Question: The rms and the average value of an ac voltage $V = V_0 \sin \omega t$ volt over a cycle respectively will be:

Step 1 (Formula / Governing Law & Given Data): [0.5 Mark]

• The rms (root mean square) value of an AC voltage is defined as the square root of the mean of the squares of the voltage over one complete cycle: $$V_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} V^2 dt}$$
• The average value of an AC voltage over one complete cycle ($T$) is defined as: $$V_{avg} = \frac{1}{T} \int_{0}^{T} V dt$$

Step 2 (Step-by-Step Mathematical Working): [0.25 Mark]

• For rms value: $$V_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} V_0^2 \sin^2(\omega t) dt} = \frac{V_0}{\sqrt{T}} \sqrt{\int_{0}^{T} \frac{1 - \cos(2\omega t)}{2} dt} = \frac{V_0}{\sqrt{2}}$$
• For average value: $$V_{avg} = \frac{1}{T} \int_{0}^{T} V_0 \sin(\omega t) dt = \frac{V_0}{T} \left[ -\frac{\cos(\omega t)}{\omega} \right]_{0}^{T}$$ Since $\omega = \frac{2\pi}{T}$, then $\omega T = 2\pi$: $$V_{avg} = \frac{V_0}{2\pi} [-\cos(2\pi) + \cos(0)] = \frac{V_0}{2\pi} [-1 + 1] = 0$$

Step 3 (Final Concluded Answer): [0.25 Mark]

The rms value is $\frac{V_0}{\sqrt{2}}$ and the average value over a complete cycle is $0$.

Correct Option: (C)

💡 EL10 Concept Insight: Think of an AC cycle like a wave that goes up and down equally; because the positive part perfectly cancels out the negative part, the average is zero, but the rms value tells us the "effective" strength by squaring the values so they are all positive before averaging them.