anonymous
  • anonymous
Capacitors...circuits?
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
|dw:1362323572542:dw|
anonymous
  • anonymous
@UnkleRhaukus My first time doing a circuit problem Let's see how much I know....
anonymous
  • anonymous
should I "add" \(C_1 \;\text{and}\; C_2\) first?

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anonymous
  • anonymous
@.Sam.
anonymous
  • anonymous
|dw:1362324461114:dw| these are in parallel...no they're in series correct?
anonymous
  • anonymous
|dw:1362324513695:dw| is my charge distribution correct so far?
anonymous
  • anonymous
is it true that the part circled doesn't have a charge? or a net charge of zero?|dw:1362324758141:dw|
anonymous
  • anonymous
does that mean \(Q_1 \ne Q_2\)
UnkleRhaukus
  • UnkleRhaukus
This is physics
anonymous
  • anonymous
I'll move my question
.Sam.
  • .Sam.
|dw:1362325488170:dw| Series
anonymous
  • anonymous
yes
anonymous
  • anonymous
so \(C_1=C_2\)
.Sam.
  • .Sam.
I don't think you can say that they're equal
anonymous
  • anonymous
are the charges equal? \[Q_1=Q_2\] |dw:1362325598105:dw|
anonymous
  • anonymous
or should they be opposite in sign?
.Sam.
  • .Sam.
The sign is correct
anonymous
  • anonymous
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|
anonymous
  • anonymous
\[C_1=\frac{Q_1}{V_1}\] \[C_2=\frac{Q_2}{V_2}\] so far \(Q_1=Q_2\) the Potential.....is that the potential across these plates? |dw:1362325884681:dw|
anonymous
  • anonymous
|dw:1362325984158:dw| The potential is probably different between \[C_1 & C_2\]
.Sam.
  • .Sam.
The charges must be the same, otherwise they won't come up with a formula for seires: \[V=V_1+V_2\\\ \\ \text{From} Q=CV \\ \\ \frac{Q}{C}=\frac{Q}{C_1}+\frac{Q}{C_2} \\ \\ \frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2} \]
anonymous
  • anonymous
makes sense. Does this mean that the Potential difference between (parallel plates 1) and (parallel plates 2) is \(\frac 1 C\) where \(\frac 1 C =\frac 1{C_1}+\frac 1{C_2}\)
anonymous
  • anonymous
I mean it makes sense mathematically, I'm trying to conceptually understand it
.Sam.
  • .Sam.
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
anonymous
  • anonymous
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
anonymous
  • anonymous
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 \[\frac 1 C +C_3+C_4=C_{net}?\]
anonymous
  • anonymous
Let's see....
anonymous
  • anonymous
are \[\frac 1 C+C_3=\text{to some other C}\] since they're within the same enclosed circuit
.Sam.
  • .Sam.
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, \[C=C_1+C_2\]
anonymous
  • anonymous
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
anonymous
  • anonymous
oh I see...you're just giving me a general statement...sorry :S
anonymous
  • anonymous
so yes, for "all the capacitors in series" the Voltages add up
.Sam.
  • .Sam.
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, \[C_{T}=C+C_3 \\ \\ C_{T}=\frac{C_1 C_2}{C_1+C_2}+C_3\]
.Sam.
  • .Sam.
|dw:1362328282074:dw|
.Sam.
  • .Sam.
Then we reduced the circuit into |dw:1362328321279:dw| Series, \[C_{Total}=\frac{1}{C_T}+\frac{1}{C_4} \\ \\ C_{Total}=\frac{1}{\frac{C_1 C_2}{C_1+C_2}+C_3}+\frac{1}{C_4}\]
.Sam.
  • .Sam.
Whops, that should be \[\frac{1}{C_{Total}}=\frac{1}{C_T}+\frac{1}{C_4} \\ \\ \frac{1}{C_{Total}}=\frac{1}{\frac{C_1 C_2}{C_1+C_2}+C_3}+\frac{1}{C_4}\]
anonymous
  • anonymous
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?
anonymous
  • anonymous
wait....are C_T and C_4 in parallel? If we line them up it looks like they're in series |dw:1362329073958:dw|
anonymous
  • anonymous
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
.Sam.
  • .Sam.
That's series, there's no obvious branching lines.
anonymous
  • anonymous
oops I forgot something
anonymous
  • anonymous
\[\frac 1 {C_{Total}}=\frac{1}{\frac{C_1C_2}{C_2+C_1}+C_3}+\frac{1}{C_4}\]
anonymous
  • anonymous
did we just achieve the goal of this problem? Is that it?
anonymous
  • anonymous
why do you have C_Total instead of 1/C_total?
.Sam.
  • .Sam.
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 :)
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anonymous
  • anonymous
oh I see....
anonymous
  • anonymous
one more question |dw:1362329932879:dw| would this be a parallel series?
.Sam.
  • .Sam.
You need to know where's the power coming from..
anonymous
  • anonymous
|dw:1362330036313:dw|
anonymous
  • anonymous
that's the power right?
.Sam.
  • .Sam.
Series, you can ignore that middle line|dw:1362330096032:dw|
anonymous
  • anonymous
can you draw capacitors in parallel for me
.Sam.
  • .Sam.
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
anonymous
  • anonymous
oh so does parallel mean parallel to the power supply?
anonymous
  • anonymous
|dw:1362330342861:dw| C_5 and C_6 would be capacitors in parallel , whereas C_T and C_3 are capacitors in series correct?
anonymous
  • anonymous
I mean C_T and C_4....typo :S
.Sam.
  • .Sam.
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
anonymous
  • anonymous
oh I see.... so for capacitors in parallel \[V=V_1+V_2\] still holds, correct? but the Q's are not equal?
.Sam.
  • .Sam.
Parallel capacitors is \[V=V_1=V_2\]
anonymous
  • anonymous
oh interesting Let's see \[V=\frac Q C\] \[\frac Q C=\frac{Q_1}{C_1}=\frac{Q_2}{C_2}\]
anonymous
  • anonymous
Q's the same again?
.Sam.
  • .Sam.
For Parallel, Q is not the same, but voltage is the same, so, example: \[Q=Q_1+Q_2+Q_3\] \[CV=C_1V+C_2V+C_3V\] \[C=C_1+C_2+C_3\]
.Sam.
  • .Sam.
That's in parallel, you have to remember.
.Sam.
  • .Sam.
Series= Charges equal Parallel= Voltage equal
anonymous
  • anonymous
Yes! Makes sense :)
anonymous
  • anonymous
You are awesome! Thank you soo much for teaching me capacitors in parallel and series...Thank You @.Sam.
.Sam.
  • .Sam.
hahaha, welcome :)

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