Q107 / 180 · JIPMER 2018 · Physics

Find i in shown figure Figure

  • A
    $0.2 A$
  • Correct answer
    B
    $0.1 A$
  • C
    $0.3 A$
  • D
    $0.4 A$

Solution

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

Hence, the correct answer is option: 'B'.

Relevant Questions