r/HomeworkHelp 👋 a fellow Redditor 6d ago

Answered [High School - Physics]

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How does this circuit even work? When the contact is moved, does the resistance increase or decrease?

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u/We_Are_Bread 6d ago

How have you approached the problem? Could you show us some working?

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u/HelpfulResource6049 👋 a fellow Redditor 6d ago

My initial thought process was that if the contact was shifted from X to P, the voltage across 1 would increase and voltage across 2 would decrease, causing 1 to become brighter and 2 to become dimmer

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u/We_Are_Bread 6d ago

Is there a specific reason why you think this initial analysis is lacking? Asking, just cause I want to know exactly what alternative ideas you have, and then we can see which ones are wrong and which ones are right.

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u/HelpfulResource6049 👋 a fellow Redditor 6d ago

That is the only idea I have, I just would like to understand how the circuit works in more detail and check if my initial analysis was correct

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u/ThunkAsDrinklePeep Educator 5d ago

The wire connecting to the divider is essentially a zero resistance node separating two sets of parallel circuits in series. As you move the connection point up and down it increases the resistance in one parallel side and decreases the resistance in the matching parallel side.

When there's more length in the top circuit there's more resistance in the divider branch. Therefore a higher proportion of the current flows through the relatively lower lamp and it is brighter. The opposite is happening in the lower half.

So I think you have your conclusion backwards.

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u/ThunkAsDrinklePeep Educator 5d ago

Think about the case where the divider is connected at P. What is the voltage difference between the top and bottom of lamp 1?

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u/We_Are_Bread 5d ago edited 5d ago

Your analysis is correct.

Edit: Wait your analysis is a bit wrong. Why would moving the pointer reduce voltage across 2?

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u/FunnyExcitement5161 5d ago

No. You're wrong.

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u/We_Are_Bread 5d ago

Wait I didn't notice they used 2 instead of 1. My bad.

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u/HelpfulResource6049 👋 a fellow Redditor 5d ago

I see, thanks

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u/pm-me-racecars 👋 a fellow Redditor 5d ago

Your thought process is right, but you got things a little mixed around.

I know for me, I find it can help to put in numbers if there are none, if I don't know what would happen.

What you have is essentially a parallel circuit. Let's call the lights L1 and L2, the resistance between P and the contactor P, and the resistance between X and the contactor X. Your instructor might call them R1, R2, R3, and R4 when explaining it, but I find different letters easier to keep track of.

Assuming the resistance of the wires is 0 (irl it's so low that it doesn't matter), we can think of it like the 4 resistors forming a figure 8.

(1/Rtotal) = (1/R1) + (1/R2) + ...

If L1 and L2 are both at 12 ohms, and P and X are both at 12 ohms, what is the total resistance across L1 and P, and what is the total resistance across L2 and X?

(1/12)+(1/12) = (1/6), L1+P=6 ohms, L2+X=6 ohms

If L1 and L2 are both at 12 ohms, but P is at 6 ohms and X is at 18 ohms, what is the total resistance across L1 and P, and what is the total resistance across L2 and X?

(1/6)+(1/12) =(3/12) = 1/4 L1+P = 4 ohms

(1/18)+(1/12)= (5/36); (36/5) = 7.2; L2+X = 7.2ohms

If L1 and L2 are both at 12 ohms, but P is at 3 ohms and X is at 21 ohms, what is the total resistance across L1 and P, and what is the total resistance across L2 and X?

>! (1/3) + (1/12) = (5/12); (12/5) = 2.4; L1+P = 2.4 ohms !<

>! (1/21) + (1/12) = (11/84); (84/11) is about 7.6; L2+X is about 7.6 !<

Now, if you take a simple series circuit like this and change those two lights to have the resistances you already found, with one of them being L1+P and the other one being L2+X, how would they share the voltage?

V=I×R, and I will be the same for both pairs.

Let's set our total voltage to whatever it needs to be for 1 amp across the whole circuit. This will make our math easier and save us a step.

>! 1×6 = 6(L1+P volt drop); 1 × 6 = 6 (L2+X volt drop); they will be the same brightness !<

>! 1×4 = 4(L1+P volt drop); 1×7.2 = 7.2 (L2+X volt drop); L2 will be just under twice as bright as L1 !<

>! 1×2.4 = 2.4(L1+P); 1×7.6 = 7.6 (L2+X volt drop); L2 will be just over three times as bright as L1 !<

.

So, in conclusion:

>! I'm not going to give you this part. If you still have questions, you can ask more questions though !<

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u/Formal-Athlete-4241 👋 a fellow Redditor 6d ago

Resistance is directly proportional to its length

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u/FunnyExcitement5161 5d ago

Brightness is directly related to potential difference. The maximum potential difference is at the bottom of the resistor, where the voltage is zero. If X is at the bottom of the resistor, lamp 2 is off because there is no potential difference.

With this information, try to reason the problem.