Which statement about RMS voltage is true?

Which statement about RMS voltage is true?

• A. It is the instantaneous voltage
• B. It is the average voltage over a cycle
• C. It is the square root of the mean of the squares of voltages over a cycle
• D. It is the maximum voltage

Answer: (C)

The main idea is that RMS voltage represents the effective heating value of an AC signal. It is defined as the square root of the average of the voltage squared over one cycle. In mathematical terms, for a periodic waveform v(t) with period T, the RMS is V_rms = sqrt( (1/T) ∫_0^T [v(t)]^2 dt ). This directly ties to power in a resistor, since P = V_rms^2 / R, so RMS tells you the DC-equivalent voltage that would dissipate the same power. This is why the statement is true: taking the mean of the squares over a cycle and then the square root yields the value that matches the heating effect of the waveform. For a sinusoidal input, the RMS is V_peak divided by sqrt(2), illustrating how RMS can differ from the peak value. The other ideas don’t describe RMS accurately. Instantaneous voltage is simply the momentary value v(t), not an average measure. The average over a cycle can be zero for a pure AC waveform, so it doesn’t reflect the heating effect. The maximum voltage is the peak value, not the effective RMS value.

The main idea is that RMS voltage represents the effective heating value of an AC signal. It is defined as the square root of the average of the voltage squared over one cycle. In mathematical terms, for a periodic waveform v(t) with period T, the RMS is V_rms = sqrt( (1/T) ∫_0^T [v(t)]^2 dt ). This directly ties to power in a resistor, since P = V_rms^2 / R, so RMS tells you the DC-equivalent voltage that would dissipate the same power.

This is why the statement is true: taking the mean of the squares over a cycle and then the square root yields the value that matches the heating effect of the waveform. For a sinusoidal input, the RMS is V_peak divided by sqrt(2), illustrating how RMS can differ from the peak value.

The other ideas don’t describe RMS accurately. Instantaneous voltage is simply the momentary value v(t), not an average measure. The average over a cycle can be zero for a pure AC waveform, so it doesn’t reflect the heating effect. The maximum voltage is the peak value, not the effective RMS value.