Physics - Current Electricity Question with Solution | TestHub

As$3 \mathrm{\Omega }$and$6 \mathrm{\Omega }$resistances are

In parallel, their equivalent resistance will be 2$\mathrm{\Omega }$. Here 2$\mathrm{\Omega }$and 4$\mathrm{\Omega }$are in series, their equivalent resistance will be 6$\mathrm{\Omega }$. From current distribution law



$i_1=\frac{12\times 18}{18}=\mathrm{12 }\mathrm{A}$

$i_2=\frac{6\times 18}{18}=\mathrm{6 }\mathrm{A}$







Now, 12 A current is entering in the parallel combination of$3 \mathrm{\Omega }$and$\mathrm{6 }\mathrm{\Omega }$again from current distribution law

$i_1^{\prime \prime }=\frac{6\times 12}{9}=\mathrm{8 }\mathrm{A}$

$i_2^{\prime \prime }=\frac{3\times 12}{9}=\mathrm{4 }\mathrm{A}.$

∴ The potential difference across$3 \mathrm{\Omega }$resistance

$=8\times 3=24 \mathrm{V}$