Here, `L_(1) = 12 mH = 12 xx 10^(-3) H`
`L_(2) = 30 mH = 30 xx 10^(-3) H`
(a) As current is increasing at the same rate
`(dI // dt)`, e.m.f. induced opposes the increase.
`e_(1) = L_(1) (dI)/(dt) = 12 xx 13^(-3) (dI // dt)`
`e_(2) = L_(2) (dI)/(dt) = 30 xx 10^(-3) (dI // dt)`
The variation of e eith `dI / dt` is as shown in fig.
(b) Energy stored `E = (1)/(2) LI^(2)`
For given L, `E prop I^(2)`, therefore, variation of energy stroed with the current is as shown in fig..
Power dissipated,
`P = (dE)/(dt) = (d)/(dt) ((1)/(2) LI^(2))`
`= LI (dI)/(dt) = ((L dI)/(dt)) I = e. I =` const.
As `e_(2) gt e_(1)`
`:. I_(2) lt I_(1)`
`:. (E_(2))/(E_(1)) = ((1)/(2) L_(2) I_(2)^(2))/((1)/(2) L_(1) I_(1)^(2) ) = (L_(2))/(L_(1)) ((I_(2))/(I_(1)))^(2) = (30)/(12 ) ((12)/(30))^(2)`
`= (12)/(30) = (2)/(2) = 0.4`
![image](https://learnqa.s3.ap-south-1.amazonaws.com/images/16109389284519103191610938928.png)