How Do I Calculate Voltage Drop Without Knowing Current?

[scarecrow75]

Hi all

Wondering if anyone can help. Really don't know where to start

I know Z_s = Z_e + Z₁ + Z₂

and Z₁ = PCR 8.9 milli-ohms/meter
and Z₂ = CPC 14.5 milli-ohms/meter
which equates to 0.0234X milli-ohms/meter over X meters

but where does my volt drop come in if I don't know my current?

Regards

 

[CWatters]

but where does my volt drop come in if I don't know my current?

If the fuse is 32A the current could be any value <32A ??

 

[scarecrow75]

Homework Statement

230 V, 50 Hz supply (TN)
Z_e = 0.8 ohms
Z₁ = 8.9 milli-ohms/meter
Z₂ = 14.5 milli-ohms/meter
Volt-drop/Amp/meter = 18 mV/A/m

Homework Equations

Z_s = Z_e + Z₁ + Z₂
V / Z_s = I ?

The Attempt at a Solution

Z_s = Z_e + Z₁ + Z₂
Zs = 0.8 + (0.0089 + 0.0145)X......where X is the distance in meters
Zs = 0.8 + 0.0234X

Then
230V / (0.8 + 0.0234X) = I

Not sure if this is correct!
Also is this relevant:
TN system... I know that for final circuits up to 32A disconnection time should not exceed 0.4secs
Voltage Drop, lighting circuits = 3%max and other circuits 5% max
if this is correct then 230V * 0.05 = 11.5V

Not sure if this is right approach

Any help?

Cheers

 

[scarecrow75]

Anyone?

 

[CWatters]

As you haven't had an answer I've posted a link to this thread on another forum..

http://www.askthetrades.co.uk/cgi-bin/yabb/YaBB.pl?num=1459595632/0#0