Question

The current through two inductors of self inductance 12 mH and 30 mH is increasing with time at the same rate. Draw graphs showing the variation of the (a ) e.m.f. induced with the rate of change of current in each inductor. (b) enargy stored in each inductor with the current flowing through it.
Compare the energy stored in the coils if power dissipated in the coil is same.

Answer 1

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`