Two circuits have coefficient of mutual induction of `0.09` henry. Average e.m.f. induced in the secondary by a change of current from `0` to `20` ampere in `0.006` second in the primary will be

Options

A. 120 V
B. 80 V
C. 200 v
D. 300 V

Explanation

To find the average e.m.f. induced in the secondary circuit due to a change in current in the primary circuit, we can use the formula for mutual induction. The average e.m.f. (ε) induced in the secondary can be calculated using the formula:

\[ \varepsilon = -M \frac{\Delta I}{\Delta t} \]

Where:
- \( \varepsilon \) = average e.m.f. induced in the secondary (in volts)
- \( M \) = mutual inductance (in henries)
- \( \Delta I \) = change in current in the primary (in amperes)
- \( \Delta t \) = time interval for the change in current (in seconds)

### Step 1: Identify the given values
- Mutual inductance \( M = 0.09 \, \text{H} \)
- Initial current \( I_1 = 0 \, \text{A} \)
- Final current \( I_2 = 20 \, \text{A} \)
- Time interval \( \Delta t = 0.006 \, \text{s} \)

### Step 2: Calculate the change in current \( \Delta I \)
\[ \Delta I = I_2 - I_1 = 20 \, \text{A} - 0 \, \text{A} = 20 \, \text{A} \]

### Step 3: Calculate the rate of change of current
\[ \frac{\Delta I}{\Delta t} = \frac{20 \, \text{A}}{0.006 \, \text{s}} = \frac{20}{0.006} \approx 3333.33 \, \text{A/s} \]

### Step 4: Substitute the values into the e.m.f. formula
\[ \varepsilon = -M \frac{\Delta I}{\Delta t} = -0.09 \, \text{H} \times 3333.33 \, \text{A/s} \]

### Step 5: Calculate the average e.m.f.
\[ \varepsilon = -0.09 \times 3333.33 \approx -300 \, \text{V} \]

Since we are interested in the magnitude of the e.m.f., we can ignore the negative sign:
\[ \varepsilon \approx 300 \, \text{V} \]

### Conclusion
The average e.m.f. induced in the secondary circuit is approximately **300 volts**.

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