JEE MAIN - Physics (2019 - 10th January Evening Slot - No. 20)
A current of 2 mA was passed through an unknown resistor which dissipated a power of 4.4 W. Dissipated power when an ideal power supply of 11 V is connected across it is -
11 $$ \times $$ 10ā5 W
11 $$ \times $$ 10ā3 W
11 $$ \times $$ 105 W
11 $$ \times $$ 10ā4 W
Explanation
P = I2R
4.4 = 4 $$ \times $$ 10$$-$$6 R
R = 1.1 $$ \times $$ 106 $$\Omega $$
P' = $${{{{11}^2}} \over R}$$ = $${{{{11}^2}} \over {1.1}}$$ $$ \times $$ 10$$-$$6 = 11 $$ \times $$ 10$$-$$5 W
4.4 = 4 $$ \times $$ 10$$-$$6 R
R = 1.1 $$ \times $$ 106 $$\Omega $$
P' = $${{{{11}^2}} \over R}$$ = $${{{{11}^2}} \over {1.1}}$$ $$ \times $$ 10$$-$$6 = 11 $$ \times $$ 10$$-$$5 W
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