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Two inductors `L_(1)` and `L_(2)` are connected in parallel and a time varying current flows as shown.
the ratio of current `i-(1)//i_(2)`
image
A. `L_(1)//L_(2)`
B. `L_(2)//L_(1)`
C. `(L_(1)^(2))/((L_(1)+L_(2)^(2))`
D. `(L_(2)^(2))/((L_(1)+L_(2)^(2))`

1 Answer

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Best answer
Correct Answer - B
The inductors are in parallel. Therefore, potential difference across them is same,
hence, `V_(1)=V_(2)`
or `L_(1)((di_(1))/(dt))=L_(2)((di_(2))/(dt))`
or `L_(1)(di_(1))=L_(2)(di_(2))`
integrating, we get
`L_(1)i_(1)=L_(2)i_(2)`
or `(i_(1))/(i_(2)=(L_(2))/(L_(1))`

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