- #1
The-Steve
- 8
- 1
Calculating current in an unbalanced wheatstone bridge?
- Thread starter The-Steve
- Start date
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Discussion Overview
The discussion revolves around calculating the current at a sensor resistor (RSensor) in an unbalanced Wheatstone bridge circuit, focusing on the application of circuit analysis techniques such as Ohm's law and Kirchhoff's laws. Participants explore different methods and equations to derive the current values.
Discussion Character
- Technical explanation
- Mathematical reasoning
- Debate/contested
Main Points Raised
- One participant expresses difficulty in calculating the current at RSensor when its resistance is 110 Ohms.
- Another participant suggests a procedure for calculating total resistance in the circuit, indicating a parallel combination of resistances.
- A different participant describes the circuit's operation with a potential difference and applies Kirchhoff's loop equation to relate currents in the circuit, noting that the difference in current due to the small resistance difference is expected to be minor.
- One participant proposes a specific circuit configuration and outlines five equations involving various currents and resistances, indicating a more complex analysis of the Wheatstone bridge.
- Another participant simplifies the analysis by neglecting one resistance (Rg) and calculates the current through RSensor, providing a specific value.
- A later reply mentions using a loop method to derive equations in matrix form, yielding different current values based on the presence or absence of the ammeter resistance.
Areas of Agreement / Disagreement
Participants present multiple approaches and calculations, with no consensus on a single method or final answer. Different values for the current through RSensor are proposed, indicating ongoing debate and exploration of the problem.
Contextual Notes
Some calculations depend on assumptions about the resistances in the circuit, and participants express uncertainty regarding the interpretation of the original problem. The presence of different configurations and simplifications leads to varying results.
- #2
- 11,326
- 8,755
Here is your procedure. ((100+100) in parallel with (110+100))+270+270 is the total resistance. Can you do the rest?
- #3
drvrm
- 960
- 214
In your circuit the current is being fed from two terminals kept at say +V and -V volts so effectively you are driving the circuit by a potential difference of say 2V volts - the you apply the Ohms law and calculate the current in the two arms of a loop which does not have a source of EMF-so you can apply Kirchhoff's loop equation and can get a relation between the different current in the two arms. as the R-sensor differs by 10 ohms only from the other arm the difference in current will come out to be pretty small.
well i do not guess where lies the problem of calculation-
you have to use one relation P.D. (2V) = sum of Current X Resistances if you are traversing a path from one terminal to another;
and the other relation is in a looop sum of currentX resistances = the EMF of source =0
The two arms of the loop gives you equivalent resistance 1/Req = 1/200 + 1/ 210'
if you move on those lines the current in Rsensor comes to approx. 2/55 amp. and in other arm about 21/550 amp.
well i do not know whetheri could interpret your problem correctly!
well i do not guess where lies the problem of calculation-
you have to use one relation P.D. (2V) = sum of Current X Resistances if you are traversing a path from one terminal to another;
and the other relation is in a looop sum of currentX resistances = the EMF of source =0
The two arms of the loop gives you equivalent resistance 1/Req = 1/200 + 1/ 210'
if you move on those lines the current in Rsensor comes to approx. 2/55 amp. and in other arm about 21/550 amp.
well i do not know whetheri could interpret your problem correctly!
- #4
Babadag
- 614
- 162
In my opinion, the actual Wheatstone Bridge circuitry is as in attached sketch.
In this case, the unknown 5 parameters will be I1,I2,Ix[through Rsensor],I4,Ig[through Ammeter].
The 5 equations are:
I1-Ig-I2=0
Ix+Ig-I4=0
I1*R1+Ig*Rg-Ix*Rsensor=0
I2*R2-I4*R4-Ig*Rg=0
I1*R1+I2*R2+(I1+Ix)*RX+(I2+I4)*RY=12
- #5
Babadag
- 614
- 162
Neglecting Rg[Rg=0] Ix=0.008895 A
- #6
The Electrician
Gold Member
- 1,402
- 210
Using the loop method, only 3 equations are needed. Choosing the loop currents I1, I2 and I3 like this:
we then set up the 3 equations in matrix form like this, including the resistance of the ammeter Rg to obtain a more general solution:
I2 (green) is the same as Ix, and I3 (blue) is the current through the ammeter. Setting Rg = 0, we obtain Ix = 8.895 mA.
If the ammeter is missing (Rg = ∞), then Ix = 9.112 mA.
we then set up the 3 equations in matrix form like this, including the resistance of the ammeter Rg to obtain a more general solution:
If the ammeter is missing (Rg = ∞), then Ix = 9.112 mA.
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