In a Wheatstone bridge, $P = 90\,\Omega $, $Q = 110\,\Omega $ , $R = 40\,\Omega $ and $S = 60\,\Omega $ and a cell of $4\,V\,emf$. Then the potential difference between the diagonal along which a galvanometer is connected is ............. $V$
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$\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=\mathrm{V}_{\mathrm{AB}}=\frac{4}{90+110} \times 90=\frac{9}{5}$ ............$(i)$
$\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{C}}=\mathrm{V}_{\mathrm{AC}}=\frac{4}{40+60} \times 40=\frac{8}{5}$ .........$(ii)$
eq. $(i)$ $-$ eq. $(ii)$
$\Rightarrow \mathrm{V}_{\mathrm{c}}-\mathrm{V}_{\mathrm{B}}=\frac{9}{5}-\frac{8}{5}=\frac{1}{5}=0.2 \mathrm{\,V}$
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