JEE MAIN - Physics (2021 - 25th July Morning Shift - No. 24)
An electric bulb rated as 200 W at 100 V is used in a circuit having 200 V supply. The resistance 'R' that must be put in series with the bulb so that the bulb delivers the same power is _____________ $$\Omega$$.
Answer
50
Explanation
Power, $$P = {{{V^2}} \over {{R_B}}}$$
$${R_B} = {{{V^2}} \over P} = {{100 \times 100} \over {200}}$$
$${R_B} = 50\Omega $$
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To produce same power, same voltage (i.e. 100 V) should be across the bulb.
Hence, R = RB
R = 50 $$\Omega$$
$${R_B} = {{{V^2}} \over P} = {{100 \times 100} \over {200}}$$
$${R_B} = 50\Omega $$
To produce same power, same voltage (i.e. 100 V) should be across the bulb.
Hence, R = RB
R = 50 $$\Omega$$
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