Fuse protects circuit by: MCQ with Answer and Explanation

Fuse protects circuit by:
A. Amplifying signal
B. Melting on overcurrent
C. Storing energy
D. Limiting voltage
Answer: Option B
Solution (By JKSSB Mock Tests)
Fuse wire melts when current exceeds rating (Joule heating I²R), breaking circuit to prevent damage. Sacrificial overcurrent protection. Memory aid: 'Fuse = thermal cutoff; rated current slightly above normal'. Safety device function frequently tested in competitive electricity sections.

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Practice More Physics Questions

Question #1
A ray of light is incident on a glass slab at angle i. The emergent ray is:
A. Perpendicular to incident ray
B. Bent towards the normal
C. Bent away from the normal
D. Parallel to incident ray but laterally displaced

Correct Answer: Option D


Explanation:
In a parallel-sided glass slab, refraction at first surface bends ray towards normal; at second surface (glass to air), it bends away by equal angle, making emergent ray parallel to incident ray but shifted laterally. No net deviation, only displacement. Memory aid: 'Slab: emergent ray parallel to incident; prism: emergent ray deviated'. This ray optics concept is frequently tested in competitive exams. Always distinguish slab (parallel faces) from prism (non-parallel faces) behavior: slab causes lateral shift, prism causes angular deviation.

This question belongs to: Science Physics
Question #2
The magnetic field at the center of a circular coil of radius R carrying current I is:
A. μ₀NI/(4πR)
B. μ₀NI/(2R)
C. μ₀I/(2R)
D. μ₀I/(4R)

Correct Answer: Option B


Explanation:
For a circular coil with N turns, magnetic field at center B = μ₀NI/(2R). For single turn (N=1), B = μ₀I/(2R). This derives from Biot-Savart law. Option A is for single turn; B and D have incorrect constants. Memory tip: 'Coil center: B = μ₀NI/(2R); straight wire: B = μ₀I/(2πr)'. This formula application is frequently tested in magnetism sections of competitive exams. Always note number of turns N; competitive exams often include it to test attention to detail.

This question belongs to: Science Physics
Question #3
If three resistors of 2 Ω, 3 Ω, and 6 Ω are connected in parallel, the equivalent resistance of the combination is:
A. 11 Ω
B. 1 Ω
C. 2 Ω
D. 0.5 Ω

Correct Answer: Option B


Explanation:
For resistors in parallel, the reciprocal of the equivalent resistance (1/Rp) is the sum of the reciprocals of individual resistances. 1/Rp = 1/R1 + 1/R2 + 1/R3 = 1/2 + 1/3 + 1/6. Finding a common denominator (6): 1/Rp = 3/6 + 2/6 + 1/6 = 6/6 = 1. Therefore, Rp = 1 Ω.

This question belongs to: Science Physics