Question

X and Y are two circuits having coefficient of mutual inductance 3 mH and resistances 10 \( \Omega \) and 4 \( \Omega \) respectively. To have induced current 60 \( \times 10^{-4} \, {A} \) in circuit Y, the amount of current to be changed in circuit X in 0.02 sec is:

• A. \( 1.6 \, {A} \)
• B. \( 0.16 \, {A} \)
• C. \( 0.32 \, {A} \)
• D. \( 3.2 \, {A} \)

Hint:

To find the amount of current changed in one circuit based on mutual inductance, use the relation \( I_Y = \frac{M dI_X}{R_Y} \) and apply Ohm’s law.

Answer 1

The Correct Option is B

We are given two circuits, X and Y. The coefficient of mutual inductance \( M = 3 \, {mH} = 3 \times 10^{-3} \, {H} \), resistances \( R_X = 10 \, \Omega \) and \( R_Y = 4 \, \Omega \), and the induced current in circuit Y is \( I_Y = 60 \times 10^{-4} \, {A} \). We need to find the amount of current to be changed in circuit X in 0.02 sec. The mutual inductance relationship is given by:

\( {Induced EMF in circuit Y} = M \frac{dI_X}{dt}. \)

The induced current \( I_Y \) in circuit Y is related to the induced EMF by Ohm's law:

\( I_Y = \frac{{Induced EMF in circuit Y}}{R_Y}. \)

Thus, we have:

\( I_Y = \frac{M \frac{dI_X}{dt}}{R_Y}. \)

Step 1:
Substitute the given values into the equation:

\( 60 \times 10^{-4} = \frac{(3 \times 10^{-3}) \frac{dI_X}{dt}}{4}. \)

Step 2:
Solve for \( \frac{dI_X}{dt} \):

\( \frac{dI_X}{dt} = \frac{60 \times 10^{-4} \times 4}{3 \times 10^{-3}} = \frac{240 \times 10^{-4}}{3 \times 10^{-3}} = 0.08 \, {A/sec}. \)

Step 3:
The amount of current changed in 0.02 sec is:

\( \Delta I_X = \frac{dI_X}{dt} \times 0.02 = 0.08 \times 0.02 = 0.16 \, {A}. \)

Thus, the amount of current to be changed in circuit X in 0.02 sec is \( 0.16 \, {A} \).