JEE Main201910 Jan 2019Evening ShiftPhysicsCurrent ElectricityActual
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:
Options
- A11 × 10 5   W
- B11 × 10 - 3   W
- C11 × 10 - 5   W
- D11 × 10 - 4   W
Correct answer
C. 11 × 10 - 5   W
Step-by-step solution
P = V 2 R = i 2 R R e s i s t a n c e   R = V 2 P = P i 2 Here, Power P = 4.4   W       a n d   i = 2   mA     so R = 4.4 4   × 10 - 6 Ω When 11   V power supply is connected across R P ′ = 11 2 4.4 × 4 × 10 - 6 = 11 × 10 - 5   W .