Question

When a resistor of $11\Omega$ is connected in series with an electric cell, the current flowing in it is $0.5A$ . Instead when a resistor of $5\Omega$ is connected to the same electric cell in series, the current increases by $0.4A$ . The internal resistance of the cell is:

• A. $1.5 \, \Omega$
• B. $2 \, \Omega$
• C. $2.5 \, \Omega$
• D. $3.5 \, \Omega$

Answer 1

Answer: (C)

$i=\frac{E}{R + r}$
$0.5=\frac{E}{11 + r}$
$E=5.5+0.5r$
$0.9=\frac{E}{5 + r}$
$E=4.5+0.9r$
On solving these equation, $r=2.5 \, \Omega$