The coefficients of self induction of two coils are $L_{1} =$ 8 mH and $L_{2} =$ 2 mH respectively. The current rises in the two coils at the same rate. The power given to the two coils at any instant is same. The ratio of energies stored in the coils will be :

\frac{W_{1}}{W_{2}} = 4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{W_{1}}{W_{2}} = \frac{1}{4}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{W_{1}}{W_{2}} = \frac{3}{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{W_{1}}{W_{2}} = \frac{4}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{W_{1}}{W_{2}} = \frac{1}{4}$
Power to 1st coil = $V_{1} i_{1} = L_{1} \frac{d i_{1}}{d t} i_{i}$

Power to 2nd coil = $V_{2} i_{2} = L_{2} \frac{d i_{2}}{d t} i_{2}$

As the powers are same,

$L_{1} \frac{d i_{1}}{d t} i_{i} = L_{2} \frac{d i_{2}}{d t} i_{2}$

rate of change of current is given to be same,
$L_{1} i_{i} = L_{2} i_{2}$

or, $\frac{i_{1}}{i_{2}} = \frac{L_{2}}{L_{1}} = \frac{1}{4}$
Ratio of energy stored = $\frac{W_{1}}{W_{2}} = \frac{L_{1} i_{1}^{2}}{L_{2} i_{2}^{2}} = \frac{L_{1}}{L_{2}} (\frac{i_{1}}{i_{2}})^{2} = 4 \times \frac{1}{4^{2}} = \frac{1}{4}$

Suggest Corrections

Similar questions

Two different coils have self-inductance $L_{1}$ = 8mH and $L_{2}$ = 2mH respectively. The current in both the coils is increased at a same rate. At a certain instant of time, the power being supplied to both the coils is equal. At that instant, the current, the induced voltage, and the energy stored in $L_{1}$ and $L_{2}$ are $I_{1}$ and $I_{2}$ , $V_{1}$ and $V_{2}$ , and $W_{1}$ and $W_{2}$ respectively. Then