Wheatstone Bridge and battery

In summary, the conversation is about a problem involving a wheatstone bridge circuit with a 40 volt battery, two unknown resistors, and an ammeter showing a current of 3 mA. The participants discuss various methods for solving the problem, including using Kirchoff's rules and breaking down the circuit into series and parallel components. Ultimately, they determine that the key to solving the problem is realizing that it is a combination series parallel circuit.
  • #1

wjf0

We have been doing a lot of work with wheatstone bridges in class lately and our teacher gave us a problem to try and I'm totally lost. The diagram that we were given has a 40 volt battery hooked to a wheatstone bridge circuit. The positive power flows into two resistors, one is 9 kiloohms and its opposite is unknown. The 9 kiloohm side then goes to a 5 kiloohm resistor and the unknown flows into a 12 after which the two sides rejoin and return to the 40 V battery. There is an ammeter across the bridge showing a current of 3 mA and we are asked to find the unknown resistance.

I've been working on this for about two hours and have come up with a lot of ideas that just don't work out. First I tried using kirchhof's rules but found that I had not voltage across the ammeter part of the bridge, so I tried assuming zero volts and came up short. Using my matrix math I came up with nothing.

I also tried solving the bridge as though it was balanced and figured out the overall total current and tried to use the 3 mA given to try and figure out what the four currents would be when its not balanced.

All the work we have done in class so far has been with balanced bridges or with a voltage across the bridge. This is the first time we have seen an unbalanced bridge with a current value. I'm totally lost and any push in the right direction would be great. Thanks for taking the time to read about my problem.
 
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  • #2
wjf0,
Welcome to PF !

Do you have a preferred method for dealing with this circuit?
Is the 3mA flowing from the unknown side to the junction of the 9k and 5k resistors? ...and if so;

Have you considered that the entire 3mA would be passing through the 5k resistor? Do you think this might mean the junction on the right hand side must be 15 volts higher than the one on the left, regardless of whatever the actual values might be?
 
  • #3
3.22 K?
 
  • #4
Thanks for your response. One the diagram that we were given, we were not told the direction of the current. I had thought about what you said before, but I found myself wondering if there was any current coming from the 9K resistor that needed to be taken into account. As for a preferred method, I really don't have any because at this point I'm simply looking for any new ideas.
 
  • #5
Thanks for your response. One the diagram that we were given, we were not told the direction of the current.
Are you certain there was not so much as a circle with and arrow inside (denoting a current flow and the direction thereof)?
I had thought about what you said before, but I found myself wondering if there was any current coming from the 9K resistor that needed to be taken into account.
Yes, you are correct. It is being fed by two paths. My hint was only to show what the current from one of the paths would mean.
As for a preferred method, I really don't have any because at this point I'm simply looking for any new ideas.
You can't break down this circuit into a simple parallel or series diagram but there is more than one way to work it. I'm pressed for time today but will try to walk you through it later in the day if I'm able...assuming someone else doesn't beat me to it.
Good luck
 
  • #6
WJ, this prob is solvable using Kirchoff rules (Sum(I)=0 in any junction , and Sum(V)=0 around any closed loop). Label currents in each branch (and you have 6 branches), get system of 7 equations and it is not hard to solve. Note that voltage on ammeter is zero. Don't worry about actual current directions - in your answers some of them will have negative sign if you labeled directions opposite, no big deal.
 
  • #7
Alexander,
I tried to use kirchhof's rules, but I only see four branch points (the four corners of the wheatstone bridge). Where are you getting the other 2 from? Using the four branch points I developed four equations to solve for the four unknown currents in the resistors. I assumed zero volts in the ammeter and used matrix algebra to simultaneously solve the equations. All the calculator gave me was three equations in terms of the current going through the unknown resistor and there was no way to simplify it down to get the unknown current.
 
  • #8
You have 4 nodes and 6 branches (4 sides of bridge and 2 diagonals).
 
  • #9
We were taught in class that a branch point occurs where three conductors intersect. This only happens at four points on the diagram we are given. What do you mean by a node, we've never discussed them and I don't see what you mean when I look at the diagram.
 
  • #10
Node = junction of branches (where current splits). Branch = elements connected in series from one node to another (same current via all elements).
 
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  • #11
It has been a long time since I messed with the wheatstone bridge. I do remember though that the key that helped me was to realize that it is a combination series parallel circuit. You have to break both down and then recombine them to find the difference of potential in the middle.
http://www.broadcast.net/hallikainen/theory6.html [Broken]
 
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1. What is a Wheatstone Bridge and how does it work?

A Wheatstone Bridge is a circuit used to measure an unknown electrical resistance by balancing two legs of a bridge circuit. It works by using a known resistance and an adjustable resistor to create a balanced circuit, which indicates that the two unknown resistances are equal.

2. How is a Wheatstone Bridge used to measure battery voltage?

A Wheatstone Bridge can be used to measure battery voltage by connecting the positive and negative terminals of the battery to two opposite corners of the bridge, and connecting a known resistance and an adjustable resistor to the other two corners. When the bridge is balanced, the ratio of the known resistance to the adjustable resistor can be used to calculate the battery voltage.

3. Can a Wheatstone Bridge be used with any type of battery?

Yes, a Wheatstone Bridge can be used with any type of battery as long as the voltage of the battery is within the range of the known resistance used in the circuit. However, the bridge may need to be adjusted for different battery chemistries, as they may have different internal resistances.

4. What are the advantages of using a Wheatstone Bridge to measure battery voltage?

One advantage of using a Wheatstone Bridge is that it provides a more accurate measurement of battery voltage compared to other methods, which may be affected by the internal resistance of the battery. It also allows for easy adjustment and calibration of the circuit to ensure accurate readings.

5. Are there any limitations to using a Wheatstone Bridge to measure battery voltage?

One limitation of using a Wheatstone Bridge is that it requires a known resistance, which may not always be readily available. Additionally, the accuracy of the measurement may be affected by external factors such as temperature and the quality of the components used in the circuit.

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