DC Varley Bridge: Solving "Show That" Problems

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In summary: DaveIn summary, the conversation involved working on a module for the practical application of DC bridges. The problem at hand was related to "show that" questions in a Varley Bridge connected to a faulty three core copper cable by two identical copper leads of resistance ##R_l##. The first question had three parts, with the first part involving showing that ##2R_x = 2R_c - R_i## and the second part involving showing that ##2R_c = R_f - 2R_l##. The third part of the question required showing that ##R_x + R_l = \frac{R_f - R_i}{R_f}*(R_c + R_l)##. To solve this, it was necessary to substitute
  • #1
David J
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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
 
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  • #2
First make a schema like this:
upload_2018-7-18_21-40-30.png
 

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  • #3
Sorry, this could be better:
upload_2018-7-18_22-44-4.png
 

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  • #4
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
 
  • #5
You almost found it:
From equation 2*Rc=Rf−2RL you'll get (Rc+RL)/Rf=1/2
Find now Rx + RL
 
  • #6
2*Rx=2*Rc-Ri ?
2*RL=Rf-2*Rc ?
 
  • #7
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,
 
  • #8
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)
 
  • #9
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)##
 
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  • #10
##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
 
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  • #11
Rx=(Rf-Ri)/Rf*Rc+[(Rf-Ri)/Rf-1]*Rl
 
  • #12
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:
  • #13
Rx=(Rf-Ri)/Rf*Rc+(Rf-Ri)/Rf*Rl-Rl=(Rf-Ri)/Rf*Rc+[(Rf-Ri)/Rf-1]*Rl
 
  • #14
I'm sorry I'm late. It seems your result is very correct.:smile:
 
  • #15
Hello again, thanks for your help with this. I have submitted my assignment today and will close this post.

Much appreciated

thanks
 
  • #16
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.

1637676328410.png


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
 
  • #17
SamC789 said:
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?
 
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  • #18
berkeman said:
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.
 
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1. What is a DC Varley Bridge?

A DC Varley Bridge is a type of electrical circuit used to measure unknown resistances by comparing them with known resistances.

2. How does a DC Varley Bridge work?

The DC Varley Bridge works by balancing the unknown resistance with a known resistance in a Wheatstone bridge configuration. The bridge is balanced when there is no current flowing through the galvanometer, indicating that the two resistances are equal.

3. What are "show that" problems in relation to DC Varley Bridge?

"Show that" problems in relation to DC Varley Bridge refer to mathematical equations and calculations that can be solved using the principles of the bridge, such as finding the value of an unknown resistance.

4. What are some tips for solving "show that" problems using DC Varley Bridge?

Some tips for solving "show that" problems using DC Varley Bridge include carefully setting up the circuit, ensuring all connections are secure, and using the correct formula for calculating the unknown resistance. It can also be helpful to double-check calculations and use multiple trials to ensure accuracy.

5. What other applications does DC Varley Bridge have besides solving "show that" problems?

DC Varley Bridge has several other applications, including measuring the resistance of conductors and insulators, testing the accuracy of resistors, and determining the conductivity of materials. It is also used in industries such as electronics, telecommunications, and engineering.

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