1. D - connection DELTA 3-phase distribution system
400 V (= nominal voltage), 50 Hz N L2 L3
Phase Voltage = 230 V
Phase to Phase Voltage = √3 * 230 V = 400 V L1
Peak voltages = √2 * Nominal voltages
Y - Connection
WYE
D - connection
DELTA
Symmetric 3-phase load:
P = √3 * Uphase-phase * I * cosΦ S = √3 * Uphase-phase * I
= 3 * Uphase * I * cosΦ S = 3 * Uphase * I
Q = √3 * Uphase-phase * I * sinΦ S2 = P2 + Q2
= 3 * Uphase * I * sinΦ

2. Single Phase 50 Hz Supply Voltage
Vpeak = √2 * 230 V
Amplitude of the voltage
Voltage
Current
T = 1/f = 1/50 Hz = 20 ms

3. Resistive - Inductive load: Voltage and current
Inductive load, Voltage * Current S = U * I -
Negative Power
Power/Watts

4. N L1 L2 L3 PE Phase Voltage = 230 V Ground SOCKET
Phase wires: L1 L2 L3
Neutral wire: N
Protective wire: PE N L2 L3 L1 PE
SOCKET
230 V
FUSE
Ground

5. Power triangel Inductive load Apparent Power (VA) S = U * I
Reactive Power (VAr) Q
phase angle φ
True Power (W) P
sine φ = Q/S
Power factor = cos φ = P/S
S2 = P2 + Q2
P = U * I * cos φ
Power factor = 1...0
Smaller power factor means more
reactive power.

8. Example

9. Solution: Powers: cosφ = 0,7 φ = 45,57 0
S = 2200 VA/cos45,570 = 3143 VA
Q = 3143 * sin45, = 2244 Var
Solution: φ
P = 2200 W
Q = 2244 Var
S = 3143 VA
Current:
S = U * I
I = 3143 VA/230 V = 13,66 A
Impedances:
Z = U/I = 230 V/13,66 A = 16,83 Ω
R = 0,7 * 16,83 Ω = 11,78 Ω
XL = sin 45,570 * 16,83 Ω = 12,02 Ω
Capacitor to correct power factor:
QC = QL = U2/XC
XC = 1/(2∏*f*C)
C = Q/(2∏*f*U2) = 2244 VA/(2∏*50s-1*(230 V)2) = 135 uF
Current with capacitor:
I = P/U = 2200 VA/230 V = 9,57 A
Hence the current is reduced 4,09 A

10. Detection of insulation faults
An insulation fault may be the consequence of
insulation deterioration:
c between two live conductors,
c between a conductor and the ECPs or the
protective conductor,
c on a single live conductor, making the
conductor accessible to touch.
An insulation fault between live conductors
becomes a short-circuit.
In all other cases, a fault (in common mode)
causes current to flow to earth. This current,
which does not flow back via the live conductors,
is called the earth-fault current. It is the algebraic
sum of the instantaneous values of the currents
flowing in the live conductors, hence the name
"residual current".
Remark: If the currents are sinusoidal, Fresnel
vector representation may be used and it is
possible to speak of the "vector sum" of the
currents. However this representation is not
relevant in the presence of harmonic currents
and the term "algebraic sum" is therefore more
generally applicable.
This current may be due to an insulation fault
between a live conductor and the ECPs (risk of
indirect contact) or to a failure of the measures
used to insulate or isolate live parts (risk of direct
contact). These situations are shown in figure 7
next page.
Similarly, in all cases of direct contact, the fault
current is low and cannot be detected and
cleared by standard overcurrent protective
devices. This is also the case for leakage
currents that constitute fire hazards.
Under these conditions, the fault current must be
detected and cleared by a special device, i.e. a
residual-current device (RCD), discussed in the
next section.