SSC · Circuit Theory and Networks practice

Correct answer: B

The source voltage is the open-circuit voltage and the resistance is that seen into the terminals with the sources deactivated.

Correct answer: A

Maximum power occurs at RL = 4 Ω: P = V²/(4R) = 144/16 = 9 W.

Correct answer: C

In the delta-star conversion with equal resistances, R_star = R_delta/3 = 3 Ω.

Correct answer: B

f0 = 1 / (2π√(LC)) = 1 / (2π × √(10⁻⁹)) = 1/(2π × 3.162 × 10⁻⁵) = 5033 Hz.

Correct answer: B

|Z| = √(9 + 16) = 5 Ω and cos φ = R/|Z| = 3/5 = 0.6.

Correct answer: A

The line voltage is the phasor difference of two phase voltages 120° apart, which gives √3 times the phase voltage.

Correct answer: A

P = √3 VL IL cos φ = 1.732 × 400 × 10 × 0.8 = 5542 W.

Correct answer: A

Vrms = Vm/√2 = 325/1.414 = 229.8 V.

Correct answer: A

Form factor = RMS value / average value = 0.707 Vm / 0.637 Vm = 1.11.

Correct answer: A

With zero initial voltage and a changing current, the capacitor initially passes current freely, like a short circuit.

Correct answer: C

The algebraic sum of voltage rises and drops around a closed loop is zero.

Correct answer: B

Node voltages are the unknowns, and KCL is written at each node.

Correct answer: B

It requires linearity. Power cannot be superposed because it is not linear in the voltage or current.

Correct answer: B

W1 = W2 = VL IL cos 30° when the power factor is unity. At 0.5 lagging, one wattmeter reads zero.

Correct answer: D

X = 70 − 30 = 40 Ω. |Z| = √(30² + 40²) = 50 Ω. I = 100/50 = 2 A.

Correct answer: C

Q = V I sin φ = 200 × 10 × sin 30° = 1000 VAr. The real power is 200 × 10 × cos 30° = 1732 W.

Correct answer: C

Zd = L / (C R) = 10⁻³ / (10⁻⁶ × 10) = 100 Ω.

Correct answer: B

v(t) = V (1 − e^(−t/τ)). At t = τ this is V (1 − 0.368) = 0.632 V.

Correct answer: B

Reciprocity requires the determinant of the transmission matrix to be one.

Correct answer: C

Q = (1/R) √(L/C) = (1/10) × √(10⁻³/10⁻⁶) = 31.62/10 = 3.16.