Q.When a voltage measuring device is connected to AC mains, the meter shows the steady input voltage of $220\ \text{V}$. This means

An ac voltmeter cannot track the instantaneous mains voltage, which swings sinusoidally between $+311\text{ V}$ and $-311\text{ V}$ fifty times every second. What the meter actually responds to is the mean of the square of the voltage, and it is calibrated to display the square root of that mean — the root-mean-square (rms) value:

A steady reading of $220\text{ V}$ therefore means the rms voltage of the supply is $220\text{ V}$, which corresponds to a peak voltage

When a voltmeter is plugged into an AC socket it gives one steady number, $220\text{ V}$. Yet the mains voltage is never steady: it follows $v(t)=V_0\sin(2\pi f t)$, rising to a positive maximum, falling through zero, dropping to a negative maximum, and back — fifty complete cycles each second. So what does that single displayed number mean?

1. The meter cannot show the instantaneous value

The instantaneous voltage changes far too fast for any pointer or display to follow. Averaging $v(t)$ over a full cycle is useless too, because a symmetric sine wave spends equal time positive and negative, giving a plain average of exactly zero.

2. Squaring first solves the problem

To get a meaningful measure, the instrument responds to $v^2(t)$, which is always positive, takes its time-average over a cycle, and then takes the square root. This is precisely the root-mean-square value:

The rms value is physically meaningful because an AC supply of rms voltage $V_{rms}$ delivers the same average power to a resistor as a steady DC source of that same voltage.

3. Reading off the numbers

The meter is calibrated to display $V_{rms}$ directly, so

The corresponding peak voltage is

and the instantaneous voltage swings between $+311\text{ V}$ and $-311\text{ V}$.

Method: Reasoning About What an AC Meter Actually Displays

This is a reasoning-pattern method for any conceptual question about what a voltmeter or ammeter reading "means" when connected to an AC supply.

Steps

Step 1: Rule out the instantaneous value first.

The true instantaneous voltage $v(t) = V_0\sin(\omega t)$ changes far too fast (e.g. 50 times a second) for any mechanical or electronic display to track directly — so the reading cannot be an instantaneous snapshot.

Step 2: Rule out the plain time-average too.

A symmetric sine wave spends equal time positive and negative over a full cycle, so its plain average over a cycle is exactly zero — a meter reading zero would be useless and is not what is observed, so the meter cannot be reporting a simple average either.

Step 3: Recognise that squaring first fixes the sign-cancellation problem.

Since $v^2(t)$ is always non-negative, averaging the square over a cycle gives a meaningful non-zero number. Taking the square root of that average brings the units back to volts (or amps) — this is exactly the root-mean-square construction:

Step 4 (Applying to this problem): State that the meter is calibrated to display this rms quantity, and connect it to the peak.

A "steady" reading of $220\ \text{V}$ therefore means $V_{\text{rms}} = 220\ \text{V}$, corresponding to a peak of $V_0 = V_{\text{rms}}\sqrt{2}$. This same reasoning applies to any AC voltmeter/ammeter reading quoted without qualification — by convention it is always the rms value.