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.