Calculating Volt Drop including cable resistances (real world application)

Thread Starter

fiftyhertz

Joined Mar 30, 2019
4
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 –
 

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Thread Starter

fiftyhertz

Joined Mar 30, 2019
4
I think Ive got it minutes after posting. Doh! Is it because we have to add the resistances if the circuit is in series for a segment.

Ie A1 +D + A2
A1 + B1 + E + B2 + A2
A1 + B1 + C1 + F + C2 + B2 + A1

Please can someone confirm but looks good to me.
 

wayneh

Joined Sep 9, 2010
17,152
You've added in the cable resistances as if they are in parallel. Only the loads are in parallel, the cables are in serial. So the combined load is much lower (resistance higher).
 
there should be formulas and calculators online. Electricians in the US, generally use <3% voltage drop to the circuit breakers. You could use less. That's basically how the wire gets sized. See: http://www.electrician2.com/calculators/vd_calculator_initial.html

It basically uses the formula R=pL/A where A is the cross-sectional area, L the length and p is the resistivity (copper/aluminum), 2x the length.
A can be gotten from the wire tables. You just need to get the units right. ohm-cm is the SI unit for resistivity, Tables can have ohms/1000 feet too.
 

wayneh

Joined Sep 9, 2010
17,152
I think Ive got it minutes after posting. Doh! Is it because we have to add the resistances if the circuit is in series for a segment.

Ie A1 +D + A2
A1 + B1 + E + B2 + A2
A1 + B1 + C1 + F + C2 + B2 + A1

Please can someone confirm but looks good to me.
Yes, those are the 3 paths that current will flow. They're in parallel, so you can add them up and start calculating the current in each loop. All the current is in A, less in B, and the least in C.
 

Thread Starter

fiftyhertz

Joined Mar 30, 2019
4
Thats great, thankyou for your help.

Calculated as below:

R1 = 0.2 ohms
R2= 0.15 ohms
R3 = 0.3 ohms
R4 = 529 ohms
R5 = 600 Ohms
R6= 480 ohms

R7 = 0.3 ohms
R8= 0.15 ohms

R9 = 0.2 ohms
Total Circuit Resistance:

Work out parallel Resistance

1/R4) + (1/R5) + (1/R6) = 1/RT

(1/529) + (1/600) + (1/480) = 1/RT

1/RpT = 1 / 5.64x10-3
RpT = 177.29 ohms

Work out series Resistance:

R1 + R2 + R3 + R7 + R8 + R9 = 1.3 ohms
RSt = 1.3 ohms
RT = 1.3 + 177.29

RT= 178.59 ohms


Total current = I = V/R

230/178.59 = 1.287A

Calculate the volt drop for Path 1, to establish the voltage & Current at R4

V = I x (R1 + R9)
R1 = 1.2 x (0.2 + 0.2)= 0.48v

Voltage at R4 = 230 – 0.48 = 229.52V
Therefore R4, 229.52 / 529 = 0.43A

Calculate the volt drop for path 2, to establish the voltage and current at R5

Within path 2, R1 & R9 Full circuit current, however R2 & R8 lower current as some has been diverted through path 1.
Therefore:

V = I x (R2 + R9)

= (1.287 – 0.43) x (0.15 + 0.15) = 0.2571v
0.48v + 0.2571v = 0.7371v

Voltage at R5 = 230 - 0.7371 = 229.2629v
229.2629/600 = 0.382A


Calculate the volt drop for path 3, to establish the voltage and current at R6

Within path 3, R1 & R9 Full circuit current, R2 & R8 lower current as some has been diverted through path 1, and R3 & R7 lowest circuit current as the majority of current has passed through path 1 & 2.
Therefore:

V = I x (R3 + R7)

= (1.287 – 0.43 – 0.382) x (0.3 + 0.3) = 0.285v
0.48v + 0.2571v + 0.285v = 1.0221

Voltage at R6 = 230 – 1.0221 = 228.9779v
229.2629/480= 0.477A
 

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wayneh

Joined Sep 9, 2010
17,152
Here's my simulation of your system (using 230V DC for simplicity).
Screen Shot 2019-03-30 at 1.03.20 PM.png

Currents:
R4 430 mA 434
R5 390 382
R6 475 477

R1 and R9 1,290 1,293
R2 and R8 859 859
R3 and R7 477 (and R6) 477

Voltages
across R4 224.48 229.48
across R5 229.22 229.23
across R6 228.94 228.94
 
Last edited:

bertus

Joined Apr 5, 2008
21,360
Hello,

@wayneh, there are some things that do to fit.
You say that IR6 is 475, but IR3 and IR7 is 477.
The voltage on R6 is higher as on R5, wich also can not be.
Also the voltage on R5 is higher as on R4.

Bertus
 

Thread Starter

fiftyhertz

Joined Mar 30, 2019
4
Here's my simulation of your system (using 230V DC for simplicity).
View attachment 173716

Currents:
R4 430 mA
R5 390
R6 475

R1 and R9 1,290
R2 and R8 859
R3 and R7 477 (and R6)

Voltages
across R4 224.48
across R5 229.22
across R6 228.94
Thankyou for taking the time to do that, but as the above comment. The amperages look the same as mine other than rounding errors - but shouldn’t the voltage drop up to the R6 resistor be the highest as the voltage drop is compounding as the resistances are in series.

Cheers
 

wayneh

Joined Sep 9, 2010
17,152
Hello,

@wayneh, there are some things that do to fit.
You say that IR6 is 475, but IR3 and IR7 is 477.
The voltage on R6 is higher as on R5, wich also can not be.
Also the voltage on R5 is higher as on R4.

Bertus
Some errors corrected above. The small differences before (475 vs 477) were due to not plotting the results so that I could read the values more precisely. The bigger error was a typo for the voltage across R4. Note that the voltages across the resistors are not the voltages on top the resistor, but the top minus the bottom. In case that wasn't clear.
 
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