When a resistor of $11 \Omega$ is connected in series with an electric cell, the current flowing in it is 0…
When a resistor of $11 \Omega$ is connected in series with an electric cell, the current flowing in it is 0.5 A. Instead, when a resistor of $5 \Omega$ is connected to the same electric cell in series, the current increases by $0.4 \mathrm{~A}$. The internal resistance of the cell is
$1.5 \Omega$
$2 \Omega$
$2.5 \Omega$
$3.5 \Omega$
Solution
Current taken from the cell, $i=\frac{E}{R+r}$
where $R=$ external resistance and
$r=$ internal resistance
$\begin{aligned} \text { Ist Case } i_1 & =\frac{E}{R_1+r} \\ 0.5 & =\frac{E}{11+r}\end{aligned}$
$\begin{aligned} & \text { IInd Case } 0.5+0.4=\frac{E}{5+r} \\ & \Rightarrow \quad 0.9=\frac{E}{5+r}\end{aligned}$
Dividing Eq. (ii) by (i),
$\begin{aligned} \frac{0.5}{0.9} & =\frac{\frac{E}{(11+r)}}{\frac{E}{(5+r)}} \\ \frac{5}{9} & =\frac{5+r}{11+r} \\ 55+5 r & =45+9 r \\ 10 & =4 r \Rightarrow r=2.5 \Omega\end{aligned}$