When a current of 4 A through primary give rise to a flux of magnitude 1.35 Wb through secondary. What is the coefficient off mutual induction (M)?

Options

A. 2.96 H
B. 5.4 H
C. 0.34 H
D. Zero

Answer

To find the coefficient of mutual induction (M) when a current of 4 A through the primary coil gives rise to a flux of 1.35 Wb through the secondary coil, we can use the formula for induced electromotive force (emf) in the secondary coil:

1. **Understanding the relationship**: The induced emf (E) in the secondary coil is related to the mutual inductance (M) and the rate of change of current (di/dt) in the primary coil. The formula is given by:
\[ E = M \frac{di}{dt} \]

2. **Identifying the values**: From the question, we know:
- E = 1.35 Wb (the induced emf in the secondary)
- di/dt = 4 A (the change in current through the primary)

3. **Rearranging the formula**: We need to solve for M, so we rearrange the formula:
\[ M = \frac{E}{\frac{di}{dt}} \]

4. **Substituting the values**: Now we can substitute the known values into the equation:
\[ M = \frac{1.35 \, \text{Wb}}{4 \, \text{A}} \]

5. **Calculating M**: Performing the division gives:
\[ M = 0.3375 \, \text{H} \]

6. **Rounding off**: The value can be rounded to two decimal places:
\[ M \approx 0.34 \, \text{H} \]

Thus, the coefficient of mutual induction (M) is approximately **0.34 Henry**.