AC circuit: Frequency of the generator

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 generator.

Answer:
The voltage varies with time, so
$$V = V_m \sin \omega t$$
where $V_m$ is the peak voltage. From the condition, we can express
$$0.5V_m = V_m \sin \left(\omega\cdot \frac{1}{720}\right)$$
$V_m$ is cancelled out. Then take the arcsine.
$$\frac{\omega}{720} = \sin^{-1}0.5$$
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.

Powered by Hirophysics.com