Loop Impendance question

[duff030985]

Hi,

I am currently studying to become an electrician and keep seeing this question (normally different values) and I am not sure how to work this out. Would someone be able to let me know how you work this out?

The measured value of loop impedance for a circuit is 2.4Ω. If the temperature at the time of the test was 20°C and the cable is 70°C (factor 1.2) what is the corrected value. Ze = 0.4Ω:
a) 2.88 Ω
b) 2.8 Ω
c) 2.4 Ω
d) 2.0 Ω

 

[Julie.]

Removed as I realised it may be a test question

 

[timhoward]

@Julie is it me - but I don't like the question much.....!

As far as I can see, either
a) they want you to incorrectly assume that the supply will be rising from 20 degrees to 70 degrees too, and apply the correction factor to the whole measured Zs.
or
b) they want you to incorrectly assume that Measured Zs-Ze = R1+R2 , so deduct the Ze and then apply the correction factor. Obviously you'd never do this due to parallel paths.

I'll get my coat....

 

[westward10]

@Julie is it me - but I don't like the question much.....!

As far as I can see, either
a) they want you to incorrectly assume that the supply will be rising from 20 degrees to 70 degrees too, and apply the correction factor to the whole measured Zs.
or
b) they want you to incorrectly assume that Measured Zs-Ze = R1+R2 , so deduct the Ze and then apply the correction factor. Obviously you'd never do this due to parallel paths.

I'll get my coat....

You are correct in my opinion with b).

 

[Julie.]

@Julie is it me - but I don't like the question much.....!

As far as I can see, either
a) they want you to incorrectly assume that the supply will be rising from 20 degrees to 70 degrees too, and apply the correction factor to the whole measured Zs.
or
b) they want you to incorrectly assume that Measured Zs-Ze = R1+R2 , so deduct the Ze and then apply the correction factor. Obviously you'd never do this due to parallel paths.

I'll get my coat....

No I don't either, it can be read in two ways they say loop - are they meaning the 2.4 represents R1 + R2 ? In which case the answer is just looking to add R1+R2 +Re (and they are looking to compare the measured "loop" cold to cold max Zs of the protective device),

Or as you say, are they saying this is the Zs, so subtract the Re, to estimate R1 + R2, recalculate this for temperature change then add the Re back on?
(In which case they will have Zs[hot] and need to compare against max Zs [hot] of the protective device.)

It's one thing that annoys me about C&G - their questions aren't actually clear.

There is a reason we should use correct terminology and nomenclature.

If the question was: measured Zs at 20 degrees was 2.4 ohm, assuming Ze is 0.4 ohm, what would the corrected Zs be for a temperature of 70 degrees (1.2 factor)

Then it would determine if the person understands the actual point of the question without stupid ambiguity purely due to poor question writing.

 

[westward10]

@duff030985 has your tutor ever gone through this.

 

[Lucien Nunes]

And also @duff030985, why not post your working or your take on the question, so we can critique and point you in the right direction.

The working-out is easy if you understand why it needs doing. Why is temperature important? Why is it likely to be different during testing compared to operating conditions? What could be wrong with the figure for Zs if temperature is not taken into account?

 

[duff030985]

@duff030985 has your tutor ever gone through this.

Hi,

No the tutor hasn't gone through this with me. Its just a question I keep see coming up on different papers we get to test ourselves each week and I don't know how to work it out

 

[duff030985]

Hi @Lucien Nunes

And also @duff030985, why not post your working or your take on the question, so we can critique and point you in the right direction.

The working-out is easy if you understand why it needs doing. Why is temperature important? Why is it likely to be different during testing compared to operating conditions? What could be wrong with the figure for Zs if temperature is not taken into account?

Hi @Lucien Nunes,

I dont know where to start as I dont think i have enough infromation to work it out. The only thing i could think of was 2.4Ω - 0.4Ω = 2.0Ω but i dont think this is correct

 

[timhoward]

It's a poorly worded question. You are right that you would ideally have more information.

My understanding to hopefully set you in the right direction is:

1) They want you to understand that Zs = Ze + (R1+R2)
That is: the loop impedance measured at the end of a final circuit consists of the live coming from the supply transformer into the consumer unit, the live from the CU to the end of the circuit (R1), the CPC back to the main earth terminal (R2) and the path through real earth back to the supply transformer (which will vary according to the earthing system in use).

2) They have given you the measured value at the end (Zs) and the supply portion (Ze).
So exactly as you said, taking one from the other gives you the theoretical impedance of the final circuit only.
(there are reasons you wouldn't ideally do it like this in many real life situations, I'm ignoring them for now!)

