A 50 Hz a.c. current of crest value 1A flows through the primary of a transformer. If the mutual inductance between the primary and secondary be `1.5 H` , the crest voltage induced in secondary is-

Options

A. 75`pi` volt
B. 150`pi` volt
C. 225 volt
D. 300 volt

Answer

To solve the problem step by step, we will follow the concepts of alternating current (AC), mutual inductance, and differentiation.

### Step 1: Understand the Given Information
We have:
- Frequency (f) = 50 Hz
- Crest (peak) value of current (I₀) = 1 A
- Mutual inductance (M) = 1.5 H

### Step 2: Write the Expression for AC Current
The expression for the alternating current can be written as:
\[ I(t) = I_0 \sin(\omega t) \]
where:
- \(\omega = 2\pi f\) is the angular frequency.

Calculating \(\omega\):
\[ \omega = 2\pi \times 50 = 100\pi \, \text{rad/s} \]
Thus, the current expression becomes:
\[ I(t) = 1 \sin(100\pi t) \]

### Step 3: Differentiate the Current Expression
To find the induced voltage in the secondary coil, we need to differentiate the current with respect to time (t):
\[ \frac{dI}{dt} = \frac{d}{dt} [1 \sin(100\pi t)] = 100\pi \cos(100\pi t) \]

### Step 4: Use the Mutual Inductance Formula
The induced voltage (E) in the secondary coil can be expressed as:
\[ E_s = -M \frac{dI}{dt} \]
Substituting the values we have:
\[ E_s = -1.5 \times (100\pi \cos(100\pi t)) \]
This simplifies to:
\[ E_s = -150\pi \cos(100\pi t) \]

### Step 5: Determine the Crest Voltage
The crest (peak) voltage induced in the secondary is the coefficient of the cosine function:
\[ E_0 = 150\pi \, \text{volts} \]

### Final Answer
The crest voltage induced in the secondary is:
\[ E_0 = 150\pi \, \text{volts} \]

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