PlacementPrepThree resistors of 6 Ω each are connected in parallel. The equivalent resistance is:
Solution
For n equal resistors in parallel, R/n = 6/3 = 2 Ω.
Answer: (B) 2 Ω
Technical · Electrical and Electronics (EEE)
Network Theory MCQs for placement tests
20 technical questions of moderate difficulty, as asked in campus placement and company tests. Each has a short explanation.
Resistors of 4 Ω and 12 Ω are in parallel. The equivalent resistance is:
Solution
R = (4 × 12)/(4 + 12) = 48/16 = 3 Ω.
Answer: (A) 3 Ω
Kirchhoff's current law is based on the conservation of:
Solution
Current entering a node equals current leaving it.
Answer: (D) Charge
Kirchhoff's voltage law is based on the conservation of:
Solution
The sum of voltages around a closed loop is zero.
Answer: (C) Energy
Thevenin's theorem replaces a linear network by:
Solution
Norton's theorem uses a current source in parallel.
Answer: (B) A voltage source in series with a resistance
Maximum power is transferred to a load when the load resistance equals:
Solution
The maximum power transfer theorem.
Answer: (D) The Thevenin resistance of the source
The time constant of an RC circuit with R = 10 kΩ and C = 100 µF is:
Solution
τ = RC = 10⁴ × 10⁻⁴ = 1 s.
Answer: (C) 1 s
At resonance in a series RLC circuit:
Solution
XL = XC cancel, leaving only R.
Answer: (B) Impedance is minimum and equals R
The resonant frequency of an LC circuit is:
Solution
f₀ = 1/(2π√(LC)).
Answer: (D) 1/(2π√(LC))
In a purely inductive AC circuit, the current:
Solution
Inductors oppose changes in current, so current lags.
Answer: (C) Lags the voltage by 90°
In a purely capacitive AC circuit, the current:
Solution
The capacitor's current leads its voltage by 90°.
Answer: (C) Leads the voltage by 90°
The RMS value of a sinusoidal voltage with peak 100 V is about:
Solution
Vrms = Vm/√2 ≈ 70.7 V.
Answer: (D) 70.7 V
A 100 W, 230 V lamp has a resistance of about:
Solution
R = V²/P = 230²/100 = 529 Ω.
Answer: (A) 529 Ω
In a balanced star connection, line voltage equals:
Solution
VL = √3 Vph; line current equals phase current.
Answer: (A) √3 times the phase voltage
The power in a balanced three-phase load is:
Solution
P = √3 × line voltage × line current × power factor.
Answer: (D) √3 VL IL cos φ
Two capacitors of 4 µF and 4 µF in series give:
Solution
Series capacitors combine like parallel resistors: 4/2 = 2 µF.
Answer: (A) 2 µF
The energy stored in an inductor carrying current I is:
Solution
W = ½ L I².
Answer: (C) ½ L I²
The superposition theorem applies to:
Solution
It needs linearity: responses add.
Answer: (D) Linear circuits
A current of 2 A flows through a 5 Ω resistor. The power dissipated is:
Solution
P = I²R = 4 × 5 = 20 W.
Answer: (C) 20 W
The quality factor Q of a series RLC circuit is:
Solution
Q = ω₀L/R = (1/R)√(L/C).
Answer: (D) (1/R)√(L/C)
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