Question

An alternating voltage is represented by \( V = 80 \sin(100\pi t) \cos(100\pi t) \) volt. The peak voltage is

• A. 20 V
• B. 40 V
• C. 30 V
• D. 50 V

Hint:

Whenever you see a product of sine and cosine with the same argument, use the double‑angle identity to simplify and find amplitude.

Answer 1

The Correct Option is B

Step 1: See what we must find.
The voltage is $V = 80 \sin(100\pi t)\cos(100\pi t)$. We need its peak value, that is the biggest value it can reach.

Step 2: Spot the handy identity.
There is a neat trig rule: $2 \sin A \cos A = \sin 2A$. So $\sin A \cos A = \frac{1}{2}\sin 2A$.

Step 3: Apply it here.
Here $A = 100\pi t$, so $\sin(100\pi t)\cos(100\pi t) = \frac{1}{2}\sin(200\pi t)$.

Step 4: Rewrite the voltage.
Then $V = 80 \times \frac{1}{2}\sin(200\pi t) = 40 \sin(200\pi t)$.

Step 5: Read off the peak.
A sine wave swings between plus one and minus one, so the largest value of $V$ is just the number in front, which is $40$.

Step 6: State it clearly.
The peak voltage is $40\ \text{V}$.
\[ \boxed{40\ \text{V}} \]