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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:

Option 1 -

0.50 W

Option 2 -

0.10 W

Option 3 -

0.125 W

Option 4 -

0.072 W

0 2 Views | Posted a month ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    a month ago
    Correct Option - 2


    Detailed Solution:

    ↑? (1) ε = 3
    (2) ε – Ir = 2.5 V
    ⇒ Ir = 0.5
    Now, IR = 2.5
    ⇒ R/r = 5.
    ⇒ PR/Pr = I²R/I²r = R/r = 5
    ⇒ Pr = 0.5/5 = 0.1

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