I'm calculating the energy losses of using AC to power rectified (DC) devices.
Rectified loads have capacitor filters that demand sharp peaks of current. This increases conduction losses across the entire AC electrical installation.
We can model all rectified loads in an installation as one large rectifier, for simplicity:
The current in this rectifier may be 20 A (or ~5 kW).
The ripple may be 5% (Vr = 16 V).
The capacitor will then be around 0.01 F:
C = I / (2f * Vr) = 20 A / (2 * 50 Hz * 16 V) = 0.01 F
In a classical AC load, the RMS current is the average current. But in a rectified load, the RMS current demanded is much higher than the average current, because of the sharp peaks. This causes conduction losses.
To calculate the RMS current powering this ~5kW rectifier, I did:
The capacitive reactance is:
Xc = 1 / (2 * π * f * C) = 0.3 Ω
The RMS current is:
I_rms = V_rms / Xc = 230 V / 0.3 Ω = 770 A
Which is obviously wrong!
Can you help me understand what I should have done differently?