In the adjoining circuit shown the potential difference between points $A$ and $B$ will be
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Current $\mathrm{I}$ through $\mathrm{CBD}=(2 / 15)$ $\mathrm{amp}$
Current $\mathrm{I}$ through $\mathrm{CDA}=(2 / 15)$ $\mathrm{amp}$
$\mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{B}} =\frac{2}{15} \times 10 \mathrm{\,volt} $
$\mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{A}} =\frac{2}{15} \times 5 \mathrm{\,volt} $
$\therefore \mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}} =\left(\mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{B}}\right)-\left(\mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{A}}\right) $
$=\frac{2}{15}[10-5]=\left(\frac{2}{3}\right) \mathrm\,{volt}$
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