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In the circuit shown the value of $I$ in ampere is
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Verified Answer
The correct answer is:
$0.4$
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, $\quad i=I+I^{\prime}=I+\frac{3}{2} I=\frac{5}{2} I$
$\begin{array}{lc}\therefore & 1=\frac{5}{2} I \\ \Rightarrow & I=\frac{2}{5}=0.4 \mathrm{~A}\end{array}$
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