At time t = 0 second, voltage of an A.C. Generator starts from 0V and becomes 2V at time `t = 1/(100 pi)` second. The voltage keeps on increasing up 100 V, after wihich it starts to decrease. Find the frequency of the Generator.

Options

A. 100 Hz
B. 1 Hz
C. 2 Hz
D. 5 Hz

Answer

To find the frequency of the A.C. generator, we can follow these steps:

### Step 1: Understand the A.C. Voltage Equation
The voltage produced by an A.C. generator can be expressed as:
\[ E(t) = E_0 \sin(\omega t) \]
where:
- \( E(t) \) is the instantaneous voltage at time \( t \),
- \( E_0 \) is the maximum voltage (amplitude),
- \( \omega \) is the angular frequency in radians per second.

### Step 2: Identify Given Values
From the problem, we know:
- At \( t = 0 \) seconds, \( E(0) = 0 \) V.
- At \( t = \frac{1}{100\pi} \) seconds, \( E\left(\frac{1}{100\pi}\right) = 2 \) V.
- The maximum voltage \( E_0 = 100 \) V.

### Step 3: Substitute Known Values into the Voltage Equation
Substituting the known values into the voltage equation:
\[ 2 = 100 \sin\left(\omega \cdot \frac{1}{100\pi}\right) \]

### Step 4: Simplify the Equation
Dividing both sides by 100:
\[ \sin\left(\omega \cdot \frac{1}{100\pi}\right) = \frac{2}{100} = 0.02 \]

### Step 5: Use Small Angle Approximation
Since \( \omega t \) is small (as \( t = \frac{1}{100\pi} \) is small), we can use the small angle approximation:
\[ \sin(x) \approx x \]
Thus, we can write:
\[ \omega \cdot \frac{1}{100\pi} \approx 0.02 \]

### Step 6: Solve for Angular Frequency \( \omega \)
Rearranging the equation gives:
\[ \omega \approx 0.02 \cdot 100\pi \]
\[ \omega \approx 2\pi \]

### Step 7: Find the Frequency \( f \)
The relationship between angular frequency \( \omega \) and frequency \( f \) is given by:
\[ \omega = 2\pi f \]
Substituting the value of \( \omega \):
\[ 2\pi = 2\pi f \]
Dividing both sides by \( 2\pi \):
\[ f = 1 \text{ Hz} \]

### Conclusion
The frequency of the A.C. generator is:
\[ \boxed{1 \text{ Hz}} \]

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