WheatStone bridge

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Dabu WhiteHacker

Joined Sep 5, 2017
65
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.

IMG_20180819_160419.jpg
 
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Jony130

Joined Feb 17, 2009
5,089
But the loop currents (mesh current) is not a "real" current that flows in the circuit. What you are doing here is that you apply a Kirchhoff's voltage law (KVL) around the loop. And this is why you go start at point A and goto DB and back to the A point again.
 

WBahn

Joined Mar 31, 2012
24,974
hi,
The currents I1 and I2 are stated as zero, so where do you see a 'higher potential'.?
The currents I1 and I2 are NOT stated as zero and, in general, they won't be. In order for zero current to flow in the galvanometer, all that is required for I1 and I2 is for them to be equal.

If you were to solve for I1 and I2, you would find that they are positive and equal (they have to be positive to give the correct direction of current in the top two resistors) but that I3 will be positive and larger so that the total current in the bottom two resistors is also in the correct direction.
 

ericgibbs

Joined Jan 29, 2010
9,113
Hi WB,
Perhaps a small difference in interpretation of the way the problem is worded.
The I1 and I2 current arrows, in his diagram, show them to be equal and opposite in their direction.
ie: the bridge is in balance and therefore Nodes 'A' and 'D' are at the same potential, no Galvo current.

Node 'C' is at the positive potential with respect to Node 'A' , so I3, in this example, is in same direction for all four resistors.

Eric
 

WBahn

Joined Mar 31, 2012
24,974
Hi WB,
Perhaps a small difference in interpretation of the way the problem is worded.
The I1 and I2 current arrows, in his diagram, show them to be equal and opposite in their direction.
ie: the bridge is in balance and therefore Nodes 'A' and 'D' are at the same potential, no Galvo current.

Node 'C' is at the positive potential with respect to Node 'A' , so I3, in this example, is in same direction for all four resistors.

Eric
The I1, I2, and I3 arrows are merely the defined loop currents (and the problem explicitly states that this is what they are) for applying mesh analysis. Nothing about their magnitude or polarity is dictated by how they are drawn.

The requirement that no galvanometer current flow is what requires the currents I1 and I2 to be equal (the problem also explicitly states this).

The loop current I3 only flows in the lower two resistors.

The currents in the four resistors, flowing right-to-left, are

I_R1 = I1
I_R2 = I2
I_R3 = I3 - I1
I_R4 = I3 - I2

When the galvanometer current is zero

I1 = I2 = Vbattery / (R1 + R2) and these are NOT zero.
 

ericgibbs

Joined Jan 29, 2010
9,113
hi,
In the special case of the bridge being in balance,
ie: no potential difference between Nodes 'B' and 'D', the I3 current draw from the excitation voltage across Nodes 'A' and 'C' can be ignored.

Considering the 'ADBA' resistance path, the algebraic sum of voltages around that loop is zero, therefore I1 current is zero.
The same applies for the 'CBDC' path, so I2 current is zero.

E
 

WBahn

Joined Mar 31, 2012
24,974
hi,
In the special case of the bridge being in balance,
ie: no potential difference between Nodes 'B' and 'D', the I3 current draw from the excitation voltage across Nodes 'A' and 'C' can be ignored.

Considering the 'ADBA' resistance path, the algebraic sum of voltages around that loop is zero, therefore I1 current is zero.
The same applies for the 'CBDC' path, so I2 current is zero.

E
If I1 and I2 are zero, that means that the total current in R1 and R2 is zero. The only way that is possible is if BOTH R1 and R2 are open circuits (infinite resistance).

Let's put some real numbers in here, okay.

Let's say that the battery voltage is 10 V and that R1 and R2 are each 500 Ω and R3 and R4 are each 5000 Ω.

Assuming, for the moment (we'll verify it later) that the bridge is balanced, then no current flows in the galvanometer and the circuit is reduced to two parallel branches, each with 10 V applied across them. The top branch contains R1 and R2 which total 1000 Ω so a current of 10 mA flows, right to left, through each of them. The middle branch contains R3 and R4, which total 10,000 Ω so a current of 1 mA flows, right to left, through each of them. The bottom branch is just the battery and the closed switch and a current of 11 mA flows left to right though them.

Since the ONLY loop current that flows through R1 is I1 and since I1 flows right to left through R1, that means that I1 = 10 mA.

Since the ONLY loop current that flows through R2 is I2 and since I1 flows right to left through R2, that means that I2 = 10 mA.

