A community for students.
Here's the question you clicked on:
 0 viewing
JenniferSmart1
 2 years ago
Capacitors...circuits?
JenniferSmart1
 2 years ago
Capacitors...circuits?

This Question is Closed

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1362323572542:dw

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0@UnkleRhaukus My first time doing a circuit problem Let's see how much I know....

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0should I "add" \(C_1 \;\text{and}\; C_2\) first?

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1362324461114:dw these are in parallel...no they're in series correct?

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1362324513695:dw is my charge distribution correct so far?

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0is it true that the part circled doesn't have a charge? or a net charge of zero?dw:1362324758141:dw

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0does that mean \(Q_1 \ne Q_2\)

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0I'll move my question

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

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

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0are the charges equal? \[Q_1=Q_2\] dw:1362325598105:dw

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0or should they be opposite in sign?

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0is 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

JenniferSmart1
 2 years ago
Best 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

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1362325984158:dw The potential is probably different between \[C_1 & C_2\]

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

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0makes sense. Does this mean that the Potential difference between (parallel plates 1) and (parallel plates 2) is \(\frac 1 C\) where \(\frac 1 C =\frac 1{C_1}+\frac 1{C_2}\)

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0I mean it makes sense mathematically, I'm trying to conceptually understand it

.Sam.
 2 years ago
Best ResponseYou've already chosen the best response.3That 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

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0I 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

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0so 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 \[\frac 1 C +C_3+C_4=C_{net}?\]

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0are \[\frac 1 C+C_3=\text{to some other C}\] since they're within the same enclosed circuit

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

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0but 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

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0oh I see...you're just giving me a general statement...sorry :S

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0so yes, for "all the capacitors in series" the Voltages add up

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

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0ok 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?

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0wait....are C_T and C_4 in parallel? If we line them up it looks like they're in series dw:1362329073958:dw

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0oh 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.
 2 years ago
Best ResponseYou've already chosen the best response.3That's series, there's no obvious branching lines.

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0oops I forgot something

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0\[\frac 1 {C_{Total}}=\frac{1}{\frac{C_1C_2}{C_2+C_1}+C_3}+\frac{1}{C_4}\]

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0did we just achieve the goal of this problem? Is that it?

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0why do you have C_Total instead of 1/C_total?

.Sam.
 2 years ago
Best ResponseYou've already chosen the best response.3The 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 :)

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0one more question dw:1362329932879:dw would this be a parallel series?

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

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1362330036313:dw

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0that's the power right?

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

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0can you draw capacitors in parallel for me

.Sam.
 2 years ago
Best ResponseYou've already chosen the best response.3Lets 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

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0oh so does parallel mean parallel to the power supply?

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1362330342861:dw C_5 and C_6 would be capacitors in parallel , whereas C_T and C_3 are capacitors in series correct?

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0I mean C_T and C_4....typo :S

.Sam.
 2 years ago
Best ResponseYou've already chosen the best response.3no, 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

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0oh I see.... so for capacitors in parallel \[V=V_1+V_2\] still holds, correct? but the Q's are not equal?

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

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0oh interesting Let's see \[V=\frac Q C\] \[\frac Q C=\frac{Q_1}{C_1}=\frac{Q_2}{C_2}\]

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0Q's the same again?

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

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

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0Yes! Makes sense :)

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0You are awesome! Thank you soo much for teaching me capacitors in parallel and series...Thank You @.Sam.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.