A current of 2 A flows through a $2 \Omega$ resistor when connected across a battery. The same battery…
A current of 2 A flows through a $2 \Omega$ resistor 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
$1 / 3 \Omega$
$1 / 4 \Omega$
$1 \Omega$
$0.5 \Omega$
Solution
$R=\frac{E}{z+r}$
$\begin{aligned}
& 2=\frac{E}{2+r} \\
& 0.5=\frac{E}{9+r}
\end{aligned}$
From Eqs. (i) and (ii), We have
$\begin{aligned}
\frac{2}{0.5} & =\frac{9+r}{2+r} \\
4 & =\frac{9+r}{2+r} \\
3 r & =1 \\
r & =\frac{1}{3} \Omega
\end{aligned}$