Voltage drop help please

helio007 · Jan 7, 2024

all currently working towards my 2391 c&g I&T . I know how to calculate v.d. but the mock test I found online seems a bit out of date or I'm potentially doing something wrong. Here comes de Q.

Voltage drop of a single-phase distribution circuit supplying a power distribution board in a remote building is to be verified as part of the periodic inspection and testing within a workshop complex. The installation forms part of a public 400/230 V TN-S system. The circuit has a measured R1+Rn value of 0.15 Ω and an Ib of 60 A. The circuit protective device has an In of 80 A. What is the voltage drop for this distribution circuit?
mv/A/m × ib × l
1000
According to the mock test the right answer is 10.8V however with result was a little bit higher .

2.8x60x65/1000= 10.92v

Table 4D1B osg
16mm
Mvam 2.8
Could anyone point out where and what I'm doing wrong . TIA

foresite123 · Jan 9, 2024

helio007 said:
all currently working towards my 2391 c&g I&T . I know how to calculate v.d. but the mock test I found online seems a bit out of date or I'm potentially doing something wrong. Here comes de Q.

Voltage drop of a single-phase distribution circuit supplying a power distribution board in a remote building is to be verified as part of the periodic inspection and testing within a workshop complex. The installation forms part of a public 400/230 V TN-S system. The circuit has a measured R1+Rn value of 0.15 Ω and an Ib of 60 A. The circuit protective device has an In of 80 A. What is the voltage drop for this distribution circuit?
mv/A/m × ib × l
1000
According to the mock test the right answer is 10.8V however with result was a little bit higher .

2.8x60x65/1000= 10.92v

Table 4D1B osg
16mm
Mvam 2.8
Could anyone point out where and what I'm doing wrong . TIA

Hi helio007 , I think they have used a different calculation but your answer would also be right.
They seem to have used the 0.15 value,

0.15 x 60 x 1.2 (70 degrees ) = 10.8v,

pc1966 · Jan 9, 2024

helio007 said:
According to the mock test the right answer is 10.8V however with result was a little bit higher .

As you have actually measured the cold resistance (R1 + Rn = 0.15) then you have the voltage drop at the expected current (60A) so you can compute the drop cold as 60 * 0.15 = 9V

You are then assuming this will run hot at the usual CCC rating of 70C so you have a further 20% rise in R1+Rn so you get 9V * 1.2 = 10.8V

helio007 said:
2.8x60x65/1000= 10.92v

Table 4D1B osg
16mm
Mvam 2.8
Could anyone point out where and what I'm doing wrong . TIA

If you are planning the circuit then you would use this, but the fact is you have a measured value, which allows for cable length and tolerance more precisely, so you should use that instead. Of course you should see if it is close the the expected value in case of some error (wrong calculation, length longer/shorter than planned, poor junction if cable not continuous, etc) but as this is only about 1.1% difference you can safely assume it is OK.

Remember also that the value of 2.8 mV/A/m looks rounded to 1 digit, so it could be 2.8 +/- 0.05 which is +/- 1.8% on the tabulated value, greater than the difference in answers. Of course, your meter is probably only 3% + couple of digits, so unlikely to be much better!