Need Explanation of Voltage Drop Please...

[eli]

The calculations for Voltage drop as described in the IEE Regulations book is shown below:

The actual voltage drop is given by:

mV x Ib x l
1000

=
7.3 x 37.4 x 24.5
1000

= 6.7 V (well below the maximum of 11.5V)

This voltage drop, whilst not causing the kiln to work unsafely, may mean inefficiency, and it is perhaps better to use a 10.0mm2 cable.

For a 10.0mm2 cable, the voltage drop is checked as

4.4 x 37.4 x 24.5
1000

= 4.04V

If the 6.7 V voltage drop using a 6mm2 cable is deemed sufficient as it is less than 11.5V (i.e. 5% of 230 V) for a power system, why and what are the reasons for using a 10.0mm2 cable even with a voltage drop of only 4.04V. There is not that much difference between the values of the two voltage drop (6.7V vs. 4.04V).

In simple terms can you describe what is voltage drop?

thanks

 

[telectrix]

basically it's ohms law. when you pass a current through a resistance you get a volt drop across that resistance. ideally, the resistance of your circuit would be purely the load so that the whole supply voltage was dropped across the load. in practice, however, the cable has a finite resistance, and so will drop some of the supply votage along it' s length. thereby reducing the voltage drop across the load.

 

[eli]

thanks