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In the circuit shown, switch's is closed...


In the circuit shown, switch's is closed at t=0. then the ratio `(U_(1))/(U_(2))` of potential energy stored in the inductors `L_(1)` and `L_(2)` in steady state is

A

`(L_(1))/(L_(2))`

B

`(L_(2))/(L_(1))`

C

`((L_(1))/(L_(2)))^(2)`

D

`((L_(2))/(L_(1)))^(2)`

Text Solution

Verified by Experts

The correct Answer is:
B
`L_(1)(di_(1))/(dt)=L_(2)(di_(2))/(dt)` ltbr. `L_(1)underset(0)overset(i_(1))intdi_(1)=L_(2)underset(0)overset(i_(2))intdi_(2)`
`L_(1)i_(1)=L_(2)i_(2)` .(i)
`i_(1)+i_(2)=i` …(ii)
`i_(1)(1+(L_(1))/(L_(2)))=i`
`impliesi_(1)=((L_(2))/(L_(1)+L_(2)))i` and `i_(2)=((L_(1))/(L_(1)+L_(2)))i` ltbr. `therefore(U_(1))/(U_(2))=((1)/(2)L_(1)i_(1)^(2))/((1)/(2)L_(2)i_(2)^(2))=(L_(1)L_(2)^(2))/(L_(2)L_(1)^(2))=(L_(2))/(L_(1))`
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