# Calculating current in an unbalanced wheatstone bridge?

1. Oct 25, 2015

### The-Steve

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

2. Oct 25, 2015

### Staff: Mentor

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

3. Oct 25, 2015

### drvrm

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!

4. Oct 29, 2015

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. Oct 29, 2015

Neglecting Rg[Rg=0] Ix=0.008895 A

6. Oct 30, 2015

### The Electrician

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.

Last edited: Oct 30, 2015