The measured value of loop Impedance for a circuit is 2.4 Ohms

[Rather be Fishing]

Hello trainee forum thank you very much for letting me in here and as a consequence hopefully helping me to develop my skills and knowledge.

I'm currently focusing on the 2391 and have enrolled myself on a course with no previous Inspection and testing experience or Knowledge so I do feel kind of up against it but ill give it my best shot and see how many re sits it takes me.

First of all I have to pass a multiple choice that requires you to get 75% and I'm currently doing some homework for it but my notes are leaving me unable to answer this question.

If anyone could help me it would be much appreciated.

The measured value of loop impendence for a circuit is 2.4 ohms. If the temperature at the time of the test was 20 degrees C and the cable is 70 degrees C (factor 1.2) what is the corrected value. Ze =0 .4ohms

Here are the possible answers they give you to choose from

A)2.88ohms
B)2.8 ohms
C)2.4 ohms
D)2.0 ohms


Ill show you what I have written down

Zs = 2.4
Ze = 0.4
R1 + R2
R1 + R2 x 1.2 = 2.4ohms
=0.4ohms

= 2.8ohms


Does anyone know if my math's is right as it doesn't feel right to me as when I apply it to similar questions the answer is always wrong.

 

[westward10]

Your answer is correct but your maths isn't clear.
Zs - Ze = 2.0
2.0 × 1.2 = 2.4
2.4 +Ze = 2.8

 

[Rather be Fishing]

Thank you @westward10 much appreciated mate.

I thought I would apply it to another question that I got wrong

The measured value of loop impedance for a circuit is 0.83ohm. If the temperature at the time of the test was 20C and the cable is 70C (factor 1.2) what is the corrected value. Ze = 0.4ohm:

a) 0.43 ohm
b) 0.86 ohm
c) 0.996 ohm
d) 0.916 ohm

Zs-Ze
= 0.83 - 0.4
= 0.43

0.43 x 1.2 = 0.516

0.516 + 0.4

= 0.916 ohms

which is what they said the answer was


so thank you very much you've just taught me how to do something

 

[westward10]

No worries if you are stuck on anything post it up with your workings and answer.

 

[Rather be Fishing]

You are performing the initial verification on an installation but are concerned with the length of the lighting circuit conductors. Because of your concerns you have decided to verify the circuits voltage drop. The circuit is wired in 1.5mm2 conductors and runs for a circuit length of 48m.

You have measured the cold R1+Rn value for the circuit and found it to be 1.18ohms.

The designer has advised the calculated voltage drop for the circuit is 4.3V

If the design current is 3.04A, does the circuit comply with the design requirements?

Here is what I seem to have written down

R1 Rn = 1.18ohms x 1.04 ohms = 1.2272 ohms

voltage drop = 3.04 x 1.2272 = 3.730688

yes it complys

Wrong

R1 Rn = 0.63 ohms x 1.2 = 0.756
16 x 0.756 = 12.096V

5% = 11.5 (230/100=2.3V)

2.3 x 5% = 11.5

No it does not comply

I'm not sure if I have the right answer there or not its just what I wrote down but now I'm going back over it even if I did have the right answer I don't really follow it.








3.04it complys

 

[westward10]

Where have you got 1.04 and 0.63 ohm from.
You need to verify your measured R1 + Rn is correct in relation to the design length of 48m.

 

[Lister1987]

Where have you got 1.04 and 0.63 ohm from.
You need to verify your measured R1 + Rn is correct in relation to the design length of 48m.

The 1.04 and 0.63 are the poster doubling the R1+Rn value, presumably thinkibg it was a single conductor reading (R1 or Rn)

 

[Rather be Fishing]

Where have you got 1.04 and 0.63 ohm from.
You need to verify your measured R1 + Rn is correct in relation to the design length of 48m.

I have no Idea I was wondering that to

 

[westward10]

I could be wrong here but your known design facts are length, volt drop and design current. R1 + Rn is your measured figure so I believe that once you have established this is correct you can double check it all meets the volt drop criteria (3%).

 

[Rather be Fishing]

Where have you got 1.04 and 0.63 ohm from.
You need to verify your measured R1 + Rn is correct in relation to the design length of 48m.

I have no Idea I think I might have just been having ago at it having been given the question in a lesson

Looking at my notes it gives me a couple of worked out example's. One using method1 and the other using method2

I seem to remember being told method 1 is the easier of the two methods so I'm going to go with that one

After linking the line and neutral in the distribution board, a cold continuity reading of 0.78ohms was obtained from a circuit with a design current (Ib)of 14A, wired using standard twin and earth cable.

Using table B3 the following calculation is performed

R1=Rn = 0.78 ohms x 1.20 = 0.936 ohms

using ohms law the voltage drop is established as

Voltage drop = 14 x 0.96 = 13.10v

 

[Rather be Fishing]

You are performing the initial verification on an installation but are concerned with the length of the lighting circuit conductors. Because of your concerns you have decided to verify the circuits voltage drop. The circuit is wired in 1.5mm2 conductors and runs for a circuit length of 48m.

You have measured the cold R1+Rn value for the circuit and found it to be 1.18ohms.

The designer has advised the calculated voltage drop for the circuit is 4.3V

If the design current is 3.04A, does the circuit comply with the design requirements?

So

R1 Rn = 1.18ohms x 1.04 (because they are singles so not incorporated in a cable) = 1.2272

voltage drop = 3.04 x 1.2272 = 3.730688V

The designer said volt drop allowed is 4.3V

so yes it complies as its 3.730688V

 

[westward10]

Or use vd = (R1 +Rn ) × Ib × multiplier.

 

[Rather be Fishing]

Or use vd = (R1 +Rn ) × Ib × multiplier.

Thank you @westward10 going to jot that down

 

[Rather be Fishing]

Im trying to use my regs book to find the answer to this question but I'm struggling to hone in on it. Would anyone be able to point me in the right direction so I can find the answer for myself?

Periodic Inspection and testing of a mobile catering unit is to be carried out as a requirement of the clients insurer?

The supply is from a 3KVA 230V single - phase portable generator and is connected as a TN-S system with fault protection provided by BS EN 61008 30mA RCD

What is the maximum test current applied to the RCD to confirm fault protection is provided.

a)15 mA
b)30 mA
c)60 mA
d)300 mA

I would guess d but I would prefer to see it in the reg's book instead of going with a number that's in my head for some reason. If 150 mA would have been an option id have been in a dilemma if I was to guess at it.

 

[westward10]

You need to understand with regard to a 30ma rcd the difference between fault and additional protection. Knowing this will give you the answer.