FEEE Unit-1 MCQ & Short Questions

This collection of FEEE (Fundamentals of Electrical & Electronic Engineering) Unit-1 MCQs and Short Questions is carefully compiled based on previous years’ questions of WBSCTE. It is designed to help polytechnic students understand the exam pattern, important topics, and frequently asked concepts.

1. Which component opposes any change in current? a) Resistor b) Capacitor c) Inductor d) Diode -JAN 2022

Answer: c) Inductor

Explanation: According to Lenz’s Law, an inductor develops an induced electromotive force (EMF) that opposes the change in magnetic flux. Mathematically, v=Ldidtv = L \frac{di}{dt}, meaning the voltage across the inductor is proportional to the rate of change of current, effectively resisting any sudden variations.

2. In case of ideal current sources, they have ___________ a) zero internal resistance b) low value of voltage c) large value of current d) infinite internal resistance –JAN 2022

Answer: d) infinite internal resistance

Explanation: An ideal current source must provide a constant current regardless of the load connected to it. To ensure that no current is diverted internally, its internal resistance (connected in parallel) must be infinite (Rint=R_{int} = \infty).

3. For a DC voltage, an inductor (a) is virtually a short circuit. (b) is an open circuit. (c) Depends on polarity. (d) Depends on voltage value. –NOV 2022

Answer: (a) is virtually a short circuit.

Explanation: In steady-state DC, the frequency is zero (f = 0). Since inductive reactance is XL=2πfLX_L = 2\pi fL, the reactance becomes zero. Therefore, the inductor offers no opposition to DC current and behaves like a short circuit (ideal wire).

4. Energy is stored by the capacitor in the form of ___________. -NOV 2022

Answer: Electrostatic Field.

Explanation: When a voltage is applied across the plates of a capacitor, an electric field is established between the plates. The energy stored is given by the formula E=12CV2E = \frac{1}{2}CV^2.

5. If two capacitance of 10μF & 15 μF are connected in parallel then the equivalent capacitance will be _________ μF. -NOV 2022

Answer: 25 μF.

Explanation: For capacitors connected in parallel, the total capacitance is the sum of individual capacitances: Ceq=C1+C2C_{eq} = C_1 + C_2. Therefore, 10 + 15 = 25 μF.

6. What is the frequency of a D.C signal? –NOV 2022

Answer: Zero (0 Hz).

Explanation: Frequency is defined as the number of cycles per second. Since a Direct Current (DC) signal maintains a constant magnitude and direction over time, it does not complete any cycles, making its frequency zero.

7. What is a dependent Source? –NOV 2022

Answer: A source whose output (voltage or current) is controlled by another voltage or current existing elsewhere in the circuit.

Explanation: Unlike independent sources, dependent (or controlled) sources are represented by diamond shapes and are categorized into four types: VCVS, VCCS, CCVS, and CCCS.

8. What do you mean by capacitance of a capacitor? What is its unit? -2023

Answer: Capacitance is the ability of a capacitor to store electric charge per unit of potential difference. Its SI unit is the Farad (F).

Explanation: It is defined by the ratio C=Q/VC = Q/V. One Farad is the capacitance of a conductor which has a potential of one volt when it carries a charge of one coulomb.

9. An ideal current source has (a) zero internal resistance (b) infinite internal resistance (c) low value of voltage (d) large value of current –2024

Answer: (b) infinite internal resistance

Explanation: High internal resistance ensures that the source current does not branch off internally, maintaining a constant flow to the external load. (Repeated from Jan 2022).

10. The resistance of a conductor increases if its (a) length increases (b) area of cross-section increases (c) length as well as area of cross-section increases (d) value of specific resistance is kept constant. –2024

Answer: (a) length increases

Explanation: Resistance follows the formula R=ρlAR = \rho \frac{l}{A}. Thus, R is directly proportional to the length (l) and inversely proportional to the cross-sectional area (A). Increasing length adds more collisions for electrons, increasing resistance.

11. A wire has resistance of 16Ω. It is bent in the form of a circle. The effective resistance between two points on any diameter of the circle is a) 32Ω, b) 16Ω, c) 8Ω, d) 4Ω –2025

Answer: d) 4Ω

Explanation: When the 16Ω wire is bent into a circle, a diameter divides it into two equal semicircular paths. Each path has a resistance of 16/2=816/2 = 8Ω. These two 8Ω paths are effectively in parallel. Thus, Req=8×88+8=4R_{eq} = \frac{8 \times 8}{8 + 8} = 4Ω.

12. If 110V is applied across 220V, 100W bulb, the power consumed by it will be a) 100W, b) 50W, c) 25W, d) 12.5W. –2025

Answer: c) 25W

Explanation: Resistance of the bulb is R=2202100=484R = \frac{V^2}{P} = \frac{220^2}{100} = 484Ω. When 110V is applied, the new power consumed is P=1102484=12100484=25P’ = \frac{V’^2}{R} = \frac{110^2}{484} = \frac{12100}{484} = 25W. Alternatively, since voltage is halved, power (which is V2\propto V^2) becomes (12)2=14(1/2)^2 = 1/4th of original.

13. A current of 10A enters a parallel combination of two resistances of 2Ω and 3Ω. Then the currents through resistances will be a) 4A and 6A, b) 6A and 4A, c) 2A and 8A, d) 5A and 5A –2025

Answer: b) 6A and 4A

Explanation: Using the Current Division Rule: I1=I×R2R1+R2=10×35=6I_1 = I \times \frac{R_2}{R_1 + R_2} = 10 \times \frac{3}{2+3} = 6A. I2=I×R1R1+R2=10×25=4I_2 = I \times \frac{R_1}{R_1 + R_2} = 10 \times \frac{2}{2+3} = 4A.

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