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Question: Two different coils have self inductance \[{L_1} = 8mH\] , \({L_2} = 2mH\). The current in the secon...

Two different coils have self inductance L1=8mH{L_1} = 8mH , L2=2mH{L_2} = 2mH. 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 i1{i_1} , V1{V_1} and W1{W_1} ​ respectively. Corresponding values for the second coil at the same instant are i2{i_2} , V2{V_2} and W2{W_2} ​ respectively. Then
A.i1i2=14\dfrac{{{i_1}}}{{{i_2}}} = \dfrac{1}{4}
B.i1i2=4\dfrac{{{i_1}}}{{{i_2}}} = 4
C.W2W1=4\dfrac{{{W_2}}}{{{W_1}}} = 4
D.V2V1=14\dfrac{{{V_2}}}{{{V_1}}} = \dfrac{1}{4}

Explanation

Solution

This question has multiple correct answers. First find the voltage induced in both the coils, with that find the power in both the coils. Take the ratio of their power and it will give us the ratio between the currents in the coils. Then find the ratio of their voltages and finally use the ratio of current to find the ratio of the energy stored in the coils.

Complete answer:
We have been given self inductance for the first coil L1=8mH=8×10−3H{L_1} = 8mH = 8 \times {10^{ - 3}}H and self inductance for second coil L2=2mH=2×10−3H{L_2} = 2mH = 2 \times {10^{ - 3}}H
For first coil let the rate of change of current i1{i_1} will be mm that is
di1dt=m\dfrac{{d{i_1}}}{{dt}} = m
We know the power in a circuit is the product of current and voltage in the circuit
P=VIP = VI
For first coil as voltage of an inductor is given as
V1=−L1di1dt{V_1} = - {L_1}\dfrac{{d{i_1}}}{{dt}}
⇒P1=V1i1\Rightarrow {P_1} = {V_1}{i_1}
⇒P1=−8×10−3×m×i1\Rightarrow {P_1} = - 8 \times {10^{ - 3}} \times m \times {i_1}
Similarly, we will find the rate of change of current for i2{i_2} . as The current in the second coil is also increased at the same constant rate.
Therefore, di2dt=m\dfrac{{d{i_2}}}{{dt}} = m
So V2=−L2di2dt{V_2} = - {L_2}\dfrac{{d{i_2}}}{{dt}}
⇒P2=V2i2\Rightarrow {P_2} = {V_2}{i_2}
⇒P2=−2×10−3×m×i2\Rightarrow {P_2} = - 2 \times {10^{ - 3}} \times m \times {i_2}
Now, P1P2=−8×10−3×m×i1−2×10−3×m×i2\dfrac{{{P_1}}}{{{P_2}}} = \dfrac{{ - 8 \times {{10}^{ - 3}} \times m \times {i_1}}}{{ - 2 \times {{10}^{ - 3}} \times m \times {i_2}}}
As the power given to the two coil is the same, therefore P1=P2{P_1} = {P_2} so we get,
i1i2=14\dfrac{{{i_1}}}{{{i_2}}} = \dfrac{1}{4}
Hence, option A) is the correct answer.
Now as we found out our voltage for both the coil above so,
V1V2=−8×10−3×m−2×10−3×m\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{ - 8 \times {{10}^{ - 3}} \times m}}{{ - 2 \times {{10}^{ - 3}} \times m}}
⇒V1V2=4\Rightarrow \dfrac{{{V_1}}}{{{V_2}}} = 4
Hence option D) is also correct.
Now finding the energy stored in an inductor which is given as W=12×L×I2W = \dfrac{1}{2} \times L \times {I^2}
Therefore, W1W2=12×8×10−3×i1212×2×10−3×i2\dfrac{{{W_1}}}{{{W_2}}} = \dfrac{{\dfrac{1}{2} \times 8 \times {{10}^{ - 3}} \times {i_1}^2}}{{\dfrac{1}{2} \times 2 \times {{10}^{ - 3}} \times {i_2}}}
⇒W1W2=4(i1i2)2\Rightarrow \dfrac{{{W_1}}}{{{W_2}}} = 4{\left( {\dfrac{{{i_1}}}{{{i_2}}}} \right)^2}
We have found the ratio of current for both the coils above so,
⇒W1W2=4×(14)2\Rightarrow \dfrac{{{W_1}}}{{{W_2}}} = 4 \times {\left( {\dfrac{1}{4}} \right)^2}
⇒W1W2=14\Rightarrow \dfrac{{{W_1}}}{{{W_2}}} = \dfrac{1}{4}
⇒W2W1=4\Rightarrow \dfrac{{{W_2}}}{{{W_1}}} = 4

Hence option C) and D) is also correct.

Note:
An inductor is a coil with a defined number of turns made up of an insulated wire wound over an air or iron core. When current travels through the coil, it stores energy in the magnetic field. Inductance is a quality that opposes any change in the quantity of current flowing through the coil.