ES2C6 – Power Electronics 1 – AC Power
AC Power
For a normal AC voltage:
𝑉 = 𝑉𝑝 cos 𝜔𝑡 (1)
which gives a sine wave with a constant period, 𝑇, and peak voltage, 𝑉𝑝 .
Root Mean Square
The UK uses AC power at 50𝐻𝑧 and 240𝑉𝑟𝑚𝑠 , where the root mean square gives a DC equivalent of
an AC voltage.
To find the RMS voltage from (1), 𝑉 must first be squared:
𝑉𝑝2 cos2 𝜔𝑡
The mean is then found by integrating between time 0 and 𝑇, then dividing by the period:
1 𝑇 2
∫ 𝑉 cos2 𝜔𝑡 𝑑𝑡
𝑇 0 𝑝
This gives the RMS voltage as:
1 𝑇 2
𝑉𝑅𝑀𝑆 = √ ∫ 𝑉 cos2 𝜔𝑡 𝑑𝑡
𝑇 0 𝑝
𝑉𝑝2 𝑇 1 + cos 2𝜔𝑡
=√ ∫ 𝑑𝑡
𝑇 0 2
𝑉𝑝2 𝑡 sin 2𝜔𝑡 𝑇
=√ [ + ]
𝑇 2 4𝜔 0
𝑉𝑝
= (2)
√2
This means that the peak voltage is the RMS voltage multiplied by √2, so in the UK 𝑉𝑝 = 240√2.
Similarly, with current:
𝐼𝑝
𝐼𝑅𝑀𝑆 = (3)
√2
Real Power
All electrical circuits have resistors, inductors and capacitors. Resistors dissipate real power, whilst
inductors and capacitors are reactive elements – meaning they store energy.
Resistors don’t induce a phase shift, so there is no phase shift in this circuit.
1
AC Power
For a normal AC voltage:
𝑉 = 𝑉𝑝 cos 𝜔𝑡 (1)
which gives a sine wave with a constant period, 𝑇, and peak voltage, 𝑉𝑝 .
Root Mean Square
The UK uses AC power at 50𝐻𝑧 and 240𝑉𝑟𝑚𝑠 , where the root mean square gives a DC equivalent of
an AC voltage.
To find the RMS voltage from (1), 𝑉 must first be squared:
𝑉𝑝2 cos2 𝜔𝑡
The mean is then found by integrating between time 0 and 𝑇, then dividing by the period:
1 𝑇 2
∫ 𝑉 cos2 𝜔𝑡 𝑑𝑡
𝑇 0 𝑝
This gives the RMS voltage as:
1 𝑇 2
𝑉𝑅𝑀𝑆 = √ ∫ 𝑉 cos2 𝜔𝑡 𝑑𝑡
𝑇 0 𝑝
𝑉𝑝2 𝑇 1 + cos 2𝜔𝑡
=√ ∫ 𝑑𝑡
𝑇 0 2
𝑉𝑝2 𝑡 sin 2𝜔𝑡 𝑇
=√ [ + ]
𝑇 2 4𝜔 0
𝑉𝑝
= (2)
√2
This means that the peak voltage is the RMS voltage multiplied by √2, so in the UK 𝑉𝑝 = 240√2.
Similarly, with current:
𝐼𝑝
𝐼𝑅𝑀𝑆 = (3)
√2
Real Power
All electrical circuits have resistors, inductors and capacitors. Resistors dissipate real power, whilst
inductors and capacitors are reactive elements – meaning they store energy.
Resistors don’t induce a phase shift, so there is no phase shift in this circuit.
1
