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Two inductors L(1) and L(2) are connecte...

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)`

A

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

B

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

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
A
As the inductors are in parallel, induced emf across the two inductors is the same, i.e,
`e_(1)=e_(2)`
`L_(1) ((di_(1))/(dt))= L_(2) ((di_(2))/(dt))`
On integrating both sides, we get
`L_(1) int (di_(1))/(dt) = L_(2) int (di_(2))/(dt)`
`L_(1)i_(1)= L_(2)i_(2)`
`rArr (i_(1))/(i_(2))= (L_(2))/(L_(1))`
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Knowledge Check

  • In the circuit as shown in figure, the ratio of current i_1//i_2 is

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    `4//3`
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    `1//2`
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    D
    none of these

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    `i_(1)=1.5A, i_(2)=0.5A`
    B
    `i_(1)=0.5A, i_(2)=1.5A`
    C
    `i_(1)=1A. i_(2)=3A`
    D
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    A
    `3A,(-3A)/(2)A,(9)/(2)A`
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    D
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