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 ................ $V$
Medium
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Equivalent external resistance of the given circuit ${R_{eq}} = 4\,\Omega $
Current given by the cell $i = \frac{E}{{{R_{eq}} + r}} = \frac{{10}}{{(4 + 1)}} = 2\,A$
Hence, $({V_A} - {V_B}) = \frac{i}{2} \times ({R_2} - {R_1}) = \frac{2}{2}(2 - 4) = - 2\,V.$
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