Get Expert Help with Wheatstone Bridge Circuit Analysis - 15mV Voltage Given

Click For Summary

Discussion Overview

The discussion revolves around analyzing a Wheatstone bridge circuit with a specific focus on determining the unknown resistor value (Rx) given a voltage of 15mV across the terminals. Participants explore various methods of circuit analysis, including Kirchhoff's Current Law (KCL) and Ohm's Law, while addressing the relationships between the components in the circuit.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests writing a KCL equation for the loops containing R1 and R2, and R3 and Rx, to find the voltage at the nodes.
  • Another participant calculates a current (i1) and derives an expression for Rx, but arrives at a negative value, indicating a potential error in their approach.
  • A different participant questions the origin of a specific value (24) used in the calculations and proposes an alternative KCL equation for the second loop, leading to a different value for Rx.
  • Another contribution emphasizes using Ohm's Law to relate the voltages and currents in the circuit, suggesting that the voltage at the left center point should be half the supply voltage.
  • One participant expresses confusion about writing a loop equation for the bottom two resistors, questioning the validity of their approach.

Areas of Agreement / Disagreement

Participants express differing methods and calculations for determining Rx, with no consensus on the correct approach or final value. The discussion remains unresolved as various interpretations and calculations are presented.

Contextual Notes

Some calculations appear to depend on assumptions about the circuit configuration and the application of KCL, which may not be universally accepted among participants. The origin of certain values used in the equations is also questioned, indicating potential gaps in the reasoning.

James889
Messages
190
Reaction score
1
Hi,

I have this circuit:
[PLAIN]http://img813.imageshack.us/img813/1189/wheatstone.png

All resistors are 1K except Rx which is unknown.
The voltage across the terminals are given to be 15mV.

Im assuming i have to write a KCL equation but I am not sure how.

Any hints?
 
Last edited by a moderator:
Physics news on Phys.org
Write KCL for the loop containing R1 and R2 and find the voltage at the node between the two resistors. Do the same thing for the loop containing R3 and Rx (it will be a funtion of Rx).
The voltage difference between the two nodes is 15 mV.
 
So i have
[tex]2000i_1 = 18[/tex]

Which gives i1 = 9mA

And thus

[tex]24mA(1000+R_x) = 18[/tex]

But 18/24mA = 750

Subtracting 1000 gives Rx = -250:confused:
 
I don't see where the 24? came from.

The second KCL loop would be:
18 = i2(1000 + Rx)
If you write a loop for the bottom two resistors you can say:
15 mV = 9 mA (1000) - i2(Rx)
Then 8.85 = i2(Rx) → Rx = 8.85/ i2
Substituting back into the previous equation gets you:
18 = i2(1000 + 8.85/ i2)
18 = 1000 i2 + 8.85
i2 = 9.15 mA and Rx = 967
 
In this case, you can work the answer out with Ohm's Law.

You know the voltage at the left centre point has to be half the supply voltage (because R1 and R2 are equal) and the voltage at the right centre point has to be 0.015 volts less than this (relative to the negative supply terminal).

So, you can work out the current in the top right 1K resistor and hence work out the size of the bottom right resistor, which must have the same current and you know the voltage across it.
 
Last edited:
woomerp said:
I don't see where the 24? came from.

The second KCL loop would be:
18 = i2(1000 + Rx)
If you write a loop for the bottom two resistors you can say:
15 mV = 9 mA (1000) - i2(Rx)


Hm, so you can write a loop equation for the bottom two resistors even though they really aren't forming a loop?
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 14 ·
Replies
14
Views
24K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
5
Views
5K
Replies
15
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 9 ·
Replies
9
Views
6K
  • · Replies 1 ·
Replies
1
Views
17K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K