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Two different coils have self-inductance...

Two different coils have self-inductances `L_(1) = 8 mH and L_(2) = 2 mH`. The current in one coil is increased at a constant rate. The current in the second coil is also increased at the same constant rate. At a certain instant of time, the power given to the two coil is the same. At that time, the current, the induced voltage and the energy stored in the first coil are `i_(1), V_(1) and W_(1)` respectively. Corresponding values for the second coil at the same instant are `i_(2), V_(2) and W_(2)` respectively. Then:

A

`(i_(1))/(i_(2)) = (1)/(4)`

B

`(i_(1))/(i_(2)) = 4`

C

`(W_(1))/(W_(2)) = (1)/(4)`

D

`(V_(1))/(V_(2)) = 4`

Text Solution

Verified by Experts

The correct Answer is:
A, C, D
`V_(1) = L_(1) (dI_(1))/(dt)` and `V_(2) = L_(2) (dI_(2))/(dt)`
But `(dI_(1))/(dt) = (dI_(2))/(dt)` (given) `:. (V_(1))/(V_(2)) = (L_(1))/(L_(2)) = (8)/(2) = (4)/(1)`
Again, same power is given to the two coils.
`:. V_(1)I_(1) = V_(2)I_(2)` or `(I_(1))/(I_(2)) = (V_(2))/(V_(1)) = (1)/(4)`
Again, energy `= (1)/(2) LI^(2)`
`:. (W_(2))/(W_(1)) = ((1)/(2) L_(2)I_(2)^(2))/((1)/(2) L_(1)I_(1)^(2)) = ((L_(2))/(L_(1)))((L_(2))/(I_(1)))^(2) = (2)/(8) (4)^(2) = (4)/(1)`
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  • Two different coils have self-inductance L_(1)=8 mH, L_(2)=2 mH . The current in one coil is increased at a constant rate. The current in the second coil is also increased at the same rate. At a certain instant of time, the power given to the two coils is the same. At that time the current, the induced voltage and the energy stored in the first coil are i_(1), V_(1) and W_(1) respectively. Corresponding values for the second coil at the same instant are i_(2), V_(2) and W_(2) respectively. Then

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  • Two different coils have self inductances L_(1)=8mH and L_(2)=2mH . The current in both the coil is increased at same constant rate. At a certain instant power given to two coils is same. At that time the energy stored in both the coils are V_(1) & V_(2) respectively, then (V_(1))/(V_(2)) is

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