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Two different coils have self inductance...

Two different coils have self inductances `L_1=9mH` and `L_2=2mH`. 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 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,

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

Potential difference across an inductor:
`VpropL`, if rate of change of current is constant `(V=-L(di)/(dt))`
`:. V_2/V_1=L_2/L_1=2/8=1/4`
or `V_1/V_2=4`
Power given to the two coils is same i.e.
`V_1 i_1 =V_2i_2`
or `i_1/i_2=V_2/V_1=1/4`
Energy stored `W=1/2Li^2`
`:. W_2/W_1=(L_2/L_1)(i_2/i_1)^2=(1/4)(4)^2`
or `W_1/W_2=1/4`
`:.` The correct options are a, c and d.
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Knowledge Check

  • 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:

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