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 :
Solution
P<sub>R</sub> = 0.5 W
<br><br>$\Rightarrow$ i<sup>2</sup>R = 0.5 W
<br><br>iR = 2.5
<br><br>$\Rightarrow$ i = 0.2 A & R = 12.5 $\Omega$
<br><br>Also, V = E – ir
<br><br>$\Rightarrow$ 2.5 = 3 – (0.2)r
<br><br>$\Rightarrow$ r = 2.5 $\Omega$
<br><br>Power dissipated in internal resistance
<br><br>= i<sup>2</sup>r = (0.2)<sup>2</sup>(2.5) = 0.1 W
About this question
Subject: Physics · Chapter: Current Electricity · Topic: Ohm's Law and Resistance
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