Nodal analysis has a node at 12 V linked through 2 kΩ to Vx and through 4 kΩ to ground. What is Vx?
A 8 V
B 4 V
C 6 V
D 10 V
Answer: A. 8 V. KCL gives (12−Vx)/2k = Vx/4k, so Vx = 8 V.
Clockwise mesh currents are 3 A and 1 A. What shared-branch current follows their opposition?
A 4 A toward mesh two
B 2 A toward mesh one
C 2 A toward mesh two
D 4 A toward mesh one
Answer: B. 2 A toward mesh one. Opposing mesh currents subtract in the shared branch: 3 − 1 = 2 A toward mesh one.
For a 12 V source with 3 Ω series resistance, what Norton current results?
A 36 A
B 9 A
C 3 A
D 4 A
Answer: D. 4 A. IN = V/R = 12/3 = 4 A; the parallel resistance remains 3 Ω.
After deactivating a 12 V source beside a 4 Ω series resistor, what resistance remains at its terminals?
A 0 Ω
B 3 Ω
C 4 Ω
D 12 Ω
Answer: C. 4 Ω. Deactivate an ideal voltage source by shorting it; the resistor remains.
From a Thevenin source of 18 V with 6 Ω resistance, what Norton current follows?
A 3 A
B 12 A
C 108 A
D 0.33 A
Answer: A. 3 A. Source transformation gives IN = VTH/RTH = 18/6 = 3 A.
For maximum DC load power, what relation should RL satisfy?
A RL = Rth
B RL = Rth/2
C RL = 2Rth
D RL = 0
Answer: A. RL = Rth. Maximum DC power transfer occurs when RL equals RTH.
Vth = 12 V and Rth = 6 Ω yield what maximum load power?
A 12 W
B 24 W
C 3 W
D 6 W
Answer: D. 6 W. Pmax = VTH²/(4RTH) = 144/24 = 6 W.
Independent sources contribute +8 V and −3 V separately. Superposition gives what total?
A 5 V
B 24 V
C 11 V
D −5 V
Answer: A. 5 V. Superposition adds signed contributions: 8 + (−3) = 5 V.
When a 10 V source feeds a 5 Ω resistor and 5 Ω load in series, what load current flows?
A 5 A
B 2 A
C 0.5 A
D 1 A
Answer: D. 1 A. Total resistance is 10 Ω, so current is 10/10 = 1 A.
R = 5 kΩ and C = 20 µF give what RC time constant?
A 1 s
B 0.1 s
C 100 s
D 0.01 s
Answer: B. 0.1 s. τ = RC = 5000 × 20 µF = 0.1 s.
For an RL branch with L = 0.6 H and R = 30 Ω, what is τ?
A 0.2 s
B 18 s
C 50 s
D 0.02 s
Answer: D. 0.02 s. The RL time constant is L/R = 0.6/30 = 0.02 s.
After one time constant, an initially uncharged capacitor reaches about what fraction?
A 36.8 percent
B 50 percent
C 86.5 percent
D 63.2 percent
Answer: D. 63.2 percent. Charging reaches 1 − e⁻¹, about 63.2%, after one time constant.
Immediately after switching, capacitor voltage can do what?
A Reverse instantly
B Jump freely
C Remain continuous
D Become infinite
Answer: C. Remain continuous. Capacitor voltage cannot jump instantly because that would require impulsive current.
Immediately after switching, inductor current can do what?
A Reverse instantly
B Remain continuous
C Become zero always
D Jump freely
Answer: B. Remain continuous. Inductor current cannot jump instantly because that would require impulsive voltage.
For a 50 Hz sinusoid, what is the angular frequency?
A 314 rad/s
B 628 rad/s
C 100 rad/s
D 50 rad/s
Answer: A. 314 rad/s. ω = 2πf = 2π × 50 ≈ 314 rad/s.
At 50 Hz, a 0.1 H inductor has what reactance?
A 50 Ω
B 3.14 Ω
C 31.4 Ω
D 314 Ω
Answer: C. 31.4 Ω. XL = 2πfL = 2π × 50 × 0.1 ≈ 31.4 Ω.
At 50 Hz, a 100 µF capacitor has what reactance?
A 31.8 Ω
B 318 Ω
C 3.18 Ω
D 50 Ω
Answer: A. 31.8 Ω. XC = 1/(2πfC) ≈ 31.8 Ω.
At series RLC resonance, the net reactance becomes what?
A Zero
B Inductive
C Capacitive
D Maximum
Answer: A. Zero. At resonance XL equals XC, so their series reactances cancel.
What is the impedance of a series RLC circuit at resonance?
A XL + XC
B R
C Infinite
D 0 Ω
Answer: B. R. At resonance the reactances cancel, leaving the resistance R.
For a purely resistive AC circuit, current and voltage are what?
A Opposite phase
B In phase
C 90° apart
D Random phase
Answer: B. In phase. An ideal resistor has zero phase difference between voltage and current.
For an ideal inductor, current lags voltage by how much?
A 180 degrees
B 45 degrees
C 90 degrees
D 0 degrees
Answer: C. 90 degrees. Inductor voltage leads its current by 90° in sinusoidal steady state.
With power factor 0.8, what is the phase-angle magnitude?
A 53.1 degrees
B 80 degrees
C 8 degrees
D 36.9 degrees
Answer: D. 36.9 degrees. cos⁻¹(0.8) ≈ 36.9°.
For a sinusoid with 20 V peak, what RMS voltage results?
A 28.28 V
B 14.14 V
C 20 V
D 10 V
Answer: B. 14.14 V. For a sine wave VRMS = Vpeak/√2 = 14.14 V.
For an ideal capacitor, current leads voltage by how much?
A 180 degrees
B 45 degrees
C 90 degrees
D 0 degrees
Answer: C. 90 degrees. Capacitor current leads its sinusoidal voltage by 90°.
In a first-order RC low-pass, output is usually taken across what?
A Inductor
B Source
C Capacitor
D Resistor
Answer: C. Capacitor. The standard RC low-pass takes its output across the capacitor.