Comparing AC and DC on an oscilloscope — what the traces look like

Alternating Currents — AQA A-Level Physics

A DC signal on an oscilloscope is a flat horizontal line. The height above the centre line gives the 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.. It doesn't change with time.
An AC signal is a sine wave oscillating above and below the centre line. The centre line represents zero volts.
If you connect a DC supply of 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. V to a resistor, the powerThe rate of energy transfer. Measured in watts (W). dissipated is $P = V^2/R$ continuously.
If you connect an AC supply with peak 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. V₀ to the same resistor, the powerThe rate of energy transfer. Measured in watts (W). varies between zero (when V = 0) and $V_0^2/R$ (when V = V₀). The mean powerThe rate of energy transfer. Measured in watts (W). is $V_0^2/(2R)$.
For the AC supply to deliver the same mean power as the DC supply: $V^2/R = V_0^2/(2R)$. This gives $V = V_0/\sqrt{2}$. That's the RMS value. The connection is direct: RMS is defined by power equivalence.

Practical comparison

Connect a lamp to a DC supply at voltage V. Note the brightness.
Now connect the same lamp to an AC supply. Adjust the peak voltage until the brightness matches.
When the brightness matches, the AC supply is delivering the same mean power. The peak voltage will be V × √2, and the RMS voltage will equal V.
This is a physical demonstration of the RMS definition.

Common Mistake MEDIUM

Students often: Thinking AC power is zero because the average voltage is zero.
Instead: Power depends on V² (or I²), which is always positive. The average of V² is not zero even though the average of V is zero. The mean power in an AC circuit is $P = V_{\text{rms}}^{2}/R$, which is always positive.