Question

When a battery of emf $8V$ with internal resistance $0.5\Omega$ is charged by a $120VDC$ supply using a series resistance of $15.5\Omega$ , the terminal voltage of the battery is $\frac{x}{2}V$ , find $x$ .

Answer 1

Answer: 23

Here, emf of the battery $=8V$
The internal resistance of the battery, $r=0.5\Omega$
The voltage of DC supply = $120V$
External resistance, $R=15.5\Omega$
When the battery of emf $8V$ is charged from a $DC$ supply of $120V$ , the effective emf in the circuit is
$ϵ=120-8=112V$
The total resistance of the circuit $=R+r$
$=\left(15.5+0.5\right)\Omega =16\Omega$
$\therefore$ Current in the circuit during charging
$I=\frac{ϵ}{R+r}=\frac{112}{16}=7A$
$\therefore$ The terminal voltage of the battery,
$V=\text{emf of the battery}+\text{voltage drop across the battery}$
$=8+Ir=8+7\left(0.5\right)=11.5V=\frac{23}{2}V$