# Varley Bridge

#### David J

Gold Member
Thread moved from the technical forums, so no Homework Template is shown
I am currently working on a module for the practical application of DC bridges. The particular problem I have at the minute involve "show that" questions.
The first question is in 3 parts and reads as follows: -
A Varley Bridge is connected to a faulty three core copper cable by two identical copper leads of resistance $R_l$
(a)
Show for the initial reading (connection to earth) that;

$2R_x = 2R_c - R_i$.........................(1)

Where $R_c$ is the resistance of the cable core
$R_i$ is the initial reading of the bridge
$R_x$ is the cable resistance to the fault from the bridge

Then, for the final reading show that;

$2R_c = R_f - 2R_l$...........................(2)

Where $R_l$ is the lead resistance
and $R_f$ is the final reading resistance

Then by substituting (2) into (1) and re arranging the equation, show:

$R_x +R_l = \frac {R_f - R_i}{R_f}(R_c + R_l)$

For the first part I created this answer:-

The Varley Bridge is balanced when: -

$\frac{R_a}{R_b}=\frac{(2R_c - R_x)}{R_x+R_i}$

$R_a = R_b$ so $\frac{R_ a}{R_b}$ must = 1

So 1 = $\frac{(2R_c - R_x)}{R_x + R_i}$ so $R_x +R_i = 2R_c - R_x$

So $R_x + R_x +R_i = 2R_c$ or $2R_x +R_i =2R_c$

So $2R_c =R_f + 2R_l$ hence proven

For the second part I created this answer:-

$2R_c$ + the Ohmic value of the leads, needs to = $\frac{R_a}{R_ b}$

So $\frac{R_a}{R_b} =\frac{2R_c + 2R_l}{R_f}$ So $R_f = 2R_c + 2R_l$

So $2R_c = R_f + 2R_l$ hence proven

Could someone please comment on the 2 answers I have provided above and let me know if the approach is correct. I think I have shown how both equations are created as requested in the question.

Regarding the 3rd part of the question I have struggled a bit with this and this is where I require some assistance. I have been shown an answer but I dont understand how it is achieved.

I know equation 1 = $2R_x =2R_c - R_i$
I know equation 2 = $2R_c =R_f -2R_l$

$2R_c =R_f - 2R_l \Rightarrow R_x + R_l = 2R_c - R_i - R_x + R_l$.................(3) I do not understand how this is created ???
$2R_x =2R_c - R_i \Rightarrow R_c + R_l = 2R_x + R_i - R_c + R_l$.................(4) I do not understand how this is created ???
I dont understand how $R_x + R_l$ and $R_c + R_l$ are included in equations 3 and 4. The first thing I need to understand is how we get to equations (3) and (4)

Once I understand the above I think I need to divide equation 3 by equation 4 so:-

$\frac{R_x + R_l}{R_c +R_l}=\frac{(2R_c -R_i)-R_x+R_l}{(2R_x +R_i)-R_c +R_l}$

Which can equate to $\frac{R_x + R_l}{R_c + R_l} = \frac{2R_x -R_x +R_l}{2R_c -R_c +R_l}$

At this point I am stuck and need a little bit of guidance.

Would someone be able to comment on the first 2 answers above and also advise on the answer to part 3 so far, am I on the right track or not ? If not where have I gone wrong, etc

Appreciated as always

Thanks

Dave

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First make a schema like this:

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Sorry, this could be better:

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#### David J

Gold Member
Hi, thanks for helping with this. I know that the reading on the Galvanometer needs to be zero so looking at the diagram above the variable resistance $R_i , R_f$ for reading 2 will play some part in the answer. My problem is trying to show that $R_x +R_l = \frac {R_f - R_i}{R_f}(R_c + R_l)$

This is whats confusing me

You almost found it:
From equation 2*Rc=Rf−2RL you'll get (Rc+RL)/Rf=1/2
Find now Rx + RL

2*Rx=2*Rc-Ri ?
2*RL=Rf-2*Rc ?

