Hi all,
Hope you can help as im going round in circles.
Basically, as the title im trying to calculate the volt drop of a parallel lighting circuit at 230v mains voltage (this is theoretical, for my own peace of mind)
The issue im having, is that when such low resistances are used ie 0.5 ohms for the cable, the current shoots up to an unrealistic figure. However, one of the first steps is to calculate the circuit total resistance.
A = 0.2 ohms x2 (neutral)
B = 0.15 ohms x2 (neutral)
C = 0.3 ohms x2 (neutral)
D = 529 ohms
E = 600 Ohms
F = 480 ohms
Total Circuit Resistance:
(1/A1) + (1/A2) + (1/B1) + (1/B2) + (1/C1) + (1/C2) + (1/D) + (1/E) + (1/F) = 1/RT
(1/0.2) + (1/0.2) + (1/0.15) ) + (1/0.15) ) + (1/0.3) + (1/0.3) + (1/529) + (1/600) + (1/480)
1/RT = 30
RT = 0.03 ohms
Total current = I = V/R
230/0.0330 = 696A. This is the issue from the low resistance cable being included
I then want to Calculate the volt drop for A, to establish the voltage at D
V = I x R
Voltage at D = 230 –
Hope you can help as im going round in circles.
Basically, as the title im trying to calculate the volt drop of a parallel lighting circuit at 230v mains voltage (this is theoretical, for my own peace of mind)
The issue im having, is that when such low resistances are used ie 0.5 ohms for the cable, the current shoots up to an unrealistic figure. However, one of the first steps is to calculate the circuit total resistance.
A = 0.2 ohms x2 (neutral)
B = 0.15 ohms x2 (neutral)
C = 0.3 ohms x2 (neutral)
D = 529 ohms
E = 600 Ohms
F = 480 ohms
Total Circuit Resistance:
(1/A1) + (1/A2) + (1/B1) + (1/B2) + (1/C1) + (1/C2) + (1/D) + (1/E) + (1/F) = 1/RT
(1/0.2) + (1/0.2) + (1/0.15) ) + (1/0.15) ) + (1/0.3) + (1/0.3) + (1/529) + (1/600) + (1/480)
1/RT = 30
RT = 0.03 ohms
Total current = I = V/R
230/0.0330 = 696A. This is the issue from the low resistance cable being included
I then want to Calculate the volt drop for A, to establish the voltage at D
V = I x R
Voltage at D = 230 –
Attachments

76.9 KB Views: 7