b $\mathrm{R}_{\mathrm{eq}}=\frac{2}{3}+2=\frac{8}{3}, \quad \mathrm{V}=2 \mathrm{\,V} \quad \therefore \mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}=\frac{3}{4} \mathrm{\,A}$
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The four arms of a Wheatstone bridge have resistances as shown in the figure. A galvanometer of $15\, \Omega$ resistance is connected across $BD$. Calculate the current through the galvanometer when a potential difference of $10\, V$ is maintained across $AC.$
Current density in a cylindrical wire of radius $R$ is given as $J =$ $\left\{ {\begin{array}{*{20}{c}}
{{J_0}\left( {\frac{x}{R} - 1} \right)\,\,for\,\,0 \leqslant x < \frac{R}{2}} \\
{{J_0}\frac{x}{R}\,\,\,\,for\,\,\,\frac{R}{2} \leqslant x \leqslant R}
\end{array}} \right.$The current flowing in the wire is: