Verified Solution • 2025–26 Session moderate

QJEEM20Phfe65: JEE Main 2020 Physics - Class 12 JEEM20Phfe65 Solution: A battery of 3.0 V is connected to a resistor dissipating 0.5 W …

A battery of 3.0 V is connected to a resistor dissipating 0.5 W of power. If the terminal voltage of the battery is 2.5 V, the power dissipated within the internal resistance is :

Multiple Choice Options

  • A 0.50 W
  • B 0.072 W
  • C 0.10 W
  • D 0.125 W

Step-by-Step Verified Solution

PR = 0.5 W

⇒ i2R = 0.5 W

iR = 2.5

⇒ i = 0.2 A & R = 12.5 Ω

Also, V = E – ir

⇒ 2.5 = 3 – (0.2)r

⇒ r = 2.5 Ω

Power dissipated in internal resistance

= i2r = (0.2)2(2.5) = 0.1 W

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