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Varley Bridge

  • Thread starter David J
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  • #1
David J
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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)##

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 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
 

Answers and Replies

  • #4
David J
Gold Member
137
14
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
 
  • #5
429
119
You almost found it:
From equation 2*Rc=Rf−2RL you'll get (Rc+RL)/Rf=1/2
Find now Rx + RL
 
  • #6
429
119
2*Rx=2*Rc-Ri ?
2*RL=Rf-2*Rc ?
 
  • #7
David J
Gold Member
137
14
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,
 
  • #8
429
119
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
David J
Gold Member
137
14
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:
  • #10
David J
Gold Member
137
14
##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:
  • #11
429
119
Rx=(Rf-Ri)/Rf*Rc+[(Rf-Ri)/Rf-1]*Rl
 
  • #12
David J
Gold Member
137
14
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
429
119
Rx=(Rf-Ri)/Rf*Rc+(Rf-Ri)/Rf*Rl-Rl=(Rf-Ri)/Rf*Rc+[(Rf-Ri)/Rf-1]*Rl
 
  • #14
429
119
I'm sorry I'm late. It seems your result is very correct.:smile:
 
  • #15
David J
Gold Member
137
14
Hello again, thanks for your help with this. I have submitted my assignment today and will close this post.

Much appreciated

thanks
 

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