|
This will show the effect of the resistance of wires in a circuit.
Wires are made of copper, which has a electrical resistance. This is measured as following:
The formula is; R=ρ×l÷A
R= the resistance of the wire
ρ= the electrical resistivity of the material in
ohm-meter
l= the length of the wire in metres
A= the cross-sectional area in m²
For the top circuit we take the next data:
R= ... Ω
ρ= 0,0000000175 Ω-m (1,75·10-8)
l= 250 m
A= 0,0000005 m² (5,0·10-7)
(0,5 mm² (20 AWG))
We will calculate the data:
R=ρ×l÷A = ... Ω=1,75·10-8 × 250 ÷ 5,0·10-7
This will be 8,75 Ω per wire. It will be a total of 17,5 Ω, because the current needs to flow through the + and −.
In combination with the load, the lamp which has a power of 24 W (resistance of 24 Ω), a large portion of the voltage will drop over the wires.
In the middle circuit we take the next data:
R= ... Ω
ρ= 0,0000000175 Ω-m (1,75·10-8)
l= 250 m
A= 0,0000015 m² (15,0·10-7)
(1,5 mm² (16 AWG))
We will calculate the data:
R=ρ×l÷A = ... Ω=1,75·10-8 × 250 ÷ 15,0·10-7
This will be 2,9 Ω per wire. It will be a total of 5,8 Ω.
The lamp is identical to the top one. Now there will be a smaller voltage which drops over de wires.
In the bottom circuit we take the next data:
R= ... Ω
ρ= 0,0000000175 Ω-m (1,75·10-8)
l= 250 m
A= 0,0000025 m² (25,0·10-7)
(2,5 mm² (14 AWG))
We will calculate the data:
R=ρ×l÷A = ... Ω=1,75·10-8 × 250 ÷ 25,0·10-7
This will be 1,75 Ω per wire. It will be a total of 3,5 Ω.
The lamp is identical to the others. Now there will be a even smaller voltage which drops over de wires.
This is why the cable always needs to be calculated. Not only for the current, but also for the resistance of the wires in the cable.
|
