Question

The voltage of an A.C. source varies with time according to the equation $V=50\sin 100\pi t\cos 100\pi t$, where $t$ is in seconds and $V$ is in volts. Then:

• A. the peak voltage of the source is $100V$
• B. the peak voltage of the source is $\frac{100}{\sqrt{2}}$.
• C. the peak voltage of the source is $25V$
• D. the frequency of the source is $50Hz$

Answer 1

The correct option is C the peak voltage of the source is $25V$

Peak voltage is when sin function and cos function both are maximum which is $\frac{1}{\sqrt{2}}$

So, peak voltage = $50\times \frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}$

$=\frac{50}{2}=25V$.

$V=50sin100\pi tcos100\pi t$

$=25sin200\pi t$ $(\because 2sinACosA=sin2A)$

So, frequency is $\frac{200\pi }{2\pi }=100Hz$