- #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)

##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

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

Where ##R_l## is the lead resistance

and ##R_f## is the final reading resistance

##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

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