The average of AC is zero, but power is always positive

Alternating Currents — AQA A-Level Physics

Over one complete cycle, the currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A). flows one way for half the cycle and the other way for the other half. The positive and negative contributions are symmetric, so the average currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A). is exactly zero.
The same applies to voltageThe energyThe capacity to do work. Measured in joules (J). transferred per unit chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). between two points. Measured in volts (V). Informal term for potential difference.. Average voltageThe energyThe capacity to do work. Measured in joules (J). transferred per unit charge between two points. Measured in volts (V). Informal term for potential difference. over a full cycle: zero.
But powerThe rate of energy transfer. Measured in watts (W). doesn't cancel. Here's why:

$$P = I^2 R$$

$P$: instantaneous powerThe rate of energy transfer. Measured in watts (W). (W)
$I$: instantaneous currentThe rate of flow of charge. Measured in amperes (A). (A)
$R$: resistanceThe opposition to current flow. The ratio of potential difference to current. Measured in ohms (Ω). (\Omega)

Current squared is always positive. Whether I is +3 A or -3 A, $I^{2} = 9 A^{2}. Squaring removes$ the sign.
So powerThe rate of energy transfer. Measured in watts (W). $P = I^{2} R$ is always positive, regardless of which direction the current flows. EnergyThe capacity to do work. Measured in joules (J). is always being dissipated as heat in the resistor — in both halves of the cycle.
This is why you can't just use the average current (zero) to calculate power. You'd get $P = 0^{2} R = 0$, which would mean your kettle never boils. Clearly wrong.
RMS fixes this. By squaring first (making everything positive), averaging, then square-rooting, you get a value that correctly represents the power-delivering capability of the AC signal.
The RMS value is always less than the peak value, but greater than the average absolute value. For a sine wave: $RMS = 0.707 \times peak$.

Examiner Tips and Tricks

A common 2-mark question: 'Explain why the RMS value is used rather than the peak value for AC calculations.' Answer: the RMS value is the equivalent DC value that dissipates the same power in a resistor.
The peak value overestimates the actual power delivered because the voltageThe energy transferred per unit charge between two points. Measured in volts (V). Informal term for potential difference. is only at peak for an instant.