Two different coils have self- inductances `L_(1) = 16` mH and `L_(2)= 12`mH. At a certain instant, the current in the two coils is increasing at the same rate and power supplied to the two coils is the same. Find the ratio of i) induced voltage ii) current it) energy stored in the two coils at that instant.

Explanation

i) `V_(1)=L_(1)(dI)/(dt), V_(2)=L_(2)(dI)/(dt)`
`(V_(1))/(V_(2))=(L_(1))/(L_(2))=(16)/(12)=(4)/(3)`
ii) `P=V_(1)I_(1)=V_(2)I_(2) implies (I_(1))/(I_(2))=(V_(1))/(V_(2))=(3)/(4)`
iii) `(U_(1))/(U_(2))=((1)/(2)L_(1)I_(1)^(2))/((1)/(2)L_(2)I_(2)^(2))=((L_(1))/(L_(2)))((I_(1))/(I_(2)))^(2)=(4)/(3)((3)/(4))^(2)=(3)/(4)`