Two coils have self inductances `L_(1)=4 mH and L_(2)=8mH`. Current in both the coils is increasing at same rate. At an instant, when the power given to the two coils is same, find
(i) Ratio of current in the inductors.
(ii) Ratio of potential difference
(iii) Ratio of energy stored

Explanation

`L_(1)=4mH , L_(2)=8mHrArr (L_(1))/(L_(2))=(1)/(2).`
`"Power is same "rArr V_(i)i_(1)=V_(2)i_(2)", where "V_(1) and V_(2)` are potential difference and `i_(1),i_(2)` are currents.
`V_(1)=L_(1)(di)/(dt),V_(2)=L_(2)(di)/(dt)" "(because (di)/(dt)" is same")`
`rArr" "(V_(1))/(V_(2))=(L_(1))/(L_(2))=(1)/(2)`
(i) `V_(1)i_(1)=V_(2)i_(2)rArr (i_(1))/(i_(2))=(V_(2))/(V_(1))=(L_(2))/(L_(1))=2`
(ii) `(V_(1))/(V_(2))=(1)/(2)`
(iii) Energy stored `(U_(1))/(U_(2))=(L_(1)i_(1)^(2))/(L_(2)i_(2)^(2))=((1)/(2))(2)^(2)=2.`