A 50 Hz ac current of peak value 2 A flows through one of the pair of coils. If the mutual inductance between the pair of coils is 150 mH, then the peak value of voltage induced in the second coil is

Options

A. `30piV`
B. `60piV`
C. `15piV`
D. `300piV`

Explanation

To find the peak value of the voltage induced in the second coil due to the alternating current flowing through the first coil, we can use the formula for induced electromotive force (EMF) in a coil due to mutual inductance. The formula is given by:

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

Where:
- \(E\) is the induced EMF (voltage) in the second coil,
- \(M\) is the mutual inductance,
- \(\frac{di}{dt}\) is the rate of change of current in the first coil.

### Step 1: Determine the peak current and its time derivative
Given that the current \(i\) in the first coil is alternating, it can be expressed as:

\[ i(t) = I_0 \sin(\omega t) \]

Where:
- \(I_0\) is the peak current (2 A),
- \(\omega\) is the angular frequency, which can be calculated as \(\omega = 2\pi f\) with \(f = 50 \text{ Hz}\).

Calculating \(\omega\):

\[ \omega = 2\pi \times 50 = 100\pi \text{ rad/s} \]

Now, the derivative of the current with respect to time is:

\[ \frac{di}{dt} = I_0 \omega \cos(\omega t) \]

Substituting the values:

\[ \frac{di}{dt} = 2 \times 100\pi \cos(100\pi t) = 200\pi \cos(100\pi t) \]

### Step 2: Calculate the induced voltage
Now we substitute \(\frac{di}{dt}\) back into the EMF formula:

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

Substituting \(M = 150 \text{ mH} = 150 \times 10^{-3} \text{ H}\):

\[ E = -150 \times 10^{-3} \times 200\pi \cos(100\pi t) \]

The peak value of the induced voltage \(E_{\text{max}}\) occurs when \(\cos(100\pi t) = 1\):

\[ E_{\text{max}} = 150 \times 10^{-3} \times 200\pi \]

Calculating this gives:

\[ E_{\text{max}} = 30 \pi \text{ volts} \]

### Step 3: Numerical calculation
Now, substituting \(\pi \approx 3.14\):

\[ E_{\text{max}} \approx 30 \times 3.14 \approx 94.2 \text{ volts} \]

### Conclusion
Thus, the peak value of voltage induced in the second coil is approximately:

\[ E_{\text{max}} \approx 94.2 \text{ volts} \]