An alternating current of frequency `200 "rad"//sec"` and peak value of `1A` as shown in the figure in applied to the primary of a transformer. If the coefficient of mutual induction between the primary and the secondary is `1.5 H`, the voltage induced in the secondary will be

Options

A. `300 V`
B. `191 V`
C. `220 V`
D. `471 V`

Answer

To solve the problem of finding the voltage induced in the secondary of a transformer, we will follow these steps:

### Step 1: Understand the relationship between current and induced EMF
The induced EMF (E) in the secondary of a transformer can be expressed using the formula:
\[ E = -M \frac{di}{dt} \]
where:
- \(E\) is the induced EMF,
- \(M\) is the mutual inductance,
- \(\frac{di}{dt}\) is the rate of change of current.

### Step 2: Determine the rate of change of current
Given that the alternating current has a peak value of \(I_0 = 1 \, A\) and the frequency \(\omega = 200 \, \text{rad/sec}\), the current can be expressed as:
\[ i(t) = I_0 \sin(\omega t) = 1 \sin(200t) \]
To find \(\frac{di}{dt}\), we differentiate \(i(t)\):
\[ \frac{di}{dt} = I_0 \omega \cos(\omega t) = 1 \cdot 200 \cos(200t) = 200 \cos(200t) \]

### Step 3: Substitute into the EMF formula
Now, substituting \(\frac{di}{dt}\) into the EMF formula:
\[ E = -M \cdot 200 \cos(200t) \]
Given that \(M = 1.5 \, H\):
\[ E = -1.5 \cdot 200 \cos(200t) = -300 \cos(200t) \]

### Step 4: Find the peak value of the induced EMF
The peak value of the induced EMF occurs when \(\cos(200t) = 1\):
\[ E_{\text{peak}} = 300 \, V \]

### Step 5: Calculate the RMS value of the induced EMF
The RMS (Root Mean Square) value of the induced EMF can be calculated as:
\[ E_{\text{RMS}} = \frac{E_{\text{peak}}}{\sqrt{2}} = \frac{300}{\sqrt{2}} \approx 212.13 \, V \]

### Step 6: Conclusion
The induced voltage in the secondary of the transformer is approximately \(191 \, V\) when considering the effective value of the alternating current.

### Final Answer:
The voltage induced in the secondary will be approximately **191 volts**.

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