Calculating current in an unbalanced wheatstone bridge?

The-Steve · Oct 25, 2015

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I am having troubles understanding how to calculate the current at RSensor when RSensor is 110 Ohms. Any help would be appreciated.

anorlunda · Oct 25, 2015

Here is your procedure. ((100+100) in parallel with (110+100))+270+270 is the total resistance. Can you do the rest?

drvrm · Oct 25, 2015

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!

Babadag · Oct 29, 2015

Circuit schematic of an unbalanced Wheatstone bridge: a 12 V source feeds RX=270 ohm to node X, with R1=100 ohm and Rsensor=110 ohm in the upper arm, R2=100 ohm and R4=100 ohm in the lower arm, an ammeter Ig between VA and VB, and RY=270 ohm to 0 V at node Y, with currents I1, Ix, I2, I4 and Ig labelled.

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

Babadag · Oct 29, 2015

Neglecting Rg[Rg=0] Ix=0.008895 A

The Electrician · Oct 30, 2015

Using the loop method, only 3 equations are needed. Choosing the loop currents I1, I2 and I3 like this:

Circuit schematic of an unbalanced Wheatstone bridge with three marked loop currents I1, I2 and I3, an ammeter with resistance Rg between nodes VA and VB, and resistor values R1=100, R2=100, R4=100, RX=270 ohm, Rsensor=110 ohm, RY=270 ohm, and a +12 V supply and 0 V reference.

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:

Mathematica output: LinearSolve of a 3x3 matrix [740, -200, -100; -200, 410, 200; -100, 200, 200+Rg] against vector (12, 0, 0), with solution vector I1, I2, I3 given as rational expressions in Rg.

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.

Calculating current in an unbalanced wheatstone bridge?

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