2 − 4
+ + =0 (2)
5 4 2

Before proceeding further, it is important to make sure that both equations are set up correctly.
The rest is algebra.

From (1),
5V1 − 10 + 10V1 + 6V1 − 6V2 = 0
21V1 − 10
V2 = (3)
6
From (2),

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 4V2 − 4V1 + 5V2 + 10V2 − 40 = 0
4V1 + 40
V2 = (4)
19
Equating (3) and (4),
21V1 − 10 4V1 + 40
=
6 19
399V1 − 190 = 24V1 + 240
240 + 190 430 86
V1 = = = = 1.147 V
399 − 24 375 75
From (3)
21(1.147) − 10
V2 = = 2.347 V
6

Q6. The circuit in Fig. 6 contains a dependent voltage source. Find V1 and V2.

Fig. 6

Solution:
We need two equations for two unknowns V1 and V2.

Apply KCL to node P.

V1 V2
2+ + =0 (1)
5 4
Apply KVL to the right mesh.
V1 + 4V2 − V2 = 0
V1 = −3V2 (2)
Substituting into (1),
−3V2 V2
2+ + =0
5 4
40 − 12V2 + 5V2 = 0
40
V2 = = 5.71 V
7
From (2),
40 120
V1 = −3 ( ) = − = −17.14 V
7 7

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Q7. Use superposition to find V1 in the circuit shown in Fig. 7.

Fig. 7

Solution:
A. Current source acting alone:
The voltage source is short-circuited, resulting in the circuit diagram shown below.

I2

I
I1
The 4-Ω and 6-Ω resistors are connected in parallel. They form a current divider dividing up
the incoming current I = −1 A into I1 and I2. From the current divider formula,
4
I1 = (−1) = −0.4 A
4+6
Ohm’s law then yields
V1 = 6I1 = −2.4 V

B. Voltage source acting alone:

The current source is open-circuited, resulting in the circuit diagram shown below.

I2

I1 = 0
Note that I1 = 0 (open circuit). Then the same current I2 goes through the 4-Ω and 6-Ω
resistors. Apply KVL to the bottom right mesh.
4 = I2 (4 + 6)
I2 = 0.4 A
Ohm’s law then yields
V1 = 6I1 = 2.4 V

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C. Finally, by superposition,
V1 = −2.4 + 2.4 = 0 V

ALTERNATIVE METHOD
The problem can be solved directly without using superposition by applying KCL to node P in
Fig. 7.
V1 − 4 V1
1+ + =0
4 6
12 + 3V1 − 12 + 2V1 = 0
V1 = 0 V

In the exams, you can choose whatever method you prefer.

Q8. Determine the electric power in each circuit element in Fig. 8. Specify whether each circuit
component is supplying or absorbing power (dissipating power).

P
I1 I2

I3

I1

Fig. 8
Solution:
Define the currents as shown. The usual method of applying KCL at node P fails here because the
6-V voltage source is the only circuit element in the middle branch for which the current I3 is
undetermined. However, we can use KVL to solve the problem.

Apply KVL to the left mesh.

10 − 6I1 −6 − 2I1 = 0
I1 = 0.5 A
Apply KVL to the right mesh.
6 − 4I2 − 4 = 0
I2 = 0.5 A
Finally, apply KCL to node P.
I3 = I1 − I2 = 0 A

Electric power calculations:

1) P6Ω = 6I1 2 = 6(0.5)2 = 1.5 W (absorbing power)
2) P4Ω = 4I2 2 = 4(0.5)2 = 1 W (absorbing power)
3) P2Ω = 2I1 2 = 2(0.5)2 = 0.5 W (absorbing power)
4) P10V = −10I1 = −10 × 0.5 = −5 W (supplying power)
5) P6V = 6I3 = 6 × 0 = 0 W (neither supplying nor absorbing power)
6) P4V = 4I2 = 4 × 0.5 = 2 W (absorbing power)

The net electric power is zero. Thus, electric power balance is maintained.
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