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:
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 - <p>0.50 W</p>
Option 2 - <p>0.10 W</p>
Option 3 - <p>0.125 W</p>
Option 4 - <p>0.072 W<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
4 Views|Posted 5 months ago
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5 months 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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