Question

When an ammeter of negligible internal resistance is inserted in series with circuit, it reads 1 A. When a voltmeter of very large resistance is connected across $R_1$, it reads 3 V. But when the points A and B are short circuited by a conducting wire, then the voltmeter measures 10.5 V across the battery. The internal resistance of the battery is equal to

• A. $\frac{3}{7}\omega$
• B. 5 $\omega$
• C. 3 $\omega$
• D. none of these

Answer 1

Answer: (A)

$\frac{3}{7}\omega$

$3=I{R}_1$ or $3=1\times R_1$ or $R_1=3\omega$
When points A and B are connected by a conducting wire, $r_2$ is short-circuited. Therefore,
$10.5=I^{\prime }{R}_1$ or $10.5=I^{\prime }\times 3$
or $I^{\prime }=\frac{10.5}{3}=3.5A$
But $10.5=E-I^{\prime }r$ or $10.5=12-3.5r$
$\therefore r=\frac{1.5}{3.5}=\frac{3}{7}\omega$