- #1
The-Steve
- 8
- 1
Calculating current in an unbalanced wheatstone bridge?
- Thread starter The-Steve
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Calculating the current at RSensor in an unbalanced Wheatstone bridge involves using Ohm's Law and Kirchhoff's loop equations. The total resistance is determined by combining resistances in parallel and series, leading to a potential difference of 2V across the circuit. The current through RSensor is approximately 2/55 A, while the current in the other arm is about 21/550 A. By setting up equations based on the loop currents and resistances, the current through RSensor can be calculated as 8.895 mA when the ammeter resistance is negligible. If the ammeter is absent, the current increases slightly to 9.112 mA.
- #2
- 11,326
- 8,754
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
- 611
- 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
- 611
- 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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