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Q.
In the circuit shown below, the cell has an e.m.f. of $10\, V$ and internal resistance of $1\, ohm$. The other resistances are shown in the figure. The potential difference $V _{ A }- V _{ B }$ is
Solution:
Equivalent external resistance of the given circuit $R_{eq}=4\, \Omega$
Current given by the cell $i=\frac{E}{R_{e q}+r}=\frac{10}{(4+1)}=2 A$
Hence, $\left(V_{A}-V_{B}\right)=\frac{i}{2} \times\left(R_{2}-R_{1}\right)$
$=\frac{2}{2}(2-4)=-2 V$.