Nonlinear Relation in Wheatstone Bridge Experiment

Click For Summary
SUMMARY

The discussion centers on the nonlinear relationship between current (I) and variable resistance (Rₓ) in a Wheatstone Bridge experiment. While the circuit is linear, the relationship between I and Rₓ is not, except for small variations where linear approximation can be applied. The derived formula for the galvanometer current (I_G) is $$I_G = \frac{(3.6 - 2R_X)V}{(6+a)R_X+1.2a}$$, with specific values of V = 14.4 V and a = -6.0 kΩ leading to a simplified linear relation $$I_G = 4R_x - 7.2$$. However, the negative value of a indicates inconsistencies in the provided data.

PREREQUISITES
  • Understanding of Wheatstone Bridge principles
  • Familiarity with linear and nonlinear circuit analysis
  • Knowledge of Ohm's Law and Kirchhoff's laws
  • Ability to perform algebraic manipulations with circuit equations
NEXT STEPS
  • Study the principles of linear and nonlinear circuit behavior
  • Learn about the superposition theorem in circuit analysis
  • Explore the implications of resistance in galvanometer circuits
  • Investigate common sources of data inconsistency in experimental setups
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in experimental physics or electrical measurements will benefit from this discussion.

jezza10181
Messages
13
Reaction score
1
Homework Statement
I have been trying to answer a question that I found on a schools homework sheet. I managed to answer part a), regarding the lengths L1 & L2.

As for the second half of part a), then as far as I can see, there isn't enough information there in order to produce a formula that would give 'I' in terms of Rx. I tried some Kirchoff analysis on it, but that gives equations containing values that the question doesn't supply.

Do you simply make the assumption that the two are linearly related and then find the three points in question by plugging in the other two into y = mx + c ?
Relevant Equations
V = IR
Capture.PNG
 
Physics news on Phys.org
jezza10181 said:
Do you simply make the assumption that the two are linearly related and then find the three points in question by plugging in the other two into y = mx + c ?
I think that your intuition is sound here. It's a linear circuit with linear components so I would think that the superposition principle would apply. Hence one would expect that a change in component value would have a linear effect upon other circuit values such as voltages and currents.

So yes, I would plot the given currents against Rx values and join them with a straight line, then read off the currents for the "new" Rx values (or calculate them if you want to do the algebra) .
 
  • Like
Likes jezza10181
It’s a poor question because I and Rₓ are not linearly related. Linearity is just an approximation that can be made for small variations in Rₓ.

(Note on terminology. The circuit itself is ‘linear’ or a ‘linear network’. This means the individual component values (resistances here) are constant, not functions of voltage or current. But this doesn’t mean I and Rₓ are linearly related.)

With the given data in the table, the changes in Rₓ are small. So we can approximate the relationship between I and Rₓ as being linear over the small range covered by the table.

You can fill in the current for Rₓ = 1.8kΩ straight away because it must be 0 (it’s the balance point). You can then see the current increases in steps of 0.4mA for each increase of Rₓ by 0.1kΩ.

(To fill in the blanks does not require a graph, but merely to add or subtract 0.4mA to/from values already in the table.)
 
  • Like
Likes jezza10181
I get an inconsistency with the choice of numerical values in this problem. After a significant amount of algebra, I find the following relation between the current ##I_G## in the Galvanometer and the variable resistance ##R_X##: $$I_G = \frac{(3.6 - 2R_X)V}{(6+a)R_X+1.2a}$$ where $$a = 5R_G + 2R_1$$ ##R_G## is the resistance of the galvanometer and ##R_1## is the resistance of the length ##L_1## of wire. All resistances are in kΩ, ##V## is in volts, and ##I_G## is in mA. I took positive values of ##I_G## to correspond to current in the downward direction through the galvanometer.

So, in general, the relation between ##R_X## and ##I_G## is nonlinear, as pointed out by @Steve4Physics

The values of ##V## and ##a## can be determined from the data given in the table. I find that ##V = 14.4 V## and ##a = -6.0## kΩ. These values simplify the expression for ##I_G## to $$I_G = 4R_x - 7.2$$ So, we end up with a linear relation. However, the negative value of the constant ##a## is inconsistent with the relation ##a = 5R_G + 2R_1## which certainly requires ##a## to be positive. So, unless I made a mistake somewhere, it appears that the data given in the problem is inconsistent.
 
  • Like
Likes jezza10181

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
1
Views
3K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 2 ·
Replies
2
Views
3K