AAMC Section Bank C/P #17

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sfsn

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Basic physics, my weakness..

1. What is the significance of the voltmeter reading 0? Does this mean that there is no current flowing through?

2. Why is it that when the voltmeter reads 0, we should know that the voltages through R and R1 are equal?

3. And does this imply that if the voltmeter did NOT read 0, then the voltages through R and R1 would NOT be equal?

4. I assume that the IR3 = I1R2 because the same current flows through those two resistors?

upload_2016-4-29_10-45-43.png

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Omg I remember this one, took me an hour to do lol. V = 0 means theres no I x R difference between the two connected sections of the wire. There is current going through it, I think. R and R1 does not have to be equal. IR and I1R1 does though. Yea if voltmeter is not 0, then there is a voltage difference. Not sure about the last question.
 
1. What is the significance of the voltmeter reading 0? Does this mean that there is no current flowing through?

2. Why is it that when the voltmeter reads 0, we should know that the voltages through R and R1 are equal?

3. And does this imply that if the voltmeter did NOT read 0, then the voltages through R and R1 would NOT be equal?

4. I assume that the IR3 = I1R2 because the same current flows through those two resistors?

View attachment 203044

1. If the voltmeter reads 0, there is a 0 potential difference between the two points in the circuit. That means that same voltage drop must have occurred across R1 and R. For example, if the voltage source generates 5V, some voltage drop occurs over R1 and some occurs over R (as they are different closed loops). If a different voltage drop occurred over the these resistors, the voltmeter would read non-zero. The voltmeter is comparing the potential between two points in the circuit, but there is still current flowing. If the voltage source was 0, then no current would be flowing.

2. See above, and we can represent this as I1R1 = IR since V = IR. I'm using the notation that the current splits at the junction into I1 (upwards to R1) and I (downwards to R).

3. Yes.

4. Possibly, but not necessarily. This would only be true if R2 = R3. If R2 is different than R3, but the currents are also different, then IR3 = I1R2 can still hold.
 
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4. I assume that the IR3 = I1R2 because the same current flows through those two resistors?

4. Possibly, but not necessarily. This would only be true if R2 = R3. If R2 is different than R3, but the currents are also different, then IR3 = I1R2 can still hold.
Sorry, by "same current," I meant that the one current that goes through R also goes through R2, and the other current that goes through R1 also goes through R3. Which is what you explained in your last sentence. But I just wanted to clarify for posterity!
 
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The AAMC explanation states that:

I1R1=IR and IR3=I1R2
which means the current in resistor 1 and resistor 2 is the same and the current in resistor R and resistor 3 is the same.

What I am not understanding is how the current for resistors R and R3 are being equated (there is a junction between them, so how are they considered to be in series???) and likewise for resistors R1 and R2.
 
@Formation

The voltmeter in the circuit measures the potential difference between two points. If the voltmeter reads 0, there would be a voltage of 0 between the two points. We also know that starting from the battery, within one closed loop around the circuit the voltage must drop back down to 0. When we get to the junction, we can either take the top closed loop (through R1 and R2) or the bottom closed loop (through R and R3). Let's call the current through the top part I1 and the current through the bottom part I2.

If the voltmeter reads 0, the voltage drops across R1 and R must be the same. Using V = IR, I1R1 = I2R.

It follows that the voltage drops across R2 and R3 would also be the same. I1R2 = I2R3.

Rearrange the second equation for I1 / I2 = R3 / R2, and plug into the first equation to get R = R3 * R1 / R2.
 
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The gist of this wheatstone bridge is to move the slide wire (which connects the two sides and has the voltmeter in it) so that there is no voltage difference between the sides. The voltage drop from the first junction to the slide wire is the same for both parallel pathways, meaning I1R1 = IR. The voltage drop from the slide wire to the second junction is the same for both parallel pathways, meaning I2R2 = I3R3. Current is the same in series, so I1 = I2 and I = I3. Ultimately you can work the math so that R1/R2 = R/R3. R3 is the unknown resistor, you can measure its magnitude as (R2 x R)/R1. For a good rundown on this system, see Passage V on Practice test 2.3 of Physics Book 2 (page 275).
 
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all you gotta know is that resistors in parallel have the same voltage and thus have different current. set up teh two equations and use substitution and solve for R.
 
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all you gotta know is that resistors in parallel have the same voltage and thus have different current. set up the two equations and use substitution and solve for R.

I think that is the easiest way to conceptually solve this. Thank you for that I forgot resistors in parallel have the same voltage drop across them.
 
I had a lot of trouble following what was happening.. so here is my worked out solution to the problem.. Hope this helps those who need a bit of visual guidance:

IMG_0804.jpg
 
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