Question

A battery of emf $8V$ with internal resistance $0.5\Omega$ is being charged by a $120Vd.c.$ supply using a series resistance of $15.5\Omega$ . The terminal voltage of the battery is

• A. 20.5 V
• B. 15.5 V
• C. 11.5 V
• D. 2.5 V

Answer 1

Answer: (C)

Emf of the battery $e=8V$, emf of $DC$ supply
$V=120V$
Since, the battery is bring changes, so effective emf in the circuit
$E=V-e=120-8=112V$

Current in the circuit $I=\frac{\text{Effective emf}}{\text{Total resistance}}$
$=\frac{E}{r+R}$
$=\frac{112}{0.5+15.5}$
$=\frac{112}{16}=7A$
The battery of $8V$ is being charged by $120V$, so the terminal potential across battery of $8V$ will be greater than its emf.
Terminal potential difference
$V=E+Ir$
$=8+7(.5)$
$=11.5V$