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# Capicitor Discharge Curve (URGENT!) watch

1. This question is in the sample paper for edexcel unit 4 physics:

(a) A 2200 μF capacitor is charged to a potential difference of 12 V and then discharged through an electric motor. The motor lifts a 50 g mass through a height of 24 cm.
(b) The capacitor is charged to 12 V again and then discharged through a 16 ohm resistor.
(i) Show that the time constant for this discharge is approximately 35 ms.

done

(ii) Sketch a graph of current against time for this discharge on the grid below. You should indicate the current at t = 0 and t = 35 ms.

I can't seem to do this! I know the shape, but the values they give us in the m/s aren't the same as what I get, why?
2. What values do you get?
Initial current at start of graph = V/R (V is initial pd on capacitor=12V)
Current after 35ms is V/R where V is pd after 35ms. This V is 0.37 times the start value.
3. (Original post by Stonebridge)
What values do you get?
Initial current at start of graph = V/R (V is initial pd on capacitor=12V)
Current after 35ms is V/R where V is pd after 35ms. This V is 0.37 times the start value.
I get 750ms for the initial current. The markscheme uses 1.6V for some reason instead of 12V - any idea why?
How do you know that the V is 0.37 times the start value?
4. I recognise this question. I spoke to my physics teacher about it, and it turns out the markscheme is wrong.

V = IR, I = V/R which means 12V/16 = 0.75 A

The initial value is 750mA and the current at the time constant is around 275mA. you know the graph shape
5. (Original post by DeyTa)
I recognise this question. I spoke to my physics teacher about it, and it turns out the markscheme is wrong.

V = IR, I = V/R which means 12V/16 = 0.75 A

The initial value is 750mA and the current at the time constant is around 275mA. you know the graph shape
That would make so much sense!
Thank you!

Although I don't get what you mean by the time constant being around 275mA? don't you mean that the current at 35ms is around 275mA?
6. (Original post by unamed)
How do you know that the V is 0.37 times the start value?
Because the graph is a reverse exponential graph, and according to the equation It = Io x e^(- time/time constant), at the time where t = time constant, it becomes It = Io x e^-1, which is the same as multiplying by 1/e, which is approximately 0.368.

Io is the initial current, which we now know to be 750mA,

So current at the time constant = 750 x 1/e = 750 x 0.368 = 276mA

I'll see if i can upload a graph
7. (Original post by DeyTa)
Because the graph is a reverse exponential graph, and according to the equation It = Io x e^(- time/time constant), at the time where t = time constant, it becomes It = Io x e^-1, which is the same as multiplying by 1/e, which is approximately 0.368.

Io is the initial current, which we now know to be 750mA,

So current at the time constant = 750 x 1/e = 750 x 0.368 = 276mA

I'll see if i can upload a graph
Oh, I understand now! thank you.
8. just in case :P as you can see my MS Paint skills are exquisite

attachment below (i hope)
Attached Images

9. (Original post by DeyTa)
just in case :P as you can see my MS Paint skills are exquisite

attachment below (i hope)
That's actually I good graph - considering your software!
Are you ready for the exam tomorrow?
10. (Original post by unamed)
Are you ready for the exam tomorrow?
I guess I'm pretty ready, although I still get a bit confused about electromagnetic induction sometimes As it turns out, the explanation for electromagnetic induction is just a model! So we might actually be learning wrong stuff hehe

good luck!
11. (Original post by DeyTa)
I guess I'm pretty ready, although I still get a bit confused about electromagnetic induction sometimes As it turns out, the explanation for electromagnetic induction is just a model! So we might actually be learning wrong stuff hehe

good luck!
I hope electromagnetism doesn't come up at all - I hate it with a passion. I'd be much happier with particle physics - after all, for all we know, it might be wrong!

you too!
12. It's the same as charging mate, for current anyway. (exponential decline)
13. Hi guys tomorrow i am having both phy4 and phy5
14. for this type of question, would you estimate the value of I when time is 35ms using the graph or do an actual working and multiply 750ms by 1/e?
15. (Original post by Kameo)
for this type of question, would you estimate the value of I when time is 35ms using the graph or do an actual working and multiply 750ms by 1/e?
well, considering we need the value to make the graph - you work it out. If they give us the graph, then it makes sense to use the graph.

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