JEE PYQ: Current Electricity Question 64
Question 64 - 2019 (10 Jan 2019 Shift 2)
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:
(1) $11 \times 10^{-5}$ W
(2) $11 \times 10^{-3}$ W
(3) $11 \times 10^{-4}$ W
(4) $11 \times 10^{5}$ W
Show Answer
Answer: (1)
Solution
$P = I^2 R$, $4.4 = (2 \times 10^{-3})^2 \times R = 4 \times 10^{-6} R$. $R = 1.1 \times 10^6,\Omega$. $P = \frac{V^2}{R} = \frac{11^2}{1.1 \times 10^6} = \frac{121}{1.1} \times 10^{-6} = 11 \times 10^{-5}$ W.
BETA
