Verified Solution • 2025–26 Session moderate

QJEEM19Phf861: JEE Main 2019 Physics - Class 12 JEEM19Phf861 Solution: A current of 2 mA was passed through an unknown resistor which…

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 -

Multiple Choice Options

  • A 11 ×\times 10–5 W
  • B 11 ×\times 10–3 W
  • C 11 ×\times 105 W
  • D 11 ×\times 10–4 W

Step-by-Step Verified Solution

P = I2R

4.4 = 4 ×\times 10 −- 6 R

R = 1.1 ×\times 106 Ω\Omega

P' = 112R{{{{11}^2}} \over R} = 1121.1{{{{11}^2}} \over {1.1}} ×\times 10 −- 6 = 11 ×\times 10 −- 5 W

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