Question

A cell supplies a current of $1.2A$ through two resistor each of $2\Omega$ connected in parallel.When the resistor are connected in series, it supply a current of $0.4A$. Calculate : the internal resistance.

Answer 1

Emf of cell. $\epsilon =I(R+r)$

Resistance in parallel, $R=\frac{2\times 2}{2+2}=1\Omega$

Resistance in series, $R=2+2=4\Omega$

At first event, $\epsilon =1.2(1+r)......\left(1\right)$

At second event, $\epsilon =0.4(4+r).......\left(2\right)$

From (1) and (2)

$0.4(4+r)=1.2(1+r)$

$r=0.5\Omega$

$\epsilon =0.4(4+0.5)=1.8V$

Hence, emf is $1.8V$ and internal resistance $r=0.5\Omega$