Question

Two different coils have self-inductance L₁ = 9 mH and L₂ = 3 mH. The current in first coil is increased at a constant rate. The current in the second coil is also increased at the same constant rate. At certain instant of time, the power given to the two coils is same. At that time, there was current and induced voltage in the two coils. At the same instant, the ratio of the energy stored in the first coil to that in second coil is ______.

पर्याय

• 1 : 3

• 3 : 1

• 1 : 9

• 9 : 1

Answer 1

Two different coils have self-inductance L₁ = 9 mH and L₂ = 3 mH. The current in first coil is increased at a constant rate. The current in the second coil is also increased at the same constant rate. At certain instant of time, the power given to the two coils is same. At that time, there was current and induced voltage in the two coils. At the same instant, the ratio of the energy stored in the first coil to that in second coil is 1 : 3.

Explanation:

For 1^st coil:

Induced e.m.f. (V₁) = `-L_1 (di)/dt`

= `-9 xx 10^-3 (di)/dt`    ...(i)

Power (P₁) = V₁i₁

= `-9 xx 10^-3 (di_1)/dt xx i_1`    ...(ii)

Energy (E₁) = `1/2 L_1I_1^2`

= `1/2 xx 9 xx 10^-3 xx i_1^2`    ...(iii)

For 2^nd coil,

Induced e.m.f. (V₂) = `-L_2 (di_2)/dt`

= `-L_2 (di_1)/dt`    ...[Given: `(di_1)/dt = (di_2)/dt`]

= `-3 xx 10^-3 (di_1)/dt`    ...(iv)

power (P₂) = V₂ i₂

= `-3 xx 10^-3 (di_1)/dt xx i_2`    ...(v)

Energy (E₂) = `1/2 L_2 i_2^2`

= `1/2 xx 3 xx 10^-3 xx i_2^2`    ...(vi)

From (i) and (iv),

`V_1/V_2 = 3/1`    ...(vii)

Since the power in both coils is the same at an instant,

P₁ = P₂

V₁i₁ = V₂i₂    ...[from equations (ii) and (v)]

∴ `i_1/i_2 = V_2/V_1 = 1/3`    ...[[from equation (vii)]   ...(viii)

∴ `E_1/E_2 = (1/2 xx 9 xx 10^-3 xx i_2^2)/(1/2 xx 3 xx 10^-3 xx i_2^2)`    ...[from equations (iii) and (vi)]

= `3 xx (i_1/i_2)^2`

= `3 xx (1/3)^2`    ...[(from equation (viii)]

= `1/3`