# Why the voltage acrossWheatstone's bridge isn't always zero?

1. Feb 14, 2015

### jack58

1. The problem statement, all variables and given/known data
Wheatstone bridge.

2. Relevant equations
i
Why the voltage across whetstone's bridge isn't always zero ? according to ohm's law . It must be zero (V=I*R, R=0) =>V=0 across the bridge .
3. The attempt at a solution
I think I'm missing something .
the following sentence isn't necessarily true : delta V=0 doesn't => current=0 . I could explain this according to newton's first law . because V=0=> E=0 => current could exist in a
Continuation state and flow with no help from external force( electric field).
and we actually deal with the resistance across the winston's bridge as ideal and equals to zero. and this means voltage across the bridge is always zero . By calibring the resistance . We are only trying to fix kerchiefs first rule in order to have current equal to zero across the bridge ?
I'm sorry ,I might not be clear because my english isn't good . But,in brief my problem is:

1) Do we deal with the resistance across Wheatstone bridge as ideal=0 or not?
2) In general if delta V=0 . Does this necessarily mean that current=0 ? in other words could there be a case that current flow when dela V=0 (maybe when R also equals to zero...)?

Last edited: Feb 14, 2015
2. Feb 14, 2015

### collinsmark

The voltage is only zero if the bridge is properly balanced. But if there is an imbalance, there is voltage across the center galvanometer, and thus current flowing through it.

Consider this circuit (From Wikipedia commons):

Here,
• $R_x$ is the unknown. You don't know what this resistance is, and that's the resistance you are looking for.
• $R_1$, $R_2$, and $R_3$ are known with high precision.
• $R_2$ is special among these, however, because it is variable. Even though it is variable, it is still known with high precision. You can adjust $R_2$, with a dial perhaps. But with every adjustment its resistance is still known with high precision (the dial may have high precision markings on it displaying the resistance of $R_2$).

With that you can derive a simple, high precision formula for the resistance of $R_x$, when the voltage across the galvanometer is zero. If the voltage (across the galvanometer) isn't zero, adjust $R_2$ until it is.

(I think you mean Kirchhoff, by the way.)

Assuming your component $R_x$ is linear, then sure, treat the circuit as ideal.

On the other hand, if the $R_x$ is nonlinear, and contains things like transistors and diodes, then this Wheatstone bridge isn't necessarily the best approach. It's good for linear circuits because then the results are independent of the battery voltage. But for nonlinear circuits, it's not so simple because the equivalent resistances are a function of the voltage placed across them.
It's a galvanometer across the middle. If you are dealing with DC circuits, then it can be modeled an ideal resistor. So yes, we can assume that I = 0 when V = 0.

Last edited: Feb 14, 2015
3. Feb 14, 2015

### Staff: Mentor

A galvanometer is a type of ammeter, usually a very sensitive one. As such it "looks like" a pretty good piece of wire at DC. So a Wheatstone bridge employing a galvanometer would almost by definition have zero or nearly zero volts across the bridge at all times. One balances the bridge to obtain a zero current through the galvanometer.

One can also use a voltmeter in place of the ammeter in the bridge circuit. Modern voltmeters can be even more sensitive than the old (pretty darn sensitive) galvanometers. Then the idea is to balance the bridge for zero potential difference.

The results are the same when balanced (no current, no potential difference), but the voltmeter version will show a non-zero voltage when the bridge is out of balance, while the galvanometer version will have essentially zero (very small) voltage but a non-zero current.

4. Feb 14, 2015

### jack58

Thank you for the fast replies . I read what u said quickly . I will read them more carefully when i have more time . Meanwhile i dont want the thread to die . Bare with me please : if we take Winstone circuit without galvonemeter/ammeter in the middle . (Just wire)And caculate the current in the middle . Do we get current=0? Delta V =0 also? Thanks in advance .

5. Feb 14, 2015

### Staff: Mentor

If the bridge is balanced then I = 0, otherwise there will be current if the bridge is not balanced. V must be zero always if the bridge is shorted by a wire.

6. Feb 14, 2015

### jack58

Does I(current) also equal to zero in case it is shortened by wire? If Yes , how do we explain the existence of current in any wire (that of course will have a delta V = 0 because a wire has a resistence equal to 0 ohm ) ?

7. Feb 14, 2015

### Staff: Mentor

No, as I stated. If the bridge is unbalanced then current will flow through the wire.
In the real world wires always have some resistance. So a (very tiny) potential difference will exist.

But even if that were not so, the natural tendency for like-charges to repel each other means that they want to spread out to bring a conductor to all the same potential. If there's a "drain" at one end of a wire stealing electrons, then free electrons in the wire will move to fill the vacancy and maintain an even potential. Miniscule, transient potential imbalances may pop in and out of existence as charges enter and leave due to the quantized nature of charge. But these are well under size that we would ever need to deal with in everyday circuits.

8. Feb 15, 2015

### jack58

Thank u I'm kinda satisfied (about the last reply ,I was on a rush ...) . I hoped for Newton's first law to explain the flow of current when delta V=0 (in ideal wire).