loop impedance, temperature, cable factor....

[Gary Tollison]

Hello
I'm not sure if this is the correct forum area to post this in, apologies if it isn't...

I've been working through some mock questions on-line, and found myself confused by this:

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

I just cant get my head around this!

Any help would be great!

Thanks in advance,

Gary

 

[Risteard]

Basically the temperature correction won't apply to the part external to the installation but only to your cables within the installation. So apply your temperature correction to this part and then you have your adjusted value. (I don't want to just give you the answer.)

 

[The Ghost]

Apply correction factor

 

[telectrix]

Apply correction factor

if you do, none of the answers given are correct.

 

[westward10]

They were when I did it.

 

[Ian1981]

Hello
I'm not sure if this is the correct forum area to post this in, apologies if it isn't...

I've been working through some mock questions on-line, and found myself confused by this:

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

I just cant get my head around this!

Any help would be great!

Thanks in advance,

Gary

Apply the correction factor stated, to your 0.83 value

 

[static zap]

....if you do, none of the answers given are correct.

A skilled Multichoice question writer ,has miss directed you ....
(but wasn't mean enough to give you -his trap answer !)

 

[westward10]

Apply the correction factor stated, to your 0.83 value

Not directly to it.

 

[telectrix]

They were when I did it.

show us. i'm intrigued.

 

[Ian1981]

Not directly to it.

Apply the 1.2 to the R1+R2 value and add it to the Ze assuming R1+R2 is 0.43
Naughty little question

 

[static zap]

(I'm curently humming undergroung /over ground wombling free .)
(10), 20 deg and 70 deg

 

[westward10]

show us. i'm intrigued.

You apply the factor to the circuit impedance from the board which is why you have Ze then add it back on after the calculation.

 

[The Ghost]

On my calculator 1.2 x 0.83= C

 

[Ian1981]

It's 0.43 x 1.2 = 0.516+ 0.4= 0.916 D

 

[telectrix]

Apply the 1.2 to the R1+R2 value and add it to the Ze assuming R1+R2 is 0.43
Naughty little question

yeah, but by definition, the loop impedance of a circuit includes Ze.

 

[Ian1981]

yeah, but the loop impedance of a circuit includes Ze.

The external impedance is not subject to the 1.2 correction factor
When calculating R1+R2 using 20 degrees tables during the design you apply the 1.2 correction factor to compensate for 70 degree operating temperature
External ze is not a factor during this

 

[Risteard]

yeah, but by definition, the loop impedance of a circuit includes Ze.

But the temperature correction factor does not apply to the external part of the loop.

 

[Richard Burns]

The generally accepted method of correcting for temperature is to apply it to the whole of the Zs for the circuit, however this does not , as said above provide an accurate assessment as the external supply may not be running at 70°C.
In that case the separation of Ze and R1+R2 may be undertaken and the correction factor applied to the R1+R2 only and the uncorrected Ze applied.
In the case of the question the method of calculation is not clarified.
Also the correction is from ambient (20°C) to 70°C rather than the usual correction of the tabulated value down to 20°C.
Basically you can apply the 1.2 correction factor to the 0.83Ω and get 0.996Ω
or you can apply the correction factor just to the back calculated R1+R2 (which should not be done) of 0.43Ω and add it to the Ze of 0.4Ω and get 0.916Ω.

Because the question specifies Ze you might assume they mean the more accurate correction but since it requires an unapproved calculation it seems odd.

 

[telectrix]

The generally accepted method of correcting for temperature is to apply it to the whole of the Zs for the circuit, however this does not , as said above provide an accurate assessment as the external supply may not be running at 70°C.
In that case the separation of Ze and R1+R2 may be undertaken and the correction factor applied to the R1+R2 only and the uncorrected Ze applied.
In the case of the question the method of calculation is not clarified.
Also the correction is from ambient (20°C) to 70°C rather than the usual correction of the tabulated value down to 20°C.
Basically you can apply the 1.2 correction factor to the 0.83Ω and get 0.996Ω
or you can apply the correction factor just to the back calculated R1+R2 (which should not be done) of 0.43Ω and add it to the Ze of 0.4Ω and get 0.916Ω.

Because the question specifies Ze you might assume they mean the more accurate correction but since it requires an unapproved calculation it seems odd.

that was my interpretation. correction factors are usually applied to the Zs value in BS7671 , not jusrt to the R1+R2 values.

by calcularting i the way the question implies, it would be possible to get a borderline value above the recommended value to below it.

 

[Richard Burns]

that was my interpretation. correction factors are usually applied to the Zs value in BS7671 , not jusrt to the R1+R2 values.

My first approach was to do just that as this is normally fine and is the first thing you would do and would only apply further accuracy if the value was out of range, but because they specify Ze, which is usually immaterial, I do wonder what they are going for.
Especially since they give both possible answers.