Q152 / 160 · AP EAMCET 2008 · Physics
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
- A$\frac{10}{7}$
- Correct answerB$\frac{20}{7}$
- C$\frac{25}{7}$
- D$\frac{30}{7}$
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),
$$
\Rightarrow \quad \begin{aligned}
12+\frac{12}{16} r & =11+\frac{11}{10} r \\
r & =\frac{20}{7} \Omega
\end{aligned}
$$
Hence, the correct answer is option: 'B'.