Four resistances are connected by an ideal battery of emf $15 \mathrm{V}$, the circuit is in steady-state then the current (in ampere) in wire $\mathrm{AB}$ is:

Solution

$R_{eq}=\frac{3}{4}+\frac{8}{6}=\frac{25}{12}$

$\Rightarrow i_0=\frac{V}{R_{eq}}=24\mathrm{ }\mathrm{A}$

$i_1=\frac{3}{4}\times 24=18\mathrm{ }\mathrm{A}$

$i_2=\frac{4}{6}\times 24=16 \mathrm{A}$

Current in the branch AB

$∆i=2\mathrm{ }\mathrm{A}$

Hence, the correct answer is 2.