#### David J

Gold Member
I know equation 1 = $2R_x =2R_c - R_i$
I know equation 2 = $2R_c =R_f -2R_l$

$2R_c =R_f - 2R_l \Rightarrow R_x + R_l = 2R_c - R_i - R_x + R_l$..........(3) I do not understand how this is created ???
$2R_x =2R_c - R_i \Rightarrow R_c + R_l = 2R_x + R_i - R_c + R_l$........(4) I do not understand how this is created ???
I dont understand how $R_x + R_l$ and $R_c + R_l$ are included in equations 3 and 4. The first thing I need to understand is how we get to equations (3) and (4)

Can you advise on the above please ?? I feel if I can understand this I can get this,

equation 1 = 2Rx=2Rc−Ri
equation 2 = 2Rc=Rf−2Rl then 2RL=Rf-2Rc
2(Rx+Rl)=Rf-Ri then Rx+Rl=(Rf-Ri)/2=(Rf-Ri)x1/2
But from equation 2*Rc=Rf−2RL you'll get (Rc+RL)/Rf=1/2 so:
Rx+Rl=(Rf-Ri)/2=(Rf-Ri)x1/2=(Rf-Ri)*(Rc+RL)/Rf=(Rf-Ri)/Rf*(Rc+RL)

#### David J

Gold Member
I think I have managed to understand this. This is my answer: -

equation 1 is $2R_x =2R_c -R_i$
equation 2 is $2R_c = R_f -2R_l$

Substitute 2 into 1 gives me

$2R_x =R_f -R_i -2R_l$

so $0.5(R_f - R_i) = R_x + R_l$

Now $2R_c =R_f - 2R_l$ can be re arranged as $0.5R_f = R_c + R_ l$

So $\frac{R_x + R_l}{R_c + R_l} = \frac{0.5(R_f - R_i)}{0.5 R_f} or \frac{R_f -R_i}{R_f}$

So $R_x + R_l = \frac{R_f - R_I}{R_f} * (R_c +R_l)$

Last edited:

#### David J

Gold Member
$R_x + R_l = \frac{R_f - R_i}{R_f} * (R_c +R_l)$

Moving on from the above equation I have to show that: -

$R_x = \frac{R_f -R_i}{R_f}*R_c -\frac{R_l - R_i}{R_f}$

So

$R_x + R_l = \frac{R_f - R_i}{R_f} * (R_c -R_l)$

Step 1 $R_x + R_l = \frac{(R_f R_c+R_f R_l) - (R_i R_c - R_i R_l)}{R_f}$

Step 2 $R_x + R_l = \frac{(R_f R_c+R_f R_l) - (R_i R_c - R_i R_l)}{R_f}$

Step 3 $R_x = \frac{(R_f R_c+R_f R_l) - (R_i R_c - R_i R_l)}{R_f}-R_l$

Step 4 $R_x = \frac{R_f R_c - R_i R_c - R_i R_l}{R_f}$

Step 5 $R_x = \frac{R_c(R_f - R_i)}{R_f} - \frac{R_i R_l}{R_f}$

I think this is correct, I would just like confirmation and if this is not correct please point out my mistake.

Thanks again

Last edited:

Rx=(Rf-Ri)/Rf*Rc+[(Rf-Ri)/Rf-1]*Rl

#### David J

Gold Member
what does this relate to please?

$R_x = \frac{(R_f - R_i)}{R_f}*R_c +\left[\frac{(R_f -R_i)}{R_f-1}\right]*R_l$

Should this arrangement have been included in the above answer?

Last edited:

Rx=(Rf-Ri)/Rf*Rc+(Rf-Ri)/Rf*Rl-Rl=(Rf-Ri)/Rf*Rc+[(Rf-Ri)/Rf-1]*Rl

I'm sorry I'm late. It seems your result is very correct.

#### David J

Gold Member
Hello again, thanks for your help with this. I have submitted my assignment today and will close this post.

Much appreciated

thanks

"Varley Bridge"

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