Question

A current of 2 $A$ flows through a $2\Omega$ resistance when connected across a battery. The same battery supplies a current of 0.5 $A$ when connected across a $9\Omega$ resistor. The internal resistance of the battery is

• A. $0.5\Omega$
• B. $\frac{1}{3}\Omega$
• C. $\frac{1}{5}\Omega$
• D. $1\Omega$

Answer 1

The correct option is B $\frac{1}{3}\Omega$

Let the internal resistance of the battery be $r$. Then, the current flowing through the circuit is given by
$I=\frac{E}{R+r}$
case 1: $\begin{array}{}2=\frac{E}{2+r}& \text{(i)}\end{array}$
case 2: $\begin{array}{}0.5=\frac{E}{9+r}& \text{(ii)}\end{array}$
Equating $E$ from (i) and (ii)
$4+2r=4.5+0.5r$
$1.5r=0.5$
$r=\frac{1}{3}\Omega$