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JEE Main202227 Jun 2022Morning ShiftPhysicsCurrent ElectricityActual

A 220 V , 50 Hz AC source is connected to a 25 V , 5 W lamp and an additional resistance R in series (as shown in figure) to run the lamp at its peak brightness, then the value of R (in ohm) will be

Correct answer

0

Step-by-step solution

Resistance of the bulb can be calculated as, P = V 2 R B ⇒ R B = V 2 P = 25 2 5 = 125   Ω The current through the bulb for peak brightness should be, i = 25 125 = 1 5   A Now, i rms = 1 5 = 220 R B + R R B + R = 1100 R = 1100 - 125 = 975

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