A 10 V battery connected to `5omega` resistance coil having inductance 10 H through a switch drives a constant current in the circuit. The switch is suddenly opened and the time taken to open it is 2ms. The average emf induced across the coil is

Options

A. `4xx10^(4)V`
B. `2xx10^(4)V`
C. `2xx10^(2)V`
D. `1xx10^(4)V`

Answer

D

Explanation

Here, curernt `I=(V)/(R)=(10)/(5)=2A`
Final current, when the swithc is opened is zero.
`:. (dI)/(dt)=(0-2)/(2xx10^(-3))=-1xx10^(-3) A s^(-1)`
As `epsi=-L(dI)/(dt)=-10(-1xx10^(-3))=10^(4) V`