All Free Electrical Engineering MCQs with Answers

• A. Cast iron
• B. Copper
• C. Carbon
• D. Tungsten

Explanation: Copper is an excellent conductor, so it is not a poor conductor. Cast iron, carbon, and tungsten have considerably higher resistivity than…

• A. High speed operation
• B. High rupturing capacity
• C. No ageing effect
• D. All of the above

Explanation: HRC fuses interrupt large fault currents safely, operate quickly, and are less affected by ageing than rewirable fuses.

• A. similar charges repel each other
• B. dissimilar charges attract each other
• C. repulsion is the sure test of electrification
• D. an electrically neutral body is repelled both positively and negatively charged bodies

Explanation: A neutral body is normally attracted to a charged body by electrostatic induction, because opposite charge is induced on the nearer side.

• A. zero
• B. infinite
• C. positive
• D. negative

Explanation: For a point charge, electric intensity is E = kQ/r², so it approaches zero as the distance r becomes infinitely large.

• A. positively charged
• B. electrically neutral
• C. negatively charged
• D. positively and negatively charged

Explanation: A neutral conducting ball is attracted to either charged ball because charges are induced within it, with the nearer opposite charge…

• A. 2 x 10^6m/sec
• B. 3 x 10^6m/sec
• C. 5 x 10^6m/sec
• D. 4 x 10^6m/sec

Explanation: Using ½mv² = 25 eV gives v = √(2E/m) ≈ 2.96 × 10⁶ m/s, which rounds to 3 × 10⁶ m/s. Therefore option b matches the calculated speed.

• A. parallel to one another
• B. non parallel to one another
• C. intersecting
• D. circular

Explanation: In the usual uniform-field context, equipotential planes are parallel because the electric field is perpendicular to each plane and has…

• A. charge will flow from A to B
• B. charge will flow from B to A
• C. no flow of charge will occur
• D. both a and b are possible

Explanation: The potential of an isolated charged sphere is V = kQ/R, so the smaller sphere A has the higher potential for the same charge.

• A. negative
• B. positive
• C. zero
• D. data is insufficient

Explanation: Gauss's law states that total flux through a closed surface is Qenclosed/ε₀.

• A. along the direction of force experienced by a unit positive charge
• B. opposite to the direction of force experienced by a unit positive charge
• C. perpendicular to the direction of force experienced by a uit positive charge
• D. any of above

Explanation: The electric field direction is defined as the direction of force on a positive test charge, and a unit positive charge experiences force…

• A. 459 Hz
• B. 4,591 Hz
• C. 999 Hz
• D. 2,176 Hz

Explanation: The parallel LC resonance is approximately f = 1/(2π√LC) = 1/[2π√(0.06 × 0.02 µF)] ≈ 4.59 kHz.

• A. 125 Hz
• B. 1,250 Hz
• C. 564 Hz
• D. 5,644 Hz

Explanation: For an RL high-pass circuit with output across the coil, the critical frequency is f_c = R/(2πL) = 470/(2π × 0.6) ≈ 125 Hz.

• A. 0 Hz to 3.5 kHz
• B. 0 Hz
• C. 3.5 kHz
• D. 7 kHz

Explanation: A low-pass filter passes frequencies from DC, or 0 Hz, up to its cutoff frequency. With f_c = 3.5 kHz, the pass-band is 0 to 3.5 kHz.

• A. 81 µF
• B. 161 µF
• C. 0.161 µF
• D. 220 µF

Explanation: “Ten times less than R” means the capacitive reactance is about 82 Ω, since 820/10 = 82 Ω.

• A. a higher impedance
• B. a smaller bandwidth
• C. a higher resonant frequency
• D. a larger bandwidth

Explanation: For a resonant filter, Q = fr/BW, so bandwidth is inversely proportional to Q.

• A. 10.60 V
• B. 0 V
• C. 21.21 V
• D. 15 V

Explanation: At the critical frequency of a first-order low-pass filter, the output falls to 0.707 of its maximum value. Thus, 0.707 × 15 V = 10.6 V.

• A. 18 V peak-to-peak
• B. zero
• C. 9 V peak-to-peak
• D. 12.74 V peak-to-peak

Explanation: With zero capacitive reactance, the capacitor is effectively a short circuit.

• A. 0 dB
• B. 8.30 dB
• C. -8.30 dB
• D. 0.8 dB

Explanation: Voltage gain in decibels is 20 log10(Vout/Vin) = 20 log10(0.5/1.3) ≈ -8.30 dB.

• A. 5.14 MHz
• B. 514 kHz
• C. 5.03 MHz
• D. 503 kHz

Explanation: The resonant frequency is f0 = 1/(2π√LC). Substituting L = 8 µH and C = 120 pF gives approximately 5.14 MHz; the series resistance changes…

• A. 93.8 kHz
• B. 93.8 Hz
• C. 861 Hz
• D. 86.12 kHz

Explanation: For an RL low-pass filter with output across R, the critical frequency is fc = R/(2πL).