An electric bulb rated 220 V, 100 W is operated at 110 V. The power consumed is
A. 50 W
B. 12.5 W
C. 25 W
D. 100 W
Answer: Option C
Solution (By JKSSB Mock Tests)
Resistance R = V²/P = 220²/100 = 484 Ω. At 110 V, P = V²/R = 110²/484 = 12100/484 = 25 W. Assuming resistance constant. Power reduces to quarter when voltage halved (P ∝ V² for fixed R).
Explanation:
In SHM, a = -ω²x, acceleration ∝ -displacement. Minus sign indicates direction towards mean position (restoring). Characteristic of SHM. Pendulum and spring-mass system exhibit SHM for small displacements.
Explanation:
Fuse is a safety device: thin wire melts when current exceeds safe limit due to short circuit or overload, breaking circuit. Prevents fire and appliance damage. Based on heating effect. MCB is modern alternative.
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