Question:

There is a AC circuit with a resistor \(R=50 \Omega\). When the time is zero, the voltage is equal to zero. After 1/720 s, the voltage becomes a half of the maximum voltage. Find the frequency of the voltage power supply.

Answer:

The voltage varies with time, so

\[  V = V_m \sin \omega t  \]

where \(V_m\) is the peak voltage. From the given conditions, we can have the following:

\[  0.5V_m = V_m \sin \left(\omega\cdot \frac{1}{720}\right)  \]

Then, \(V_m\) is cancelled out. Take the arcsine of both sides.

\[  \frac{\omega}{720} = \sin^{-1}0.5  \]

Therefore,

\[  \omega = 120 \pi \mathrm{rad/s}  \]

In order to get the frequency, divide it by \(2\pi\).

\[  f = \frac{\omega}{2\pi} = 60 \mathrm{Hz}  \]

Note that we don’t have to use the resistance for this problem.