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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
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Verified Answer
The correct answer is:
$1 / 3 \Omega$
$R=\frac{E}{z+r}$
$$
\begin{aligned}
& 2=\frac{E}{2+r} \\
& 0.5=\frac{E}{9+r}
\end{aligned}
$$
From Eqs. (i) and (ii), We have
$$
\begin{aligned}
\frac{2}{0.5} & =\frac{9+r}{2+r} \\
4 & =\frac{9+r}{2+r} \\
3 r & =1 \\
r & =\frac{1}{3} \Omega
\end{aligned}
$$
$$
\begin{aligned}
& 2=\frac{E}{2+r} \\
& 0.5=\frac{E}{9+r}
\end{aligned}
$$
From Eqs. (i) and (ii), We have
$$
\begin{aligned}
\frac{2}{0.5} & =\frac{9+r}{2+r} \\
4 & =\frac{9+r}{2+r} \\
3 r & =1 \\
r & =\frac{1}{3} \Omega
\end{aligned}
$$
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