A current of 2   mA was passed through an unknown resistor which dissipated a power of 4.4   W .…

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. 11×105 W
  2. 11×10-3 W
  3. 11×10-5 W
  4. 11×10-4 W

Solution

P=V2R=i2RResistance R=V2P=Pi2

Here,  Power P=4.4 W   and i=2 mA  

so R=4.44 ×10-6 Ω

When 11 V power supply is connected across R 

P′=1124.4×4×10-6

=11×10-5 W.

Asked in: JEE Main 2019 (10 Jan Shift 2)

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