## JenniferSmart1 Group Title Capacitors...circuits? one year ago one year ago

1. JenniferSmart1 Group Title

|dw:1362323572542:dw|

2. JenniferSmart1 Group Title

@UnkleRhaukus My first time doing a circuit problem Let's see how much I know....

3. JenniferSmart1 Group Title

should I "add" $$C_1 \;\text{and}\; C_2$$ first?

4. JenniferSmart1 Group Title

@.Sam.

5. JenniferSmart1 Group Title

|dw:1362324461114:dw| these are in parallel...no they're in series correct?

6. JenniferSmart1 Group Title

|dw:1362324513695:dw| is my charge distribution correct so far?

7. JenniferSmart1 Group Title

is it true that the part circled doesn't have a charge? or a net charge of zero?|dw:1362324758141:dw|

8. JenniferSmart1 Group Title

does that mean $$Q_1 \ne Q_2$$

9. UnkleRhaukus Group Title

This is physics

10. UnkleRhaukus Group Title
11. JenniferSmart1 Group Title

I'll move my question

12. .Sam. Group Title

|dw:1362325488170:dw| Series

13. JenniferSmart1 Group Title

yes

14. JenniferSmart1 Group Title

so $$C_1=C_2$$

15. .Sam. Group Title

I don't think you can say that they're equal

16. JenniferSmart1 Group Title

are the charges equal? $Q_1=Q_2$ |dw:1362325598105:dw|

17. JenniferSmart1 Group Title

or should they be opposite in sign?

18. .Sam. Group Title

The sign is correct

19. JenniferSmart1 Group Title

is it true that these two plates don't have a net charge? There is something special about these two plates, but I don't remember what it is. |dw:1362325688912:dw|

20. JenniferSmart1 Group Title

$C_1=\frac{Q_1}{V_1}$ $C_2=\frac{Q_2}{V_2}$ so far $$Q_1=Q_2$$ the Potential.....is that the potential across these plates? |dw:1362325884681:dw|

21. JenniferSmart1 Group Title

|dw:1362325984158:dw| The potential is probably different between $C_1 & C_2$

22. .Sam. Group Title

The charges must be the same, otherwise they won't come up with a formula for seires: $V=V_1+V_2\\\ \\ \text{From} Q=CV \\ \\ \frac{Q}{C}=\frac{Q}{C_1}+\frac{Q}{C_2} \\ \\ \frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}$

23. JenniferSmart1 Group Title

makes sense. Does this mean that the Potential difference between (parallel plates 1) and (parallel plates 2) is $$\frac 1 C$$ where $$\frac 1 C =\frac 1{C_1}+\frac 1{C_2}$$

24. JenniferSmart1 Group Title

I mean it makes sense mathematically, I'm trying to conceptually understand it

25. .Sam. Group Title

That C you're calculating is actually here|dw:1362326867484:dw| Then, You have to combine with C3 with C, which is in parallel, Just add them and you'll get the total capacitance for the top part without C4, as for C4 just use the formula 1/C... again

26. JenniferSmart1 Group Title

I have access to the answer. I'm trying to recreate an example by Prof. Walter Lewin....unfortunately I've been very unsuccessful so far :( https://www.youtube.com/watch?v=UFK-kof1WLQ&list=PLOggKvajo_owEtF3edwu7BYxTOFD3-CpP

27. JenniferSmart1 Group Title

so my goal is to have C4 and C3 and (C_2 and C_1) in parallel....and the only way to do it is to combine the series C_2 and C_1 We've come to the conclusion that C_1 and C_2=$\frac 1 C$ and Now I'm adding $\frac 1 C +C_3+C_4=C_{net}?$

28. JenniferSmart1 Group Title

Let's see....

29. JenniferSmart1 Group Title

are $\frac 1 C+C_3=\text{to some other C}$ since they're within the same enclosed circuit

30. .Sam. Group Title

Don't do them together, that's the biggest mistake you can do in circuit problems. For "All the capacitors" in series, remember that Voltages always add up, so: $V_{Total}=V_1+V_2+V_3\\\ \\ \text{From} Q=CV \\ \\ V=\frac{Q}{C} \\ \\ \frac{Q}{C}=\frac{Q}{C_1}+\frac{Q}{C_2}+\frac{Q}{C_3} \\ \\ \frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$ For "All the capacitors" in parallel, Voltages in Parallel are equal, $V=V_1=V_2$ So, $C=C_1+C_2$

31. JenniferSmart1 Group Title

but why are $C_1, C_2 \;\text{and}\; C_3$ in series? I agree that C_1 and C_2 are, but wouldn't the addition of C_1 and C_2 be in parallel to C_3

32. JenniferSmart1 Group Title

oh I see...you're just giving me a general statement...sorry :S

33. JenniferSmart1 Group Title

so yes, for "all the capacitors in series" the Voltages add up

34. .Sam. Group Title

Yes so for C1 and C2 we have $\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}$ $C=\frac{C_1 C_2}{C_1+C_2}$ Now we have parallel, |dw:1362328161892:dw| Then, $C_{T}=C+C_3 \\ \\ C_{T}=\frac{C_1 C_2}{C_1+C_2}+C_3$

