Question

A cell supplies a current of $1.2A$ A through two resistors each of $2\Omega$ connected in parallel. When the resistors are connected in series, it supplies a current of $0.4A$. Calculate:
(i) the internal resistance, and
(ii) electromagnetic force of the cell.

Answer 1

In parallel $R=\frac{1}{2}+\frac{1}{2}=1\Omega$
$I=1.2A$
$ϵ=I(R+r)=1.2(1+r)=1.2+1.2r$
In series $R=2+2=4\Omega$
$I=0.4A$
$ϵ=I(R+r)=0.4(4+r)=1.6+0.4r$
It means :
$1.2+1.2r=1.6+0.4r$
$0.8r=0.4$
$r=\frac{0.4}{0.8}=\frac{1}{2}=0.5\Omega$
(i) Internal resistance $r=0.5\Omega$
(ii) $ϵ=I(R+r)=1.2(1+0.5)=1.8V$