View the thread, titled "2395 estimating r1 + r2" which is posted in Periodic Inspection Reporting & Certification on Electricians Forums.

H

Hogshaw

Question in 2395 past paper need help please.

if someone can show me how they worked this out

a circuit wired in 2.5mm/1.5mm twin with CPC sheathed copper cable has an r1 + r2 reading of 1.33 ohms, the value of r2 measured separately would be

1. 0.83
2.0.5
3.1.0
4.0.33

all in ohms

thanks
 
So just using the 1.67 to 2.67 ratio in the former equation multiplied by 6 and divided by 2 to give a fraction of 5 divided by 8 to use as a factor on the given value of resistance to resolve the answer.

Sorry... just being silly!
 
Because the earth is 1.5mm² and the line 2.5mm² the earth will have 1.67 times greater resistance than the line.

The total resistance is R1+R2, but because R2 will be 1.67 * R1 the total resistance will be 2.67 * R1 (1 R1 plus 1.67 R1).

So the resistance of the earth is the total resistance multiplied by (1.67 / 2.67) and this is equivalent to 5/8 (if you multiply both sides by 3).

Maths wise

R1 = R2 / 1.67
or
R2 = R1 * 1.67

Rt = R1 + R2

Substituting for R2
Rt = R1 + (1.67* R1)
Rt = 2.67 * R1

Since R1 = R2 / 1.67

Rt = 2.67 * (R2 / 1.67) = 2.67/1.67 * R2
So R2 = 1.67/2.67 * Rt
1.67 * 3 = 5
2.67 * 3 = 8

So R2 = 5/8 *Rt
 
where 5/8 expressed as a decimal is 0.625, as in a previous post.
 
Ignore my post number 5

If you cannot see where the 2.67 comes from by the use of post 8 then I am afraid I cannot assist.

total resistance is R1 + R2
R2 is 1.67 * R1 so you have

total resistance is R1 + (1.67 * R1)

therefore you have a single R1 plus 1.67 R1
1 + 1.67 = 2.67

so you have 2.67 R1
 

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