Question

A light bulb is rated 100 W for 220 V ac supply of 50 Hz. Calculate

(a) the resistance of the bulb;

(b) the rms current through the bulb

OR

An alternating voltage given by V = 140 sin 314 t is connected across a pure resistor of 50 W. Find

(i) the frequency of the source.
(ii) the rms current through the resistor.

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Solution

straight a right parenthesis thin space straight R space equals space straight V squared subscript rms over straight P space equals space fraction numerator 220 space straight x space 220 over denominator 100 end fraction space equals space 48 space straight capital omega straight b right parenthesis thin space straight I subscript rms space equals space straight P over straight V subscript rms straight x 100 over 220 space equals space 0.45 space straight A

straight i space right parenthesis thin space straight V space equals space 140 space sin space 314 space straight t because space straight V space equals space straight V subscript straight o space sin space straight omega space straight t Hence comma space straight omega space equals space 314 space space space rightwards double arrow space 2 πν space equals space 314 space rightwards double arrow space straight nu space equals space 50 space Hz ii right parenthesis straight V subscript rms space equals space fraction numerator straight V subscript straight o over denominator square root of 2 end fraction space equals space fraction numerator 140 over denominator square root of 2 end fraction space equals space 98.9 space straight V straight I space equals space straight V subscript rms over straight R equals fraction numerator 98.9 over denominator 50 end fraction space equals space 1.97 space straight A

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