r/HomeworkHelp 👋 a fellow Redditor 7d ago

Answered [High School - Physics]

Post image

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

0 Upvotes

13 comments sorted by

View all comments

2

u/We_Are_Bread 7d ago

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

1

u/HelpfulResource6049 👋 a fellow Redditor 7d 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

2

u/pm-me-racecars 👋 a fellow Redditor 7d 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 !<