When a battery connected across a resistor of $16 \Omega$, the voltage across the resistor is $12…
When a battery connected across a resistor of $16 \Omega$, the voltage across the resistor is $12 \mathrm{~V}$. When the same battery is connected across a resistor of $10 \Omega$, voltage across it is $11 \mathrm{~V}$. The internal resistance of the battery in ohm is
$\frac{10}{7}$
$\frac{20}{7}$
$\frac{25}{7}$
$\frac{30}{7}$
Solution
$
\begin{array}{ll}
\text { Here, } & V < E \\
\therefore & E=V+I r
\end{array}
$
For first case
$
E=12+\frac{12}{16} r
$
For second case
$
E=11+\frac{11}{10} r
$
From Eqs. (i) and (ii),
$
\Rightarrow \quad \begin{aligned}
12+\frac{12}{16} r & =11+\frac{11}{10} r \\
r & =\frac{20}{7} \Omega
\end{aligned}
$