Electrical Engg (EEE) - MCQ Practice Questions

Impedance in a series RLC circuit is the vector sum of resistance and net reactance. Z = √(R² + (XL - XC)²) is the standard formula.

Power factor is the cosine of the phase angle between voltage and current, representing the fraction of apparent power that is real power.

Maximum Power Transfer Theorem states that maximum power is delivered when load resistance equals the Thevenin equivalent resistance of the source.

In a purely resistive circuit, voltage and current are in phase, so the phase difference is 0°.

At resonance, XL = XC, leading to the resonant frequency formula f₀ = 1/(2π√LC).

KCL states that the algebraic sum of currents at a node is zero, meaning currents entering equal currents leaving.

In a purely inductive circuit, current lags voltage by 90° due to self-inductance effects.

An ideal voltage source maintains constant voltage regardless of load, which requires zero internal resistance.

Thevenin equivalent voltage is the open-circuit voltage measured across the two terminals when the load is removed from the circuit.

Energy stored in capacitor = ($\frac{1}{2}$)CV² = ($\frac{1}{2}$) × 100×10⁻⁶ × 200² = 0.5 × 100×10⁻⁶ × 40000 = 2 J

In a balanced three-phase system, VL = √3 × Vph. Therefore, Vph = VL/√3 = 400/√3 = $\frac{400}{1}$.732 = 230.9 V

For parallel resistors: 1/Req = 1/R + 1/R = 2/R. Therefore, Req = R/2

Mesh analysis applies Kirchhoff's voltage law (KVL) around each independent mesh loop in the circuit

Time constant τ = L/R = 0.$\frac{5}{100}$ = 0.005 s = 5 ms

For sinusoidal current I = Im sin(ωt + φ), IRMS = Im/√2 = 5/√2 = 3.54 A

Nodal analysis uses Kirchhoff's current law (KCL) at each node to find node voltages, with one node taken as reference (ground)

Thevenin equivalent voltage is the open circuit voltage measured across the terminals, which is 8V in this case.

KVL states that the algebraic sum of all voltages in a closed loop is zero, applicable to both AC and DC circuits.

The Wheatstone bridge is balanced when R₁ × R₄ = R₂ × R₃, resulting in zero galvanometer deflection.

When considering only V₁ (V₂ is short-circuited), I₁ = V₁/R = $\frac{10}{10}$ = 1A