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

To solve the problem of determining the amount of current change required in circuit X to induce a specific current in circuit Y, we will use the formula for mutual induction:

\[ M = \frac{E}{-\frac{dI}{dt}} \]

where \( M \) is the mutual inductance, \( E \) is the induced EMF in circuit Y, and \( \frac{dI}{dt} \) is the rate of change of current in circuit X.

The induced EMF \( E \) in circuit Y causes a current \( I = 60 \times 10^{-4} \, \text{A} \) to flow through Y with a resistance of \( 4 \, \Omega \). Therefore, using Ohm's Law:

\[ E = I \times R \] \[ E = (60 \times 10^{-4}) \times 4 = 24 \times 10^{-4} \, \text{V} \]

Given that the mutual inductance \( M = 3 \, \text{mH} = 3 \times 10^{-3} \, \text{H} \), we plug these values into the induction formula:

\[ M = \frac{24 \times 10^{-4}}{-\frac{dI_x}{dt}} \] \[ 3 \times 10^{-3} = \frac{24 \times 10^{-4}}{-\frac{dI_x}{dt}} \]

Rearranging for \( \frac{dI_x}{dt} \), we have:

\[ \frac{dI_x}{dt} = -\frac{24 \times 10^{-4}}{3 \times 10^{-3}} \] \[ \frac{dI_x}{dt} = -8 \, \text{A/s} \]

The current change over the time interval \( \Delta t = 0.02 \, \text{s} \) is:

\[ \Delta I_x = \frac{dI_x}{dt} \times \Delta t \] \[ \Delta I_x = -8 \times 0.02 = -0.16 \, \text{A} \]

Since we are interested in magnitude, \( \Delta I_x = 0.16 \, \text{A} \). Thus, the amount of current change required in circuit X is \( 0.16 \, \text{A} \).

Answer 2

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 \( 0.16 \, {A} \).