C&G Level 2 Diploma Query

[KeenPensioner]

How sad am I! Here in Scotland on New Year's eve I'm wading my way through the theory of "Testing Continuity of Protective Conductors" - Method 2 (Long Lead) from the C&G Level 2 Textbook on page 384.
The question is:
"The resistance of a long lead is measured at 0.89 ohms. The low-resistance meter gives a reading of 1.63 ohms. What will the value of R2 be?"
The answer given is given as 0.9 Mega Ohms. Surely this is wrong or am I missing something?

 

[Strima]

Way off. What do we measure in Mohms?

The question is a lot easier than you think.

 

[Madocks]

Zero the leads

 

[westward10]

Your reading I assume is the product of R2 and your wander lead.
Testing continuity of protective conductors is not done using Meg ohms.

 

[Amp David]

Very wrong, just use some basic maths.

 

[polo1]

Yep, the answer given ( in your book) appears to be wrong.
It appears to suggest that the resistance of the long lead itself has not been removed from the measurement by zeroing the leads.
So answer should probably be 1.63 - 0.89 ohms.

 

[KeenPensioner]

Thank you all very much for your replies...especially on New Year's eve. To give an answer to Strima's question - we measure insulation resistance in Mega Ohms. I figured the answer to the question resistance should be the difference between the reading given for the protective conductor and the resistance of the long leads. Wouldn't you normally just "null" the leads? In any case, I'm disappointed that the book has published the wrong answer (image of the answer attached) Should we let them know?

 

[Strima]

Let them know that the answer is wrong.

So what is the correct answer?

 

[telectrix]

Let them know that the answer is wrong.

So what is the correct answer?

42

 

[Christopher Bernard]

I used a book to study for my 2394 (practical guide by Christopher Kitcher) and I came across incorrect questions and answers as well as several typos, I guess the concept of proof reading evades some.

At least you spotted the error, which is good.

 

[KeenPensioner]

In response to Strima #8, polo1 gave the answer in #6. Again, thank you all. Now I move on to Prospective Fault Current!!

 

[telectrix]

In response to Strima #8, polo1 gave the answer in #6. Again, thank you all. Now I move on to Prospective Fault Current!!

that's quite simple calc using ohm's law. I=V/R.

 

[willy fixit]

Your reading I assume is the product sum of R2 and your wander lead.
Testing continuity of protective conductors is not done using Meg ohms.

Sorry to be pendantic!