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

So, net resistance,
`R = 2.4 + 1.6 = 4.0 Omega`
Therefore, current from the battery
`i = V/R = 4/4 = 1A`
Now, from the circuit (b)
`4I. = 6I`
`implies I. = 3/2 I`
But `i = I + I. - I + 3/2 I = 5/2 I`
`therefore I = 5/2 I`
`implies I = 2/5 = 0.4 A`