Question

When a resistor of $11\Omega$ is connected in series with a electric cell. The current following 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)

$2.5\Omega$

Here, $i=\frac{E}{R+r}$
$0.5=\frac{E}{11+r}\Rightarrow E=5.5+0.5r$
$0.9=\frac{E}{5+r}$ or, $E=4.5+0.9r$
On solving we get $r=2.5\Omega$