An a.c . Of 50 Hz and 1A peak value flows in primary coil transformer whose mutual inductance is 1.5 H. Then peak value of induced emf in secondary is :-

Options

A. 150 V
B. `150 pi V`
C. `300 V`
D. `200 V`

Answer

To find the peak value of the induced EMF in the secondary coil of a transformer, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Frequency (f) = 50 Hz
- Peak current (I₀) = 1 A
- Mutual inductance (M) = 1.5 H

2. **Calculate Angular Frequency (ω)**:
The angular frequency (ω) is given by the formula:
\[ \omega = 2\pi f \]
Substituting the given frequency:
\[ \omega = 2\pi \times 50 = 100\pi \, \text{rad/s} \]

3. **Determine the Rate of Change of Current**:
The current in the primary coil can be expressed as:
\[ I(t) = I₀ \sin(\omega t) \]
To find the induced EMF, we need the derivative of the current with respect to time:
\[ \frac{dI}{dt} = I₀ \omega \cos(\omega t) \]

4. **Calculate the Induced EMF (E)**:
The induced EMF in the secondary coil is given by:
\[ E = -M \frac{dI}{dt} \]
Substituting the expression for \(\frac{dI}{dt}\):
\[ E = -M I₀ \omega \cos(\omega t) \]
The peak value of the induced EMF occurs when \(\cos(\omega t) = 1\):
\[ E_{\text{peak}} = M I₀ \omega \]

5. **Substitute Values to Find Peak EMF**:
Now substituting the known values:
\[ E_{\text{peak}} = 1.5 \times 1 \times 100\pi \]
\[ E_{\text{peak}} = 150\pi \, \text{V} \]

### Final Answer:
The peak value of the induced EMF in the secondary coil is:
\[ E_{\text{peak}} = 150\pi \, \text{V} \]