The mutual inductance between a primary and secondary circuit is `0.5H`. The resistance of the primary and the secondary circuits are `20 ohms` and `5ohms` respectvely. To genrate a current of `0.4A` in the secondary,current in the primary must be changed at the rate of

Options

A. 4 A/s
B. 2 A/s
C. 0.4 A/s
D. 0.2 A/s

Answer

A

Explanation

`e_(2)=M(dI_(1))/(dt)" "thereforeI_(2)=(e_(2))/(R_(2))=(M)/(R_(2))(dI_(1))/(dt)`
`implies (dI_(1))/(dt)=(I_(2)R_(2))/((M)=4A//s`