JEE PYQ: Current Electricity Question 21
Question 21 - 2020 (04 Sep 2020 Shift 1)
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:
(1) 0.50 W
(2) 0.072 W
(3) 0.10 W
(4) 0.125 W
Show Answer
Answer: (3)
Solution
$P_R = 0.5$ W, $E = 3$ V, $V = 2.5$ V. $V = E - ir \Rightarrow 2.5 = 3 - ir \Rightarrow ir = 0.5$. $iR = 2.5$ and $ir = 0.5$, so $\frac{R}{r} = 5$. $P_r = i^2 r = \frac{P_R}{5} = \frac{0.50}{5} = 0.10$ W.
BETA
