In the circuit shown the value of $I$ in ampere is

• A

  $1$
• B

  $0.60$
• Correct answer
  C

  $0.4$
• D

  $1.5$

Solution

So net resistance, $R=2.4+1.6=4.0 \Omega$ Therefore, current from the battery, $i=\frac{V}{R}=\frac{4}{4}=1 \mathrm{~A}$ Now from the circuit (b), $\begin{aligned} & 4 I^{\prime}=6 I \\ \Rightarrow \quad & I^{\prime}=\frac{3}{2} I\end{aligned}$ But, $\quad i=I+I^{\prime}=I+\frac{3}{2} I=\frac{5}{2} I$ $\begin{array}{lc}\therefore & 1=\frac{5}{2} I \\ \Rightarrow & I=\frac{2}{5}=0.4 \mathrm{~A}\end{array}$

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