Question

When a resistance of $\alpha\, \Omega$ is connected at the ends of a battery, its potential difference decreases from $40\, V$ to $30\, V$. The internal resistance of the battery is

• A. $3\, \Omega$
• B. $6\, \Omega$
• C. $1.5\,\Omega$
• D. $4\,\Omega$

Answer 1

Answer: (A)

$3\, \Omega$

Answer 2

$E=40\,V,\, V=30\,V,\, r=\alpha\,\Omega$
Internal resistance
$r=\left( \frac{E}{V}-1 \right)R=\left( \frac{40}{30}-1 \right)9$
$=\frac{1}{3}\times 9=3\,\Omega$