3) They want you to understand that resistance changes with temperature; the test was at 20 degrees, and the cable will be at 70 degrees.
This can sometimes matter, if you apply ohms law you will see that if the resistance changes, the current will change, and ultimately the fault current needs to be enough to operate a fuse or circuit breaker.

4) There is a factor which they give you (1.2) which comes from a calculation contained in the On Site Guide to adjust a resistance for a temperature of 70 degrees when the ambient temperature is between 10 and 20 degrees.

So I believe it comes down to find the impedance of the final circuit (which you did already) , and apply the factor.
I hope that helps - keep asking questions if not.

(I'd also reassure you that there is usually a much easier way to handle this in many real life situations!)

 

[duff030985]

It's a poorly worded question. You are right that you would ideally have more information.

My understanding to hopefully set you in the right direction is:

1) They want you to understand that Zs = Ze + (R1+R2)
That is: the loop impedance measured at the end of a final circuit consists of the live coming from the supply transformer into the consumer unit, the live from the CU to the end of the circuit (R1), the CPC back to the main earth terminal (R2) and the path through real earth back to the supply transformer (which will vary according to the earthing system in use).

2) They have given you the measured value at the end (Zs) and the supply portion (Ze).
So exactly as you said, taking one from the other gives you the theoretical impedance of the final circuit only.
(there are reasons you wouldn't ideally do it like this in many real life situations, I'm ignoring them for now!)

3) They want you to understand that resistance changes with temperature; the test was at 20 degrees, and the cable will be at 70 degrees.
This can sometimes matter, if you apply ohms law you will see that if the resistance changes, the current will change, and ultimately the fault current needs to be enough to operate a fuse or circuit breaker.

4) There is a factor which they give you (1.2) which comes from a calculation contained in the On Site Guide to adjust a resistance for a temperature of 70 degrees when the ambient temperature is between 10 and 20 degrees.

So I believe it comes down to find the impedance of the final circuit (which you did already) , and apply the factor.
I hope that helps - keep asking questions if not.

(I'd also reassure you that there is usually a much easier way to handle this in many real life situations!)

Thanks for your help, I understand it all apart from the factor (1.2). What factor am i applying?

 

[Julie.]

Thanks for your help, I understand it all apart from the factor (1.2). What factor am i applying?

Resistance goes up with temperature.

Basically although there is a proper calculation, for the 18th exam it is just a couple of rules of thumb.

The estimated Resistance at 70 degrees = 1.2 x the Resistance at 20 degrees.

The estimated Resistance at 20 degrees = 0.8 x the Resistance at 70 degrees

 

[duff030985]

Resistance goes up with temperature.

Basically although there is a proper calculation, for the 18th exam it is just a couple of rules of thumb.

The estimated Resistance at 70 degrees = 1.2 x the Resistance at 20 degrees.

The estimated Resistance at 20 degrees = 0.8 x the Resistance at 70 degrees

Hi,

I think I understand it all now. So to get the correct answer do I do:

2.4 - 0.4 x 1.2 = 2.4

 

[Julie.]

Hi,

I think I understand it all now. So to get the correct answer do I do:

2.4 - 0.4 x 1.2 = 2.4

So think about it a little more, and spell it out

Loop impedance cold(Zs) = 2.4 ohm also = (R1 + R2) + Ze

Since Ze = 0.4

Cold(R1 + R2) = cold Zs - Ze (2.4 - 0.4)

So correct for temperature
Hot(R1 + R2) = 1.2 x cold(R1 + R2)

Is this what you were asked for?

Or were you asked for hot Zs ?

Always write down what the numbers represent

Seeing (12-1.2)/0.01 = 1080

Means what?

At this point you think you have an answer, but without identifying exactly what is the question you answer the wrong question. (In this case the question was what is the power dissipated in a series resistor on 12V supply with an LED voltage drop of 1.2V with a current of 10mA)

If you add what the values represent then you would get
(V supply - V LED)/ I LED = resistance
(12V - 1.2V)/0.01A = .....

Oops, resistance is not power, therefore it cannot be correct! - I see I have done something wrong myself even before checking the answer

In your case you have identified a value for a paticular resistance - but is it the correct resistance you were asked to find?

 

[David Prosser]

My guess is they are looking for this

Zs @20c =2.4
Ze=0.4
R1+R2=2.0
(R1+R2)1.2=2.4

Zs @70c 2.4+0.4=2.8