Current Electricity

JEE Main level, single correct, medium

234 of 380

Question 234 of 380

By what percentage will the illumination of the lamp decrease if the current drops by 2020%?

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Answer: (C) 3636%

Power through the resistor is given by, P=I2RP={I}^{2}R.
Now, Pint=I2intR{P}_{int}={{I}^{2}}_{int}R
and Pfinal=(0.8Iint)2R{P}_{final}={\left(0.8{I}_{int}\right)}^{2}R
Therefore, % change in power ==
Pfinal−PintPint×100=(0.64−1)×100=−36\dfrac{{P}_{final}-{P}_{int}}{{P}_{int}}\times 100=\left(0.64-1\right)\times 100=-36%

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