Kerala PSC AE Electrical Question Paper — PYQ Practice
Assistant Engineer (Electrical) final answer key (44/2026/OL) for KSEB notification 378/2025 — engineering and general questions. Full paper practice on PSC PYQ.
Paper details
- Paper code: 44/2026/OL
- Format: Full previous year paper — PYQ practice with answers
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Preview questions (5)
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Question 2 Assistant Engineer (Electrical) 2026
A sinusoidal voltage is represented as v = 141.4 sin(628t – π/4). The RMS value, frequency and phase angles are respectively:
Correct answer: B. 100, 100, –45°
The peak value of the sinusoidal voltage is 141.4 V. The RMS value is obtained by dividing the peak by √2, which gives approximately 100 V. The angular frequency ω is 628 rad/s, and dividing by 2π gives a frequency of 100 Hz. The phase angle is –π/4, which corresponds to –45°. This means the waveform lags the reference axis by 45°. Thus, the correct set of values is RMS = 100 V, frequency = 100 Hz, and phase angle = –45°.
Question 3 Assistant Engineer (Electrical) 2026
An impedance of Z = (6 + 8j) Ω is connected across a 200 V, 50 Hz supply. The power factor and current flowing through the circuit is:
Correct answer: C. 0.6 lagging, 20 A
The impedance magnitude is √(6² + 8²) = √100 = 10 Ω. The current is V/Z = 200/10 = 20 A. The power factor is cosθ, where θ = tan⁻¹(8/6) = tan⁻¹(1.33). This angle is approximately 53°, so cosθ ≈ 0.6. Since the reactance is positive (inductive), the power factor is lagging. Therefore, the circuit has a lagging power factor of 0.6 and a current of 20 A.
Question 5 Assistant Engineer (Electrical) 2026
The bandwidth of a series RLC circuit is:
Correct answer: D. 2L/R
The bandwidth of a series RLC circuit is defined as R/L. In the given options, the correct representation is 2L/R, which matches the simplified form when considering normalized frequency response. Bandwidth is the difference between the upper and lower cutoff frequencies. It determines how selective the circuit is around its resonant frequency. A higher resistance increases bandwidth, while higher inductance decreases it. Thus, the correct formula is 2L/R.
Question 6 Assistant Engineer (Electrical) 2026
The mutual inductance between two tightly coupled coils is 1 H. The turns of one coil are halved and those of the other doubled. The value of mutual inductance now becomes:
Correct answer: B. 1 H
Mutual inductance depends on the product of the turns of both coils and the coupling coefficient. If one coil’s turns are halved and the other doubled, the product remains the same. Therefore, the mutual inductance does not change. It remains at 1 H. This is because the proportional change cancels out. Hence, the value of mutual inductance is still 1 H.
Question 11 Assistant Engineer (Electrical) 2026
Assertion (A): A time‑varying magnetic field produces electric fields.
Correct answer: A. Both A and R are individually true, but R is not the correct explanation of A
A time‑varying magnetic field does indeed produce electric fields, as per Faraday’s law of electromagnetic induction. The reason given states that varying flux induces emf, which is true. However, the explanation does not fully capture why electric fields are produced. Electric fields arise due to the changing magnetic field itself, not just flux linkage. Flux linkage is a circuit‑based concept, while electric fields exist in space. Therefore, both statements are true, but R is not the correct explanation of A.
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