Physics - Current Electricity Question with Solution | TestHub

$\because 60\Omega$ and $30\Omega$ resistor are connected in parallel. So, their net resistance $\begin{array}{rl}& {\mathrm{R}}_{\text{net }}=\frac{30\times 60}{30+60}=\frac{1800}{90}=20\Omega \\ & \therefore \mathrm{i}=\frac{2}{{\mathrm{R}}_{\text{net }}}=\frac{2}{20}=0.1\mathrm{A}\\ & \end{array}$