WheatStone bridge

[Ratch]

@Ratch I really want to say that you have too much time on your hands. You can hi-light a physics question and tell the forum they are wrong.
Current flow in the physics world is incorrect.
However, current being a quantity, can flow.
So, the current to flow (quantity) is acceptable. IMHO.

The amount of time I have available has nothing to do with the definition of current. Current IS charge flow. It does not have to be iterated twice. If you insist on doing it twice, then why not three or four times to really drive the point home? So do it twice if you must, but know that you are using technical slang and not good English when you do.

Ratch

 

[majoco]

Waaay back you said "I know that at balance point the current will be same in R1, R2 and in R3,R4" - this in reference ti this diagram...

Aaah - but it's not. As you (or somebody) said at the bottom.... the ratio of R1 to R2 is the same as the ratio of R3 to R4. The current through R1 and R2 is not the same as R3 and R4, the resistors can be totally different values and so the currents can be totally different as well BUT the voltage drop across R1 is the same as the voltage drop across R3 at balance so there is equal voltages at either end of the measuring device - therefore no current will flow - and if no current is flowing it cannot unbalance the bridge.

The function of a bridge is to identify the value of R4 by comparing the ratios of the other three resistors and knowing their values - there's another project for you!

 

[Ratch]

Waaay back you said "I know that at balance point the current will be same in R1, R2 and in R3,R4" - this in reference ti this diagram...

Aaah - but it's not. As you (or somebody) said at the bottom.... the ratio of R1 to R2 is the same as the ratio of R3 to R4. The current through R1 and R2 is not the same as R3 and R4, the resistors can be totally different values and so the currents can be totally different as well BUT the voltage drop across R1 is the same as the voltage drop across R3 at balance so there is equal voltages at either end of the measuring device - therefore no current will flow - and if no current is flowing it cannot unbalance the bridge.

The function of a bridge is to identify the value of R4 by comparing the ratios of the other three resistors and knowing their values - there's another project for you!

You do not identify the post # where the misstatement was made. That makes it hard to track down what you are talking about. You are still describing charge flow as "current will flow". As I pointed out in post 11 of this thread, "current flow" literally means "charge flow flow", which is redundant and ridiculous. You should instead say "no current will exist".

Ratch

 

[hevans1944]

Recently i studied the wheatstone bridge in me physics book in class but i could not understand it properly. I asked the teacher about my confusion but he also dont know. i have posted the picture of topic(wheatstone bridge). My question is ;
1.Why current I1 is following through point A to D in loop ADBA.
2.Why current I2 is following through point D to C in loop DCBD.

My confusion is that why current are following through these two paths. As current always flow through level of a higher potential to a level of lower potential but here it seems opposite.

When using Kirchoff's current law to analyze a circuit, you need to identify closed loops that will contain currents and (in this case) resistances to dissipate those currents. You pick arbitrary directions for the currents and let the polarity signs across the resistances be according to the (assumed) direction of the current through those devices. The proper polarity signs will appear later during your calculations. So, !1 and I2 are just two arbitrary currents that were chosen, along with current I3. Other current loops could have been selected.

You don't know a priori, when identifying which current loops to choose, what the voltage rise or fall across each resistance will be until you solve the simultaneous equations you have written from the loop equations defining each current loop. That is why it is very important to pay attention to your algebraic signs as your calculations progress.

You are not going to learn much about Wheatstone bridges by studying text books. Grab a handful of 1% tolerance film resistors and throw together a bridge on a solderless breadboard. Use an inexpensive Digital Multi-Meter on its most sensitive current range to monitor bridge balance as you substitute various value resistors for the original values.. Use a 1.5 V DC dry-cell to excite your bridge and have a "knife-like" switch to briefly and momentarily connect the dry cell to the bridge. Your goal is to get a "feel" for how the bridge becomes unbalanced and rebalanaced without sending a large slug of current through your DVM, blowing it's internal fuse (sometimes hard to come by) and sometimes damaging the meter. Start with four, say 50 kΩ, resistors and work your way up and down from there. Study some of the ways Wheatstone bridges are used today for instrumentation.

 

[majoco]

Post #9 "I know that at balance point the current will be same in R1, R2 and in R3,R4 " This is NOTtrue. The current through R1 and R2 does not have to be the same as the current through R3 and R4 TO BALANCE THE BRIDGE. Perhaps you need to understand both of Kirchoff's Laws. How can current flow from the junction of R1, R2 to or from the junction of R3, R4 when they are both at the same potential?

 

[Michael Studio1]

My first thought is: Where is a Wheatstone Bridge most commonly used? The answer, for me, is at the Primary Test Board in the Long Lines Dept of a Telephone Company. Then: Why is a Wheatstone Bridge used? The answer here is to quickly detect the precise location of a fault in a long open-wire facility in order to direct the Lineman to the fault area to rectify the fault and get the Services working as quickly as possible. In this scenario the Test Board Operator is in direct communication with the Lineman. "In using any electrical instrument it must be remembered that Ohm's Law is never failing and that it applies to every circuit branch". Here I am quoting from Chapter 5, Page 29 in my 1953 copy of the Linesman's "Green Book" "Principles of Electricity applied to Telephone and Telegraph Work, a Training Course Text by the American Telephone and Telegraph Company. Herein the Wheatstone Bridge and its uses are thoroughly explored.

 

[Ratch]

Everyone talks about the currents through the two legs of the bridge. Really, that is irrelevant. The currents can be any value whatsoever. What is important is the voltage across the path between the bridge legs. When that voltage is zero, the bridge is "balanced", and the resistances in each leg are proportional to one another. So, if you know three of the leg resistances, you can find the fourth one. You don't have to get tied up in a knot over K's law and and heavy simultaneous equation solving. It is a simple principle of voltage drop versus resistance value. If you have a specific problem, post it clearly along with your solution and let's see what your technique is.

Ratch

 

[majoco]

The only person with a problem is Dabu and he hasn't been heard since Post #1. He's probably even more confused now than he ever was at the start.