Free Electronics and Devices MCQs with Answers
758 Electronics and Devices MCQs from Electrical Engineering, each with the correct answer and a written explanation of why it is correct. Free and unlimited, with no account needed.
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- 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.
Correct answer: an electrically neutral body is repelled both positively and negatively charged bodies- 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.
Correct answer: zero- 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…
Correct answer: electrically neutral- 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.
Correct answer: 3 x 10^6m/sec- 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…
Correct answer: parallel to one another- 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.
Correct answer: charge will flow from A to B- 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.
Correct answer: zero- 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…
Correct answer: along the direction of force experienced by a unit positive charge- 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.
Correct answer: 4,591 Hz- 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.
Correct answer: 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.
Correct answer: 0 Hz 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.
Correct answer: 0.161 µF- 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.
Correct answer: a larger bandwidth- 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.
Correct answer: 10.60 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.
Correct answer: zero- 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.
Correct answer: -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…
Correct answer: 5.14 MHz- 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).
Correct answer: 93.8 kHz- A. 723 Hz
- B. 7,234 Hz
- C. 362 Hz
- D. 3,617 Hz
Explanation: For an RC high-pass filter, fc = 1/(2πRC). With R = 220 Ω and C = 0.2 µF, the result is approximately 3.617 kHz.
Correct answer: 3,617 Hz- A. 6.02 dB
- B. -6.02 dB
- C. 12.04 dB
- D. -12.04 dB
Explanation: Voltage gain in decibels is 20 log10(Vout/Vin) = 20 log10(6/12) = 20 log10(0.5) = -6.02 dB.
Correct answer: -6.02 dBElectronics and Devices MCQs: common questions
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There are 758 Electronics and Devices MCQs in the Electrical Engineering bank, shown 20 to a page with the correct answer and an explanation on each.
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