DC Varley Bridge: Solving "Show That" Problems

[David J]

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:

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

My answers

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 don't 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 don't 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

 

[Babadag]

First make a schema like this:

 

[Babadag]

Sorry, this could be better:

 

[David J]

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 what's confusing me

 

[Babadag]

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

 

[Babadag]

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

 

[David J]

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 don't 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,

 

[Babadag]

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]

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

 

[David J]

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

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

 

[Babadag]

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

 

[David J]

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?

 

[Babadag]

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

 

[Babadag]

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

 

[David J]

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

Much appreciated

thanks

 

[SamC789]

Hello,

I have also been tasked with this question for unit I am working towards. I can follow Davids calculations until step to step 5.

Would there be somebody who can explain how David has simplified Step 3 to obtain Step 4. Apologies my maths is not too strong but I am working on it. I don't expect an answer but if somebody could point me in a method I could research.

Thanks Sam

 

[berkeman]

Hello,

I have also been tasked with this question for unit I am working towards. I can follow Davids calculations until step to step 5.

View attachment 292906

Would there be somebody who can explain how David has simplified Step 3 to obtain Step 4. Apologies my maths is not too strong but I am working on it. I don't expect an answer but if somebody could point me in a method I could research.

Thanks Sam

Welcome to PF, Sam.

Convert that trailing stand-alone term to its equivalent divided by Rf so you can combine the fractions. Do you see how that cancels out one of the numerator terms in the left fraction?

 

[SamC789]

Welcome to PF, Sam.

Convert that trailing stand-alone term to its equivalent divided by Rf so you can combine the fractions. Do you see how that cancels out one of the numerator terms in the left fraction?

Thank you so much for this! I must admit I am a little annoyed I didn't see this myself, I was just not visualizing it correctly.