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JenniferSmart1Best ResponseYou've already chosen the best response.0
dw:1362323572542:dw
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
@UnkleRhaukus My first time doing a circuit problem Let's see how much I know....
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
should I "add" \(C_1 \;\text{and}\; C_2\) first?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
dw:1362324461114:dw these are in parallel...no they're in series correct?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
dw:1362324513695:dw is my charge distribution correct so far?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
is it true that the part circled doesn't have a charge? or a net charge of zero?dw:1362324758141:dw
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
does that mean \(Q_1 \ne Q_2\)
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
http://openstudy.com/study#/groups/Physics
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
I'll move my question
 one year ago

.Sam.Best ResponseYou've already chosen the best response.3
dw:1362325488170:dw Series
 one year ago

.Sam.Best ResponseYou've already chosen the best response.3
I don't think you can say that they're equal
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
are the charges equal? \[Q_1=Q_2\] dw:1362325598105:dw
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
or should they be opposite in sign?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
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
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
\[C_1=\frac{Q_1}{V_1}\] \[C_2=\frac{Q_2}{V_2}\] so far \(Q_1=Q_2\) the Potential.....is that the potential across these plates? dw:1362325884681:dw
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
dw:1362325984158:dw The potential is probably different between \[C_1 & C_2\]
 one year ago

.Sam.Best ResponseYou've already chosen the best response.3
The charges must be the same, otherwise they won't come up with a formula for seires: \[V=V_1+V_2\\\ \\ \text{From} Q=CV \\ \\ \frac{Q}{C}=\frac{Q}{C_1}+\frac{Q}{C_2} \\ \\ \frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2} \]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
makes sense. Does this mean that the Potential difference between (parallel plates 1) and (parallel plates 2) is \(\frac 1 C\) where \(\frac 1 C =\frac 1{C_1}+\frac 1{C_2}\)
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
I mean it makes sense mathematically, I'm trying to conceptually understand it
 one year ago

.Sam.Best ResponseYou've already chosen the best response.3
That C you're calculating is actually heredw: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
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
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=UFKkof1WLQ&list=PLOggKvajo_owEtF3edwu7BYxTOFD3CpP
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
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 \[\frac 1 C +C_3+C_4=C_{net}?\]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
are \[\frac 1 C+C_3=\text{to some other C}\] since they're within the same enclosed circuit
 one year ago

.Sam.Best ResponseYou've already chosen the best response.3
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, \[C=C_1+C_2\]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
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
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
oh I see...you're just giving me a general statement...sorry :S
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
so yes, for "all the capacitors in series" the Voltages add up
 one year ago

.Sam.Best ResponseYou've already chosen the best response.3
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, \[C_{T}=C+C_3 \\ \\ C_{T}=\frac{C_1 C_2}{C_1+C_2}+C_3\]
 one year ago

.Sam.Best ResponseYou've already chosen the best response.3
Then we reduced the circuit into dw:1362328321279:dw Series, \[C_{Total}=\frac{1}{C_T}+\frac{1}{C_4} \\ \\ C_{Total}=\frac{1}{\frac{C_1 C_2}{C_1+C_2}+C_3}+\frac{1}{C_4}\]
 one year ago

.Sam.Best ResponseYou've already chosen the best response.3
Whops, that should be \[\frac{1}{C_{Total}}=\frac{1}{C_T}+\frac{1}{C_4} \\ \\ \frac{1}{C_{Total}}=\frac{1}{\frac{C_1 C_2}{C_1+C_2}+C_3}+\frac{1}{C_4}\]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
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?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
wait....are C_T and C_4 in parallel? If we line them up it looks like they're in series dw:1362329073958:dw
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
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
 one year ago

.Sam.Best ResponseYou've already chosen the best response.3
That's series, there's no obvious branching lines.
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
oops I forgot something
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
\[\frac 1 {C_{Total}}=\frac{1}{\frac{C_1C_2}{C_2+C_1}+C_3}+\frac{1}{C_4}\]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
did we just achieve the goal of this problem? Is that it?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
why do you have C_Total instead of 1/C_total?
 one year ago

.Sam.Best ResponseYou've already chosen the best response.3
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 :)
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
one more question dw:1362329932879:dw would this be a parallel series?
 one year ago

.Sam.Best ResponseYou've already chosen the best response.3
You need to know where's the power coming from..
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
dw:1362330036313:dw
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
that's the power right?
 one year ago

.Sam.Best ResponseYou've already chosen the best response.3
Series, you can ignore that middle linedw:1362330096032:dw
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
can you draw capacitors in parallel for me
 one year ago

.Sam.Best ResponseYou've already chosen the best response.3
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
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
oh so does parallel mean parallel to the power supply?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
dw:1362330342861:dw C_5 and C_6 would be capacitors in parallel , whereas C_T and C_3 are capacitors in series correct?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
I mean C_T and C_4....typo :S
 one year ago

.Sam.Best ResponseYou've already chosen the best response.3
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
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
oh I see.... so for capacitors in parallel \[V=V_1+V_2\] still holds, correct? but the Q's are not equal?
 one year ago

.Sam.Best ResponseYou've already chosen the best response.3
Parallel capacitors is \[V=V_1=V_2\]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
oh interesting Let's see \[V=\frac Q C\] \[\frac Q C=\frac{Q_1}{C_1}=\frac{Q_2}{C_2}\]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
Q's the same again?
 one year ago

.Sam.Best ResponseYou've already chosen the best response.3
For Parallel, Q is not the same, but voltage is the same, so, example: \[Q=Q_1+Q_2+Q_3\] \[CV=C_1V+C_2V+C_3V\] \[C=C_1+C_2+C_3\]
 one year ago

.Sam.Best ResponseYou've already chosen the best response.3
That's in parallel, you have to remember.
 one year ago

.Sam.Best ResponseYou've already chosen the best response.3
Series= Charges equal Parallel= Voltage equal
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
Yes! Makes sense :)
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
You are awesome! Thank you soo much for teaching me capacitors in parallel and series...Thank You @.Sam.
 one year ago
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