Question

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

i) the internal resistance, and

(ii) e.m.f. of the cell.

Answer 1

In parallel:

$\frac{1}{R}=\frac{1}{2}+\frac{1}{2}\Rightarrow R=1\Omega$

I = 1.2 A

ε = I(R + r) = 1.2(1 + r) = 1.2 + 1.2 r ....(i)

In series:

$R=2+2=4\Omega$

I = 0.4 A

ε = I(R + r) = 0.4(4 + r) = 1.6 + 0.4 r ....(ii)

From (i) and (ii), 1.2 + 1.2 r = 1.6 + 0.4 r

$\Rightarrow$ 0.8 r = 0.4
$\Rightarrow \text{Internal resistance,}r=0.5\Omega$

ε = 1.2 + 1.2 $\times$ 0.5 = 1.8 V