Question

A battery of emf $ 8\,V $ 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)

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$