Question

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

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

Answer 1

The correct option is B $2.5\Omega$

Current taken from the cell $i=\frac{E}{R+r}$

Where $R=\text{external resistance}\text{and}r=\text{internal resistance}$.

First case $i_1=\frac{E}{R_1+r}$

$0.5=\frac{E}{11+r}-----\left(1\right)$

Second case:-
$0.5+0.4=\frac{E}{5+r}$

$0.9=\frac{E}{5+r}-----\left(2\right)$

$\frac{0.5}{0.9}=\frac{\frac{E}{(11+r)}}{\frac{E}{(5+r)}}$

$\frac{5}{9}=\frac{5+r}{11+r}$

$55+5r=45+9r$

$10=4r$

$r=\frac{10}{4}=2.5\Omega$