Find the voltage of a node connected to 12 V through 4 Ω and to ground through 2 Ω.
A 2 V
B 4 V
C 6 V
D 8 V
Answer: B. 4 V. KCL gives (V−12)/4 + V/2 = 0, so V = 4 V.
A 3 A source injects into a node with 2 Ω and 4 Ω paths to ground. Find the node voltage.
A 4 V
B 2 V
C 6 V
D 12 V
Answer: A. 4 V. The total conductance is 1/2 + 1/4 = 3/4 S, so V = 3/(3/4) = 4 V.
Clockwise mesh currents of 3 A and 1 A share a 5 Ω resistor. Its drop in mesh 1 is what?
A 5 V
B 15 V
C 10 V
D 20 V
Answer: C. 10 V. The shared current in mesh 1’s direction is 3−1 = 2 A, giving 10 V.
A 20 V Thévenin source has 5 Ω internal resistance and a 5 Ω load. Load current is what?
A 1 A
B 4 A
C 5 A
D 2 A
Answer: D. 2 A. The source sees 10 Ω total, so 20/10 = 2 A.
With 24 V Thévenin voltage and 6 Ω each for source and load, find load voltage.
A 12 V
B 6 V
C 18 V
D 24 V
Answer: A. 12 V. Equal series resistances split the 24 V source voltage equally.
A 18 V Thévenin source has 3 Ω resistance. What is its short-circuit current?
A 3 A
B 9 A
C 6 A
D 54 A
Answer: C. 6 A. With the output shorted, current is 18/3 = 6 A.
For Vth = 12 V and Rth = 3 Ω, what is the maximum resistive-load power?
A 3 W
B 6 W
C 12 W
D 24 W
Answer: C. 12 W. At Rload = 3 Ω, Pmax = Vth²/(4Rth) = 144/12 = 12 W.
A 4 A Norton source feeds its 3 Ω resistor and a 3 Ω load in parallel. Load current is what?
A 1 A
B 3 A
C 2 A
D 4 A
Answer: C. 2 A. Equal parallel resistances share the 4 A source current equally.
The two resistor dividers of a bridge have arm ratios 2:4 and 3:6. What is cross-branch current?
A 1 A
B 0 A
C 2 A
D 6 A
Answer: B. 0 A. The two arm ratios match, so the bridge midpoints have equal potential.
An equal-arm delta has 9 Ω in each side. Each equivalent wye arm is how much?
A 1 Ω
B 6 Ω
C 27 Ω
D 3 Ω
Answer: D. 3 Ω. For three equal delta resistances, each wye arm is one-third of an arm.
Resistors 2 Ω, 3 Ω and 6 Ω are parallel. Their equivalent resistance is what?
A 2 Ω
B 1 Ω
C 3 Ω
D 11 Ω
Answer: B. 1 Ω. 1/Req = 1/2 + 1/3 + 1/6 = 1, so Req = 1 Ω.
A 4 Ω resistor is in series with 6 Ω parallel to 3 Ω. Find the total resistance.
A 2 Ω
B 7 Ω
C 13 Ω
D 6 Ω
Answer: D. 6 Ω. 6 Ω parallel to 3 Ω is 2 Ω, then 4 + 2 = 6 Ω.
An 18 V source spans 3 Ω and 6 Ω in series. Voltage across the 6 Ω part is what?
A 12 V
B 3 V
C 6 V
D 18 V
Answer: A. 12 V. Voltage division gives 18 × 6/(3+6) = 12 V.
A 3 A current splits between 2 Ω and 4 Ω branches. Current through 2 Ω is what?
A 1 A
B 1.5 A
C 2 A
D 3 A
Answer: C. 2 A. The lower resistance gets 3 × 4/(2+4) = 2 A.
Two 10 V and 5 V nodes connect through separate 5 Ω resistors to a node with 5 Ω to ground. Find its voltage.
A 5 V
B 2.5 V
C 7.5 V
D 15 V
Answer: A. 5 V. KCL gives (V−10)/5 + (V−5)/5 + V/5 = 0, hence V = 5 V.
A dependent source obeys I = (2 S)Vx. If Vx = 3 V, what current does it supply?
A 6 A
B 1.5 A
C 3 A
D 9 A
Answer: A. 6 A. The transconductance is 2 S, so controlled current is 2 × 3 = 6 A.
A connected network graph has 7 branches and 5 nodes. How many independent loops exist?
A 3
B 2
C 5
D 12
Answer: A. 3. Independent loops equal branches minus nodes plus one: 7−5+1 = 3.
In z-parameter measurement, how is port 2 held while finding z11 = V1/I1?
A Short-circuited
B Matched to source
C Open-circuited
D Grounded through load
Answer: C. Open-circuited. With I2 = 0, port 2 is open and z11 is the input impedance.
For y11 = I1/V1, which condition is imposed at port 2?
A Open circuit
B Short circuit
C Matched load
D Current source injection
Answer: B. Short circuit. Set V2 = 0 by shorting port 2 to measure y11.
An RC network has Rth = 2 kΩ and C = 5 μF. Its time constant is what?
A 0.4 ms
B 10 ms
C 2 ms
D 100 ms
Answer: B. 10 ms. RC = 2000 × 5 microfarads = 0.01 seconds.
A source-free RL circuit has L = 4 H and R = 2 Ω. Its time constant is what?
A 2 s
B 0.5 s
C 4 s
D 8 s
Answer: A. 2 s. For an RL natural response, τ = L/R = 4/2 = 2 s.
After one time constant, an RC natural-response voltage is about what fraction of its initial value?
A 10%
B 50%
C 63%
D 37%
Answer: D. 37%. The natural response falls to e^(−1), about 36.8%, at t = τ.
At a switching instant, which capacitor quantity remains continuous for a finite-current circuit?
A Capacitor current
B Resistor voltage
C Switch resistance
D Capacitor voltage
Answer: D. Capacitor voltage. Capacitor voltage cannot jump without an impulse of current.
At a switching instant, which inductor quantity remains continuous for a finite-voltage circuit?
A Inductor voltage
B Inductor current
C Resistor current
D Source voltage
Answer: B. Inductor current. Inductor current cannot jump without an impulse of voltage.
A 4 Ω resistor spans nodes at 8 V and 2 V. What power does it absorb?
A 1.5 W
B 6 W
C 36 W
D 9 W
Answer: D. 9 W. Its voltage is 6 V, so P = V²/R = 36/4 = 9 W.