Question

A battery of emf $8V$ with internal resistance $0.5 \, \Omega$ is being charged by a $120 \,V\, d.c.$ 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)

$11.5\,V$

Answer 2

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

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