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A coil of inductance 0.20 H is connected...

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?

A

`0.050 As^(-1)`

B

`0.40 As^(-1)`

C

`0.13 As^(-1)`

D

`8.0 As^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
D

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