Two batteries $A$ and $B$ each of $e.m.f.$ $2\, V$ are connected in series to an external resistance $R = 1 \,ohm$. If the internal resistance of battery $A$ is $1.9\, ohms$ and that of $B$ is $0.9\, ohm$, what is the potential difference between the terminals of battery $A$ ............. $V$
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c
$i = \frac{{2 + 2}}{{1 + 1.9 + 0.9}} = \frac{4}{{3.8}}\,A$
$i = \frac{{2 + 2}}{{1 + 1.9 + 0.9}} = \frac{4}{{3.8}}\,A$
For cell $A$ $E = V + ir$ $ \Rightarrow $ $V = 2 - \frac{4}{{3.8}} \times 1.9 = 0$.
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