Two coils have self-inductance $L_{1} = 4$ mH and $L_{2} = 1$ mH respectively. The currents in the two coils are increased at
the same rate. At a certain instant of time both coils are given the same power. If $L_{1}$ and $L_{2}$ are the currents in the
two coils, at that instant of time respectively, then the value of $\frac{I_{1}}{I_{2}}$ is:

\frac{1}{8}

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\frac{1}{4}

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\frac{1}{2}

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Solution

The correct option is B $\frac{1}{4}$
$P_{1} = \frac{d}{d t} (\frac{1}{2} l_{1} l_{1}^{2}) = L_{1} I_{1} \frac{d I_{1}}{d t} P_{2} = \frac{d}{d t} (\frac{1}{2} L_{2} I_{2}^{2}) = L_{2} I_{2} \frac{d I_{2}}{d t} ∴ L_{1} I_{1} \frac{d I_{1}}{d t} = L_{2} I_{2} \frac{d I_{2}}{d t} \Rightarrow \frac{I_{1}}{I_{4}} = \frac{1}{4}$

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Similar questions

Q. Two different coils have self-inductance

L_{1} = 8 mH

and

L_{2} = 2 mH

. The currents in both the coils is increased at a same constant rate. At a certain instant of time, the power given to both the coils is the same. At that time the current and the induced voltage in the first coil are,

i_{1}

and

V_{1}

respectively. Corresponding values for the second coil, at the same instant are,

i_{2}

and

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respectively. Then -

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

Q. Two coils have self-inductances

L_{1} = 8 m H

and

L_{2} = 4 m H

. 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 current, the induced voltage and the energy stored in the first coil are

i_{1}, V_{1}

and

W_{1}

. Corresponding values for the second coil are

i_{2}, V_{2}

and

W_{2}

respectively. Then:
(IIT-JEE 1994)

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The correct option is B 14P1=ddt(12l1l21)=L1I1dI1dtP2=ddt(12L2I22)=L2I2dI2dt∴L1I1dI1dt=L2I2dI2dt⇒I1I4=14