The voltage of an AC source varies with time according to the relation: ` E = 120 sin 100 pi t cos 100 pi t V`. What is the peak voltage of the source?

Options

A. 60 V
B. 120V
C. 30V
D. `sqrt(2)V`

Explanation

To find the peak voltage of the AC source given by the equation \( E = 120 \sin(100 \pi t) \cos(100 \pi t) \), we can follow these steps:

### Step 1: Use the Trigonometric Identity
We can simplify the expression using the trigonometric identity for the product of sine and cosine:
\[ \sin A \cos A = \frac{1}{2} \sin(2A) \]
In our case, let \( A = 100 \pi t \). Therefore:
\[ \sin(100 \pi t) \cos(100 \pi t) = \frac{1}{2} \sin(200 \pi t) \]

### Step 2: Substitute the Identity into the Equation
Substituting this identity into the original equation gives:
\[ E = 120 \cdot \frac{1}{2} \sin(200 \pi t) = 60 \sin(200 \pi t) \]

### Step 3: Identify the Peak Voltage
The standard form of an AC voltage equation is:
\[ E = E_0 \sin(\omega t) \]
where \( E_0 \) is the peak voltage. From our modified equation:
\[ E = 60 \sin(200 \pi t) \]
we can see that the peak voltage \( E_0 \) is:
\[ E_0 = 60 \text{ volts} \]

### Conclusion
Thus, the peak voltage of the source is \( 60 \) volts.

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