Since the ONLY loop current that flows through the battery is I3 and since I3 flows left to right through the battery, that meanst that I3 = 11 mA.

The total current flowing in R3, right to left, is I3 - I1 = 11 mA - 10 mA = 1 mA, which is correct.

The total current flowing in R4, right to left, is I3 - I2 = 11 mA - 10 mA = 1 mA, which is correct.

The flaw in your reasoning is that the excitation current, I3, cannot be ignored. When you sum voltages around loop ADBA, you need to sum the voltage drops across R1, R3, and the galvanometer. But the voltage drop across R3 requires that you use the current flowing in R3, which includes both I1 and I3, so it cannot be ignored.
 

ericgibbs

Joined Jan 29, 2010
9,113
hi WB,
Would you agree that in the special case of a balanced bridge, if I removed the Galvo from the circuit, the voltages at Nodes 'B' and 'D' would not change.?

Therefore, when the Galvo is removed, I1 and I2 currents must be zero as paths 'ADBA' and 'CDBC' are open circuit.

In a balanced bridge, neither Vext or I3 will have any effect upon the current thru the Galvo, it will always be zero at balance.

The equations in your last post demonstrate this conclusion.

Eric

AA1 20-Aug-18 07.55.gif
 
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WBahn

Joined Mar 31, 2012
24,974
hi WB,
Would you agree that in the special case of a balanced bridge, if I removed the Galvo from the circuit, the voltages at Nodes 'B' and 'D' would not change.?

Therefore, when the Galvo is removed, I1 and I2 currents must be zero as paths 'ADBA' and 'CDBC' are open circuit.

In a balanced bridge, neither Vext or I3 will have any effect upon the current thru the Galvo, it will always be zero at balance.

The equations in your last post demonstrate this conclusion.

Eric

View attachment 158392
I think I see your misapprehension.

There is no ACTUAL current across the place where the Galvo was located. On this we agree. However, this does NOT mean that I1 and I2 are zero. These are VIRTUAL currents (we usually call them "loop" or "mesh" currents). The ACTUAL current through that Galvo (using a reference direction of downward) is (I2 - I1). So the only constraint that is placed by the removal of the Galvo is just that (I2-I1) must be zero, which only requires that I2=I1, it does NOT require that they each, separately, be equal to zero.

Do you see the problem that happens if you say that I1 and I2 are zero?

The ONLY current in R1 is I1 and the ONLY current in R2 is I2. So if I1 and I2 are both zero when the bridge is balanced, then you are saying that the actual currents in R1 and R2 are both zero when the bridge is balanced, and this simply is not the case.
 

ericgibbs

Joined Jan 29, 2010
9,113
hi WB
No, I am not saying the currents in R1 or R2 are zero.
I am saying that currents I1 and I2 in a balanced bridge are zero, whether the currents are considered 'real' or 'virtual' , they are zero.

It is clear that you will never agree on this particular point, so I will opt out of this Thread.

Eric
 

MrChips

Joined Oct 2, 2009
19,768
You might be confusing mesh current and branch current.
Branch current is used in KVL and KCL.

KCL does not apply to mesh current (or loop current).

If we were to assume that KVL applies, then at bridge balance (I1 + I2) = 0.
This implies that I1 = -I2
But we know that this is not true if (R1 + R2 ) ≠ (R3+R4).
 

WBahn

Joined Mar 31, 2012
24,974
hi WB
No, I am not saying the currents in R1 or R2 are zero.
I am saying that currents I1 and I2 in a balanced bridge are zero, whether the currents are considered 'real' or 'virtual' , they are zero.

It is clear that you will never agree on this particular point, so I will opt out of this Thread.

Eric
If I1 is zero. then how can the current in R1 be ANYTHING other than zero since I1 IS the current in R1?
 

ericgibbs

Joined Jan 29, 2010
9,113
hi WB,
I will give it one more shot.
Please just answer this one question.

On the balanced bridge that we discussing, what is the numerical value of current that is flowing thru the galvo.??

E
 

WBahn

Joined Mar 31, 2012
24,974
hi WB,
I will give it one more shot.
Please just answer this one question.

On the balanced bridge that we discussing, what is the numerical value of current that is flowing thru the galvo.??

E
The numerical value of the actual current that is flowing through the galvo when the bridge is balanced is zero. No one has ever disputed this.

Now answer a question in return.

If the current I1 is zero (for whatever reason), how can the current flowing in R1 be anything other than zero?
 
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