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:

Options

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`

Answer

A, C, D

Explanation

Rate of change of current `=(di_(1))/(dt)=m(say)`
Induced emf `(V_1)=-L(di_(1))/(dt)=-8xx10^(-3)xxm`
`:. (V_2)/(V-1)=1/4`
Power `P=(V_1)(i_1)=8xx10^(_3)xxmxx(i_1)`
Rate of change of current `=(di_(2))/(dt)=m(given`
Induced emf,`(V_2)=-L(di_(2))/(dt)=-2xx10^(-3)xxm`
Power `P=(V_2)(i_2)=2xx10^(_3)xxmxx(i_2)`
Since Power is equal
`:. 8xx(10^(-3))xx(mi_(1))=2xx(10^(-3))xx(mi_(2))`
`implies(i_1)/(i_2)=1/4` ...(i)
Energy `(W_1)=1/2 (L_1)(i_(1)^(2))=1/2xx8xx10^(-3)xx(i_(1)^(2))`
`:. (W_2)/(W_1)=(10^(-3)xx(i_(2)^(2)))/(4xx10^(-3)xx(i_(1)^(2)))=1/4xx4xx4=4`[(from (i))].