Find current through the branch $ad$ of the circuit shown in figure. :-............. $A$
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This is a balanced wheatstone bridge circuit. Therefore, we can drop $10 \,\Omega$ resistance across points band d for the further analysis of the bridge. In This case, combination of resistances in upper and lower branches are in series with equivalent resistances $15 \,\Omega$ and $9 \,\Omega$ respectively. These equivalent resistances are connected in parallel across junction $\mathrm{a}$ and $\mathrm{c}.$ Thus equivalent resistance of the circuit $-$
$\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{15}+\frac{1}{9}$ or $\quad \mathrm{R}_{\mathrm{eq}}=\frac{(15)(9)}{15+9}=\frac{135}{24}=5.62\, \Omega$
Current through circuit i $=\frac{10}{5.62}=1.78 \mathrm{\,A}$
If current through upper and lower branches of the network are $\mathrm{i}_{1}$ and $\mathrm{i}_{2}$, then according to Kirchoff's function law $-$
$\mathrm{i}_{1}+\mathrm{i}_{2}=1.78 \mathrm{\,A}$
and $\left(\mathrm{V}_{\mathrm{a}}-\mathrm{V}_{\mathrm{b}}\right)+\left(\mathrm{V}_{\mathrm{b}}-\mathrm{V}_{\mathrm{c}}\right)=\left(\mathrm{V}_{\mathrm{a}}-\mathrm{V}_{\mathrm{d}}\right)+\left(\mathrm{V}_{\mathrm{d}}-\mathrm{V}_{\mathrm{c}}\right)$
or $\quad \mathrm{i}_{1}(5+10)^{3 / 4}\, \mathrm{i}_{2}(3.0+6.0)$ or $(15) \mathrm{i}_{1}=(9.0) \mathrm{i}_{2}$
On solving, we get $-$ $\mathrm{i}_{1}=0.67 \mathrm{\,A}$ and $\mathrm{i}_{2}=1.1 \mathrm{\,A}$
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