35. .Sam. Group Title

|dw:1362328282074:dw|

36. .Sam. Group Title

Then we reduced the circuit into |dw:1362328321279:dw| Series, $C_{Total}=\frac{1}{C_T}+\frac{1}{C_4} \\ \\ C_{Total}=\frac{1}{\frac{C_1 C_2}{C_1+C_2}+C_3}+\frac{1}{C_4}$

37. .Sam. Group Title

Whops, that should be $\frac{1}{C_{Total}}=\frac{1}{C_T}+\frac{1}{C_4} \\ \\ \frac{1}{C_{Total}}=\frac{1}{\frac{C_1 C_2}{C_1+C_2}+C_3}+\frac{1}{C_4}$

38. JenniferSmart1 Group Title

ok I've been able to follow you up to the point where $C_T=C+C_3$ $C_T=\frac{C_1C_2}{C_2+C_1}+C_3$ that means that... |dw:1362328737355:dw| does that mean that $$C_T$$ and $$C_4$$ are in parallel? does that mean that $$V=V_T+V_4$$ does not apply anymore because it's not in series anymore? or is that always true...whether we're working with capacitors in parallel or in series?

39. JenniferSmart1 Group Title

wait....are C_T and C_4 in parallel? If we line them up it looks like they're in series |dw:1362329073958:dw|

40. JenniferSmart1 Group Title

oh that's why we're able to say $\frac 1 {C_{Total}}=\frac 1{C_T}+\frac 1{C_4}$ $\frac 1 {C_{Total}}=\frac{1}{\frac{C_1C_2}{C_2+C_1}+C_3}$ Ok I think I'm following

41. .Sam. Group Title

That's series, there's no obvious branching lines.

42. JenniferSmart1 Group Title

oops I forgot something

43. JenniferSmart1 Group Title

$\frac 1 {C_{Total}}=\frac{1}{\frac{C_1C_2}{C_2+C_1}+C_3}+\frac{1}{C_4}$

44. JenniferSmart1 Group Title

did we just achieve the goal of this problem? Is that it?

45. JenniferSmart1 Group Title

why do you have C_Total instead of 1/C_total?

46. .Sam. Group Title

The final answer should be$\frac{1}{C_{Total}}=\frac{1}{\frac{C_1 C_2}{C_1+C_2}+C_3}+\frac{1}{C_4}$ But then, I usually work out the numbers first then substitute into another equation that way is not so confusing :)

47. JenniferSmart1 Group Title

oh I see....

48. JenniferSmart1 Group Title

one more question |dw:1362329932879:dw| would this be a parallel series?

49. .Sam. Group Title

You need to know where's the power coming from..

50. JenniferSmart1 Group Title

|dw:1362330036313:dw|

51. JenniferSmart1 Group Title

that's the power right?

52. .Sam. Group Title

Series, you can ignore that middle line|dw:1362330096032:dw|

53. JenniferSmart1 Group Title

can you draw capacitors in parallel for me

54. .Sam. Group Title

Lets say there's another capacitor there |dw:1362330161795:dw| Then, You can tell C_T and C_4 is in series, but both C_T and C_4 is in parallel with C_5 So u work out the series for C_T and C_4 first, then parallel for C_5

55. JenniferSmart1 Group Title

oh so does parallel mean parallel to the power supply?

56. JenniferSmart1 Group Title

|dw:1362330342861:dw| C_5 and C_6 would be capacitors in parallel , whereas C_T and C_3 are capacitors in series correct?

57. JenniferSmart1 Group Title

I mean C_T and C_4....typo :S

58. .Sam. Group Title

no, just look at the branching first, C_T , C_6, and C_4 is in series because they're in the same line, Then C_5 is the only one parallel to all of the other 3 capacitors, you're actually relating to other capacitors

59. JenniferSmart1 Group Title

oh I see.... so for capacitors in parallel $V=V_1+V_2$ still holds, correct? but the Q's are not equal?

60. .Sam. Group Title

Parallel capacitors is $V=V_1=V_2$

61. JenniferSmart1 Group Title

oh interesting Let's see $V=\frac Q C$ $\frac Q C=\frac{Q_1}{C_1}=\frac{Q_2}{C_2}$

62. JenniferSmart1 Group Title

Q's the same again?

63. .Sam. Group Title

For Parallel, Q is not the same, but voltage is the same, so, example: $Q=Q_1+Q_2+Q_3$ $CV=C_1V+C_2V+C_3V$ $C=C_1+C_2+C_3$

64. .Sam. Group Title

That's in parallel, you have to remember.

65. .Sam. Group Title

Series= Charges equal Parallel= Voltage equal

66. JenniferSmart1 Group Title

Yes! Makes sense :)

67. JenniferSmart1 Group Title

You are awesome! Thank you soo much for teaching me capacitors in parallel and series...Thank You @.Sam.

68. .Sam. Group Title

hahaha, welcome :)