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

Options

A. 1
B. `0.60`
C. 0.4
D. 1.5

Answer

C

Explanation

We can simplify the network as shown in figure. So, net resistance,
`R=2.4+1.6=4.0Omega`
Therefore, current from the battery
`i=(V)/(R)=(4)/(4)=1A`
Now, from the circuit (b),
`4l'=6lrArrl'=(3)/(2)l`

But, `i=l+l'=l+(3)/(2)l=(5)/(2)l`
`therefore 1=(5)/(2)l`
`rArr l=(2)/(5)=0.4A`