Questions
An electric bulb, marked 40 W and 200 V is used in a circuit of supply voltage 100 V. Now its power is
detailed solution
Correct option is D
Power of bulb is given by, P=V2R⇒ P2P1 = V22V12 ∵ R is constant⇒ P2P1 = 1002002 = 14 ⇒ P2 = P14=404=10 WTalk to our academic expert!
Similar Questions
A resistor R is connected across a battery of emf 12 V. It observed that terminal potential differenceis 9 V and the power delivered to resistor is 18 W. Internal resistance of the battery ‘r’ and the resistance R are respectively
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