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

[James889]

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?

 

[CEL]

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.

 

[James889]

So i have
2000i_1 = 18

Which gives i1 = 9mA

And thus

24mA(1000+R_x) = 18

But 18/24mA = 750

Subtracting 1000 gives Rx = -250

 

[woomerp]

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

 

[vk6kro]

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.

 

[James889]

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?