A current which periodically varies between a positive to a negative value with time

The maximum value of the alternating current or voltage
peak voltage V0?= peak-to-peak voltage ÷ 2

Graph of alternating current against time showing the time period, peak current and peak-to-peak current
The variation with time?t?of the output voltage?V?of an alternating voltage supply is shown in the graph below.
Use the graph to calculate the frequency of the supply and the peak voltage.

The square root of the mean of the squares of all the values of the current in one cycle
The equivalent direct current that produces the same power

The square root of the mean of the squares of all the values of the voltage in one cycle
The equivalent dc voltage that produces the same power

The steady direct current, or voltage, that delivers the same average power in a resistor as the alternating current, or voltage

Vrms?and peak voltage. The rms voltage is about 70% of the peak voltage
Average power = Irms?× Vrms
An electric oven is connected to a 230 V root mean square (rms) mains supply using a cable of negligible resistance.Calculate the peak-to-peak voltage of the mains supply.
Step 1: Write down the Vrms?equation

Step 2: Rearrange for the peak voltage,?V0
V0?= √2 × Vrms
Step 3: Substitute in the values
V0?= √2 × 230
Step 4: Calculate the peak-to-peak voltage
Peak-to-peak voltage = (√2 × 230) × 2 = 650.538 =?651 V (3 s.f)
Remember to double-check the units on the alternating current and voltage graphs. These are often shown in milliseconds (ms) instead of seconds (s) on the x-axis.
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