What is the RMS value of an alternating current and why is it 0.707 times the peak?

In one line

The RMS (root mean square) value of an alternating current is the steady DC current that would produce the same heat in a resistor in the same time; for a sine wave it equals \( I_m/\sqrt{2} \approx 0.707I_m \).

Explained simply

An AC current keeps changing, so how do you say "how big" it is? Ask a practical question: how much does it heat a heater? If a 10 A DC current and some AC current heat the same heater equally, that AC is called "10 A RMS". That is why the 230 V written on Indian sockets is an RMS value; the actual peak is about 325 V.

Technically, heating depends on \( i^2 \). The RMS value is the square root of the mean of \( i^2 \) over one cycle.

How it works

1. Take \( i = I_m\sin\theta \). Square it: \( i^2 = I_m^2\sin^2\theta \).
2. Mean over one cycle: \( \dfrac{1}{2\pi}\int_0^{2\pi} I_m^2\sin^2\theta\,d\theta \).
3. Use \( \sin^2\theta = (1-\cos2\theta)/2 \). The cosine term integrates to zero, leaving \( I_m^2/2 \).
4. Square root: \( I_{rms} = I_m/\sqrt{2} = 0.707I_m \).

Meters (voltmeters, ammeters) and nameplates always quote RMS values unless stated otherwise.

Key formula

\( I_{rms} = \sqrt{\dfrac{1}{T}\int_0^T i^2\,dt}, \qquad \text{for a sine wave } I_{rms} = \dfrac{I_m}{\sqrt{2}}, \; V_{rms} = \dfrac{V_m}{\sqrt{2}} \)

\( I_{rms} \) = RMS current (A), \( I_m \) = peak current (A), \( T \) = time period (s), \( i \) = instantaneous current (A). Peak factor \( = I_m/I_{rms} = \sqrt{2} = 1.414 \).

Quick example

Indian mains: \( V_{rms} = 230 \) V, so \( V_m = 230\sqrt{2} \approx 325 \) V. A 1000 W heater on 230 V draws \( I_{rms} = 1000/230 = 4.35 \) A, but the current actually swings up to \( 4.35\times1.414 = 6.15 \) A at each peak. This is why insulation and capacitors are rated for the peak, not the RMS value.

Common mistakes

• Using 0.707 for non-sinusoidal waves (a square wave has RMS = peak).
• Averaging \( i \) and then squaring; you must square first, then average, then take the root.
• Thinking 230 V is the maximum voltage of the supply.

How MU asks it

"Derive the RMS value of a sinusoidal current" or "Define RMS value" (5 marks), sometimes combined with average value and form factor. For full marks: define RMS in terms of heating effect, show the integral with limits, use the \( \sin^2 \) identity, and finish with 0.707\( I_m \). A small sketch of \( i \) and \( i^2 \) helps.