Given a source with voltage varying as $v=V{m}_{0}sin\left(\omega t\right)$, what is the average voltage $v_{\mathrm{avg}}$ of the source over one complete cycle?

• A. $v_{\mathrm{avg}}=0$
• B. $v_{\mathrm{avg}}=V{m}_0$
• C. $v_{\mathrm{avg}}=\frac{2V{m}_0}{\mathrm{\pi }}$
• D. $v_{\mathrm{avg}}=\frac{V{m}_0}{2}$

Answer: (A)

Verified step by step guidance

1. Identify the given voltage function: \(v = V_{m0} \sin(\omega t)\), where \(V_{m0}\) is the peak voltage and \(\omega t\) represents the angular position in the cycle.
2. Recall that the average voltage over one complete cycle is defined as the integral of the voltage over one period \(T\), divided by the period: \(v_{avg} = \frac{1}{T} \int_0^T v(t) \, dt\).
3. Express the period \(T\) in terms of angular frequency: since \(\omega = \frac{2\pi}{T}\), then \(T = \frac{2\pi}{\omega}\).
4. Substitute \(v(t)\) and \(T\) into the average voltage formula: \(v_{avg} = \frac{1}{T} \int_0^T V_{m0} \sin(\omega t) \, dt = \frac{\omega}{2\pi} \int_0^{\frac{2\pi}{\omega}} V_{m0} \sin(\omega t) \, dt\).
5. Perform the integration by changing variables or directly integrating \(\sin(\omega t)\) over one full cycle, noting that the integral of \(\sin\) over \$0\( to \)2\pi\( is zero, which leads to the conclusion that \)v_{avg} = 0$.