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Two different coils have self inductane ...

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

Text Solution

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`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_1I_1=V_2I_2 rArr I_1/I_2=V_1/V_2=3/4`
(iii) `U_1/U_2=(1/2L_1L_1^2)/(1/2L_2I_2^2)=(L_1/L_2)(I_1/I_2)^2=4/3(3/4)^2=3/4`
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