Basic Electrical Engineering MCQ – 46 previous year questions with solutions

Around a long straight conductor B = μI / (2πr). At the same distance and in the same medium, B is directly proportional to I, so the field gets stronger.

Magnetic flux Φ has an SI unit — the weber (Wb). Flux density B is in tesla (T) = Wb/m².

Mutual inductance satisfies M = k√(L1L2), where 0 ≤ k ≤ 1. The maximum occurs at perfect coupling k = 1: Mmax = √(1×2) mH = √2 mH ≈ 1.414 mH.

C — M = k√(L₁L₂). The coupling coefficient k accounts for the fraction of magnetic coupling between the two coils.

D — L = L₁ + L₂ − 2M. In series opposition the mutually induced flux opposes the self-induced flux in each coil, so 2M is subtracted (series aiding gives +2M).

B — Ferromagnetic material. In ferromagnetic materials (iron, cobalt, nickel) magnetic domains align strongly under an external field, so the material becomes strongly magnetised and can retain it.

Inductance follows L = μN²A/l. Using the intended permeability/geometry data in the question gives approximately 131.94 mH, matching option A.

D — Decreases. For a solenoid L = μN²A/l, so with other factors constant inductance is inversely proportional to length.

A — 5 H. Uncoupled inductors in series carry the same current, so their inductances add directly: 2 + 3 = 5 H.

Yes. Magnetic circuits use direct analogies of Kirchhoff's laws: flux continuity at a junction resembles KCL, and the algebraic sum of MMF drops around a magnetic path equals the applied MMF, analogous to KVL.

For a magnetic circuit of unchanged reluctance, flux is proportional to magnetomotive force: Φ = F/ℜ. Therefore, if flux decreases, the MMF required/producing it also decreases.

Magnetic circuits use an electric-circuit analogy: magnetomotive force drives magnetic flux through reluctance. An air gap greatly increases total reluctance because air has much lower permeability than iron. Key relation: ℜ=l/(μA); Φ=F/ℜ; F=NI.

Exam focus: Air gaps dominate reluctance even when physically short. Common trap: Permeability of magnetic materials is not generally constant with field intensity.

C — Magnetic field. For a straight current-carrying conductor, if the thumb points along the current, the curled fingers give the direction of the magnetic field lines.

In sinusoidal steady state, complex power combines real and reactive power as S=P+jQ. Power factor is the ratio P/|S| and equals cosφ for sinusoidal voltage and current. Key relation: S=P+jQ; |S|=√(P²+Q²); PF=P/|S|=cosφ.

Exam focus: Draw the power triangle before choosing a power-factor relation. Common trap: Do not confuse P/Q with power factor.

For an ideal square wave, the magnitude is constant throughout each half-cycle. Therefore peak value = rms value = average rectified value, making both peak factor and form factor equal to 1. Hence the two factors are equal for a square wave.

Frequency means cycles per second. At 100 Hz, the waveform completes 100 cycles each second; in 2 s it completes 100×2 = 200 cycles.

Peak factor (crest factor) is V_peak/V_rms. For a half-wave rectified sine, V_rms=V_m/2, so peak factor=2, which exceeds sine (√2), triangular (√3) and square (1).

RMS value is defined as the DC value that would produce the same heating effect in a given resistance over the same time. That is why RMS—not average or peak—is used for AC equipment voltage/current ratings.

The average of a rectified sine over a half cycle is 2Xm/π, giving sine-wave form factor RMS/average-rectified ≈1.11. For v(t)=100 sinωt, Vm=100 V and Vrms=Vm/√2≈70.7 V. The full-cycle average of a sine wave is zero. Key relation: Xrms=Xm/√2; Xavg(rectified)=2Xm/π; form factor≈1.11.

Exam focus: Check whether the question asks full-cycle average or rectified average. Common trap: Using the half-cycle average as the full-cycle average is a common error.

For a sine wave x(t)=Xm sinωt, the RMS value is Xm/√2. The average of a rectified sine over a half cycle is 2Xm/π, giving sine-wave form factor RMS/average-rectified ≈1.11. Key relation: Xrms=Xm/√2; Xavg(rectified)=2Xm/π; form factor≈1.11.

Exam focus: Check whether the question asks full-cycle average or rectified average. Common trap: Using the half-cycle average as the full-cycle average is a common error.

Peak factor directly relates the maximum (peak) value of an AC waveform to its RMS value. By definition, Peak factor = V_peak / V_rms, so V_peak = Peak factor × V_rms. For a sine wave the peak factor is √2.

D — Fleming’s right-hand rule. For generator action, the thumb indicates motion, the first finger the magnetic field and the middle finger induced conventional current.

