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 ( in W) within the internal resistance is:

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is 0.1.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$P_{R} = 0.5 W$

$\Rightarrow i^{2} R = 0.5 W$

$Also, V = E - ir$ $\Rightarrow ir = 0.5$

$\begin{array}{l} P = (E - V) i, 0 .5 = \frac{V^{2}}{R} \Rightarrow R = 12 .5 Ω \\ i = \frac{V}{R} = \frac{1}{5} A \end{array}$

$Now iR = 2.5 and ir = 0.5$

$On dividing : \frac{R}{r} = 5$

$\frac{P_{R}}{P_{r}} = \frac{i^{2} R}{i^{2} r} \Rightarrow \frac{P_{R}}{P_{r}} = \frac{R}{r} \Rightarrow \frac{P_{R}}{P_{r}} = 5$

$\Rightarrow P_{r} = \frac{P_{R}}{5}$

$\Rightarrow P_{r} = \frac{0.50}{5} \Rightarrow P_{r} = 0.10 W$

Watch 3-min video & get full concept clarity