TBR, Physics, Section 9, Passage 4, Conceptual Understanding

[sillyjoe]

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So this passage absolutely destroyed me. I thought I had a decent understanding of circuits. Can someone please explain to me what exactly is going on in the passage? I can't seem to wrap my head around it. I attempted all of the questions and got some right. However, it would be very helpful to understand the concepts and experimental design of the wheatstone bridge

 

[sillyjoe]

This sort of helped despite the cheesy music:

 

[gettheleadout]

I remember this passage, and there's more to it (or one of the first few questions gives you more explanation in the question stem) that has to do with the attachment point z for the galvanometer in the students' apparatus being movable. Can you post the rest of it?

Also, how much of the passage can you follow from the beginning? For starters, do you understand why, if the galvanometer reads 0 V, the resistances of R1 and R2 must be equal?

 

[sillyjoe]

I remember this passage, and there's more to it (or one of the first few questions gives you more explanation in the question stem) that has to do with the attachment point z for the galvanometer in the students' apparatus being movable. Can you post the rest of it?

Also, how much of the passage can you follow from the beginning? For starters, do you understand why, if the galvanometer reads 0 V, the resistances of R1 and R2 must be equal?

My bad. I corrected it. I guess my trouble is fully understanding what is going on. I think I understand why the if it reads o V the resistance must be equal since it is not traveling through the middle. However, can you please explain it in further detail as well as explain the conceptual reasoning behind sliding the wire to adjust resistance to get the reading of O V? I think I get that since r is proportional to length you can adjust... I am just having a hard time putting it all together.

 

[gettheleadout]

My bad. I corrected it. I guess my trouble is fully understanding what is going on. I think I understand why the if it reads o V the resistance must be equal since it is not traveling through the middle.

That is correct. The galvanometer (G) in this case is serving as a voltmeter, when the G reads 0 A it follows that the potential difference between the two points of attachment is ∆V = 0 V. If there is no difference in the electric potentials at the two points of attachment, then since those two points are on wires that split off at a common junction, the current path through those wire segments must entail the same amount of resistance.

However, can you please explain it in further detail as well as explain the conceptual reasoning behind sliding the wire to adjust resistance to get the reading of O V? I think I get that since r is proportional to length you can adjust... I am just having a hard time putting it all together.

Do you follow the mathematical relationships presented in the passage? I realize they kind of throw a lot at you at once but if you can follow it then it should be clear why you can solve for the unknown resistor.

The idea behind sliding the G attachment point is indeed that varying the length of the wire segment a-z allows us to relate the resistance of R_known to the resistivity of the wire a-c. What I especially hated about this passage is that it didn't explicitly state that the wires of different metals they were using were actually in place as the segment a-c. In any case this works because the wire a-c will have a resistivity characteristic of the metal it's made of, and thus the length of wire determines the resistance that segment presents according to the relationship R = ρL / A where L is the segment length , ρ is the resistivity, and A is the cross-sectional area of the wire. By moving the attachment point z, we can change the ratio of the resistances, R_a-z : R_z-c. Recognizing that those two "halves" of the a-c wire segment represent discrete "resistors," we can draw the following analogy to the traditional Wheatstone bridge from the passage:

R1 = R_a-z
R2 = R_known
R3 = R_z-c
R4 = R_U

So, if we were to move point z until the G read 0 V, we could use the measured length of segment a-z to find the resistivity ρ, since at that point R_a-z = R_known (we'd have to be given L and A.) We could then apply that resistivity to determine R_z-c, and thus find R_U. I don't remember what the questions asked in particular though.