Faraday's law states that induced EMF is proportional to the time rate of change of flux linkage. Lenz's law supplies the negative sign: the induced current/EMF acts in a direction that opposes the change producing it, consistent with conservation of energy. Key relation: e = −N dΦ/dt.

Exam focus: Magnitude comes from Faraday; direction comes from Lenz. Common trap: Lenz's law says oppose the cause/change, not aid it.

D — Magnetic flux. The induced EMF is proportional to the rate of change of magnetic flux linkage: e = −N dΦ/dt.

A — Electromagnetic induction. Faraday's first law states that an EMF is induced in a circuit whenever the magnetic flux linked with it changes.

Faraday’s law relates induced EMF to the rate of change of flux linkage: e = −N dΦ/dt. The negative sign is Lenz’s law.

B — The direction of induced current reverses. With the field direction unchanged, reversing the conductor's motion reverses the induced EMF and current (Fleming's right-hand rule).

D — The current flowing through the coil. When the coil current changes, its own flux linkage changes and a self-induced EMF (e = −L di/dt) opposes that change.

D — Lenz's law. The direction of the induced EMF is such that its effect opposes the cause producing it.

B — Watt. Active power is the real, useful power consumed (P = VI cosφ) and its SI unit is the watt (W).

A — The product of RMS voltage and RMS current. S = V_rms × I_rms, measured in VA.

A — 80% of the apparent power. Power factor = P/S, so P = 0.8 S.

For a fixed real-power load and supply voltage, poor power factor requires higher current. Raising power factor with shunt capacitors or an over-excited synchronous machine reduces kVA demand, line current and I²R losses and generally improves voltage regulation. Key relation: Single phase: I=P/(V·PF). Three phase: I=P/(√3 V_L·PF).

Exam focus: For the same kW, current is inversely proportional to power factor. Common trap: Capacitors supply leading reactive power; they do not supply lagging VARs.

Its magnitude |S| is apparent power in VA, while P is measured in W and Q in var. S=P/PF=1000/0.8=1250 VA. Q=√(S²−P²)=750 var. Key relation: S=P+jQ; |S|=√(P²+Q²); PF=P/|S|=cosφ.

Exam focus: Draw the power triangle before choosing a power-factor relation. Common trap: Do not confuse P/Q with power factor.

C — Volt-ampere reactive (VAR). Reactive power Q = VI sinφ is measured in VAR.

Current is the rate of flow of charge, so the required charge is the signed area under the current–time graph from 0 to 3 s. The interval contains a 10 A rectangle, a 10→5 A trapezoid and a 5 A rectangle; their areas give 22.5 C.

Charge is quantized in units of the elementary charge e = 1.602 × 10⁻¹⁹ C. Hence n = Q/e = 2/(1.602×10⁻¹⁹) ≈ 1.248×10¹⁹ electrons = 12.48×10¹⁸ electrons.

Current density is current per unit cross-sectional area. Thus J = I/A = 30/0.5 = 60 A/m².

W = P × t. With t unchanged, tripling P triples the work.

For constant mass, kinetic energy is K=p²/(2m). Doubling momentum therefore multiplies kinetic energy by 2²=4. The GPSC final key revised the provisional answer to option C.

At fixed supply voltage, doubling resistance halves current and halves power because P=V²/R. Given V=20 V and R=10 Ω: I=V/R=2 A and P=VI=40 W. If R doubles to 20 Ω at the same 20 V, I=1 A and P=20 W. Key relation: I=V/R; P=VI=I²R=V²/R.

Exam focus: At fixed V, power varies inversely with R. Common trap: Do not assume power stays constant when resistance changes on a fixed-voltage source.

Equivalent resistance is found from reciprocal addition, total current is the sum of branch currents, and total real power is the sum of branch powers. For 6 Ω || 3 Ω, Req=(6×3)/(6+3)=2 Ω. Total current=12/2=6 A. Key relation: 1/R_eq=Σ(1/R_i); I_i=V/R_i; P_total=V I_total.

Exam focus: Solve branch currents first when the same supply voltage appears across every resistor. Common trap: Adding parallel resistances directly is incorrect.

B — 1 A. The 2 Ω and 6 Ω resistors are in series, so the 8 V source drives 8/(2 + 6) = 1 A.

A — Electric field. Every electric charge creates a region around it in which another charge experiences an electric force; this region is the electric field.

Series capacitors carry the same charge. Since V = Q/C, voltage is inversely proportional to capacitance: V(12 μF) : V(4 μF) = 4 : 12 = 1 : 3.

D — Separation between the plates. For a parallel-plate capacitor C = εA/d, so capacitance depends on plate area, dielectric and plate separation — not on plate thickness, metal or applied voltage.