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Question: Two coils have self-inductance \({L_1} = 4mH\) and \({L_2} = 1mH\) respectively. The currents in the...

Two coils have self-inductance L1=4mH{L_1} = 4mH and L2=1mH{L_2} = 1mH 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 i1{i_1} and i2{i_2} are currents in the two coils, at that instant of time respectively, then the value of i1i2\dfrac{{{i_1}}}{{{i_2}}} is?
A) 18\dfrac{1}{8}
B) 14\dfrac{1}{4}
C) 12\dfrac{1}{2}
D) 11

Explanation

Solution

Inductors resist change in current. If current increases, then the inductor resists an increase in current, and if the current decreases, the inductor resists decrease in current. Self-inductance occurs when current is induced in a wire or coil due to the current already passing through it by applying voltage by some external factor.

Complete step by step solution:
In a current-carrying wire, a magnetic field develops in the plane perpendicular to the wire. Change in current causes changes in the magnetic field. Conversely, a changing magnetic field also induces a current.
Moreover, a change in the current of the wire or coil causes the induction of a voltage. It appears like some potential difference is applied across the wire so that the current can travel. This induced voltage is called the electromotive force, which is in the opposite direction to the actual applied voltage on the wire. This is in accordance with Lenz’s law.
The rate of change of current is given by emf(or V)=Ldidtemf(or{\text{ }}V) = L\dfrac{{di}}{{dt}}
Here,
V=V = induced electromotive force or voltage
L=L = inductance
didt=\dfrac{{di}}{{dt}} = Rate of change of current in the coil
V1=L1di1dt{V_1} = {L_1}\dfrac{{d{i_1}}}{{dt}} and V2=L2di2dt{V_2} = {L_2}\dfrac{{d{i_2}}}{{dt}}………………..equation(1)
Since equal power is given to both coils,
i1V1=i2V2\Rightarrow {i_1}{V_1} = {i_2}{V_2}
i1i2=V2V1\Rightarrow \dfrac{{{i_1}}}{{{i_2}}} = \dfrac{{{V_2}}}{{{V_1}}}
From equation(1), we get
i1i2=L2di2dt×1L1di1dt\Rightarrow \dfrac{{{i_1}}}{{{i_2}}} = {L_2}\dfrac{{d{i_2}}}{{dt}} \times \dfrac{1}{{{L_1}\dfrac{{d{i_1}}}{{dt}}}}
It is given in the question that the current in both coils is increased at the same rate.
di1dt=di2dt\therefore \dfrac{{d{i_1}}}{{dt}} = \dfrac{{d{i_2}}}{{dt}}
i1i2=L2L1\Rightarrow \dfrac{{{i_1}}}{{{i_2}}} = \dfrac{{{L_2}}}{{{L_1}}}
Since L2L1=14\dfrac{{{L_2}}}{{{L_1}}} = \dfrac{1}{4}, we have-
i1i2=14\Rightarrow \dfrac{{{i_1}}}{{{i_2}}} = \dfrac{1}{4}

The correct answer is [B], 14\dfrac{1}{4}.

Note: Inductors are analogous to capacitors. Capacitors store energy in an electric field and resist a change in voltage, whereas inductors store energy in the magnetic field and resist a change in current. The S.I. the unit of inductance is Henry, and that of capacitance is Farad.