In the circuit shown, the value of l 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 below.

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'=6limpliesl'=(3)/(2)l`
but `i=l+l'=l+(3)/(2)l=(5)/(2)l`
`:." "1=(5)/(2)limpliesl=(2)/(5)=0.4A`