Question

Two different coils have self inductance \[{L_1} = 8mH\] , ${L_2} = 2mH$. 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 ${i_1}$ , ${V_1}$ and ${W_1}$ ​ respectively. Corresponding values for the second coil at the same instant are ${i_2}$ , ${V_2}$ and ${W_2}$ ​ respectively. Then
A.$\dfrac{{{i_1}}}{{{i_2}}} = \dfrac{1}{4}$
B.$\dfrac{{{i_1}}}{{{i_2}}} = 4$
C.$\dfrac{{{W_2}}}{{{W_1}}} = 4$
D.$\dfrac{{{V_2}}}{{{V_1}}} = \dfrac{1}{4}$

Answer 1

Hint: 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 \[{L_1} = 8mH = 8 \times {10^{ - 3}}H\] and self inductance for second coil ${L_2} = 2mH = 2 \times {10^{ - 3}}H$
For first coil let the rate of change of current ${i_1}$ will be $m$ that is
$\dfrac{{d{i_1}}}{{dt}} = m$
We know the power in a circuit is the product of current and voltage in the circuit
$P = VI$
For first coil as voltage of an inductor is given as
${V_1} = - {L_1}\dfrac{{d{i_1}}}{{dt}}$
$ \Rightarrow {P_1} = {V_1}{i_1}$
$ \Rightarrow {P_1} = - 8 \times {10^{ - 3}} \times m \times {i_1}$
Similarly, we will find the rate of change of current for ${i_2}$ . as The current in the second coil is also increased at the same constant rate.
Therefore, $\dfrac{{d{i_2}}}{{dt}} = m$
So ${V_2} = - {L_2}\dfrac{{d{i_2}}}{{dt}}$
$ \Rightarrow {P_2} = {V_2}{i_2}$
$ \Rightarrow {P_2} = - 2 \times {10^{ - 3}} \times m \times {i_2}$
Now, $\dfrac{{{P_1}}}{{{P_2}}} = \dfrac{{ - 8 \times {{10}^{ - 3}} \times m \times {i_1}}}{{ - 2 \times {{10}^{ - 3}} \times m \times {i_2}}}$
As the power given to the two coil is the same, therefore ${P_1} = {P_2}$ so we get,
$\dfrac{{{i_1}}}{{{i_2}}} = \dfrac{1}{4}$
Hence, option A) is the correct answer.
Now as we found out our voltage for both the coil above so,
$\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{ - 8 \times {{10}^{ - 3}} \times m}}{{ - 2 \times {{10}^{ - 3}} \times m}}$
$ \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 $W = \dfrac{1}{2} \times L \times {I^2}$
Therefore, $\dfrac{{{W_1}}}{{{W_2}}} = \dfrac{{\dfrac{1}{2} \times 8 \times {{10}^{ - 3}} \times {i_1}^2}}{{\dfrac{1}{2} \times 2 \times {{10}^{ - 3}} \times {i_2}}}$
$ \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,
$ \Rightarrow \dfrac{{{W_1}}}{{{W_2}}} = 4 \times {\left( {\dfrac{1}{4}} \right)^2}$
$ \Rightarrow \dfrac{{{W_1}}}{{{W_2}}} = \dfrac{1}{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.