A coil of inductance `0.20 H` is connected in series with a switch and a cell of emf `1.6 V`. The total resistance of the circuit is `4.0 Omega`. What is the initial rate of growth of the current when the switch is closed?

Options

A. `0.050 A s^(-1)`
B. `0.40 A s^(-1)`
C. `0.13 A s^(-1)`
D. `8.0 A s^(-1)`

Answer

D

Explanation

`V = RI + L(dI)/(dt)`, at `t =0, I = 0`, thus we have
`(dI)/(dt) = (V)/(L) = (1.6)/(0.2) = 8 A//s`