Alternating current of peak value `((2)/(pi))` ampere flows through the primary coil of the transformer. The coefficient of mutual inductance between primary and secondary coil is 1 henry. The peak e.m.f. induced in secondary coil is (Frequency of AC= 50 Hz)

Options

A. 100 V
B. 200 V
C. 300 V
D. 400 V

Explanation

To find the peak e.m.f. induced in the secondary coil of the transformer, we can follow these steps:

### Step 1: Identify the given values
- Peak current \( I_0 = \frac{2}{\pi} \) A
- Mutual inductance \( M = 1 \) H
- Frequency \( f = 50 \) Hz

### Step 2: Write the expression for the alternating current
The alternating current can be expressed as:
\[ i(t) = I_0 \sin(\omega t) \]
where \( \omega = 2\pi f \).

### Step 3: Calculate the angular frequency \( \omega \)
Using the frequency given:
\[ \omega = 2\pi f = 2\pi \times 50 = 100\pi \text{ rad/s} \]

### Step 4: Differentiate the current with respect to time
To find the induced e.m.f. \( e \), we use the formula:
\[ e = -M \frac{di}{dt} \]
Differentiating \( i(t) \):
\[ \frac{di}{dt} = I_0 \omega \cos(\omega t) \]
Substituting \( I_0 \) and \( \omega \):
\[ \frac{di}{dt} = \frac{2}{\pi} \cdot 100\pi \cos(100\pi t) = 200 \cos(100\pi t) \]

### Step 5: Calculate the peak e.m.f.
The peak e.m.f. is given by:
\[ e_{\text{peak}} = M \cdot I_0 \cdot \omega \]
Substituting the values:
\[ e_{\text{peak}} = 1 \cdot \frac{2}{\pi} \cdot 100\pi = 200 \text{ V} \]

### Final Answer
The peak e.m.f. induced in the secondary coil is \( 200 \) volts.

---