JEE Main 2019 Physics question
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:
Answer: (A)
Solution
R = 4.4/(4 × 10) = 1.1 × 10 Ω; P = 121/(1.1 × 10) = 1.1 × 10 W.