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 based on the information provided, we can follow these steps:

### Step 1: Understand the Voltage Function
The voltage \( V(t) \) of an A.C. generator can be expressed as:
\[ V(t) = V_0 \sin(\omega t) \]
where \( V_0 \) is the maximum voltage (amplitude) and \( \omega \) is the angular frequency.

### Step 2: Identify Maximum Voltage
From the problem, we know that the voltage increases to a maximum of 100V. Therefore, we have:
\[ V_0 = 100 \text{ V} \]

### Step 3: Determine the Time at a Given Voltage
We are given that at time \( t = \frac{1}{100\pi} \) seconds, the voltage is 2V. We can substitute this into our voltage equation:
\[ 2 = 100 \sin(\omega \cdot \frac{1}{100\pi}) \]

### Step 4: Solve for \( \sin(\omega t) \)
Rearranging the equation gives:
\[ \sin(\omega \cdot \frac{1}{100\pi}) = \frac{2}{100} = \frac{1}{50} \]

### Step 5: Use the Small Angle Approximation
Since \( \sin(\theta) \) is small, we can use the approximation \( \sin(\theta) \approx \theta \) for small angles. Thus, we have:
\[ \omega \cdot \frac{1}{100\pi} \approx \frac{1}{50} \]

### Step 6: Solve for Angular Frequency \( \omega \)
Now, we can solve for \( \omega \):
\[ \omega \approx \frac{1}{50} \cdot 100\pi = \frac{2\pi}{1} = 2\pi \text{ rad/s} \]

### Step 7: Calculate Frequency \( f \)
The frequency \( f \) is related to angular frequency by the formula:
\[ \omega = 2\pi f \]
Substituting our value of \( \omega \):
\[ 2\pi f = 2\pi \Rightarrow f = 1 \text{ Hz} \]

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