Question

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

Answer 1

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
$\epsilon =120-8=112 \, V$
The total resistance of the circuit $=R+r$
$ \, \, =\left(15.5 + 0.5\right) \, \Omega=16\Omega$
$\therefore \, \, \, $ Current in the circuit during charging
$I=\frac{\epsilon }{R + r}=\frac{112}{16}=7 \, A$
$\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)=11.5 \, V=\frac{23}{2}V$