Capacitance - How handle multiple caps in series and parallel?

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Capacitance -- How handle multiple caps in series and parallel?

I can't draw the circut for the cuestion so I attached it.

I am having trouble finding the charge across capacitor 2. I don't understand what to do with the 3 capactiors right in that group and how to figure the change in voltage
 

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  • #2
berkeman
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I can't draw the circut for the cuestion so I attached it.

I am having trouble finding the charge across capacitor 2. I don't understand what to do with the 3 capactiors right in that group and how to figure the change in voltage
Just keep using Q=CV. If two capacitors are in parallel, what is their effective total capacitance? If two caps are in series, what is their effective total capacitance?
 
  • #3
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Ok, the capacitance across the 3 would be C+ 1/C+C

then q= VC+ V/C+C or am I completly wrong?
 
  • #4
berkeman
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Ok, the capacitance across the 3 would be C+ 1/C+C

then q= VC+ V/C+C or am I completly wrong?
No, that doesn't look right to me. For one thing the units don't match. You can't add C + 1/C.

The equation for series capacitors is slightly different from what you've written. Re-check it in your book and try again?

Also, are all the capacitors supposed to be the same value? If so, you can simplify the final equation a lot...
 
  • #5
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Parallel capacitors is q/V so C1+C2+C3...
Series capacitors are 1/(1/C1+1/c2+1/C3)

Yes all the Cs are equal. I think I had my series equation wrong but I just don't understand it. I know I have to use the voltage across the capactiors but I'm also not sure how to find that.
 
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  • #6
berkeman
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Parallel capacitors is q/V so C1+C2+C3...
Series capacitors are 1/(1/C1+1/c2+1/C3)

Yes all the Cs are equal. I think I had my series equation wrong but I just don't understand it. I know I have to use the voltage across the capactiors but I'm also not sure how to find that.
Try this approach:

Combine C2 and the cap below it into one capacitance. What is that capacitance?

Now combine that series combination with the cap that is in parallel with it. What is that total combined capacitance?

Now you can figure out what the voltage is across the bottom right cap, since you know the total combined capacitance of the 3 caps above it.

Once you know the voltage of that node, you have the voltage across C2 and the cap below it in series....

Try that approach, and post your work here...
 
  • #7
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Capacitance across the series is C2*C/C+C or C2*C/2 which I assume I can make into just C^2/2Csince all the C's are equal

Capacitance across all three would be 3C/2

Then if I make it all into just 1 capactior it would be 3C/5
The current across it would be q=CV so q=10(3*10)/(5) since it's microfarods it would really be 10V*.000006F so .00006amps or 6.x10^-5 amps is the current the the bottom right of the circut basically I think.

I'm not sure how to work backwards from that. If I split it into 2 capacitors in series again, don't each have to have the same current or is that not an assumption I can make? If I can then that mean the current across the capacitor comprised of the 3 is 6.x10^5 amps.

Then would I split it into the 3 capacitors and say that since 2 are in series and 1 is parallel then the 2 series + the 1 parallel currents should equal 6.x10-5. If I am correct up to here, I don't know how to split these up to find the current across only C2
 
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  • #8
berkeman
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Capacitance across the series is C2*C/C+C or C2*C/2 which I assume I can make into just a 1 since all the C's are equal

Capacitance across all three would be The 1 from above + C so 1+C

Then if I make it all into just 1 capactior it would be (C+C^2)/(1+2C)

The current across it would be q=CV so q=10(10+100)/(21) since it's microfarods it would really be 10V*.000005F so .00005amps or 5.x10^-5 amps is the current the the bottom right of the circut basically I think.

I'm not sure how to work backwards from that. If I split it into 2 capacitors in series again, don't each have to have the same current or is that not an assumption I can make? If I can then that mean the current across the capacitor comprised of the 3 is 5.x10^5 amps.

Then would I split it into the 3 capacitors and say that since 2 are in series and 1 is parallel then the 2 series + the 1 parallel currents should equal 5.x10-5. If I am correct up to here, I don't know how to split these up to find the current across only C2
I'm sorry, but I'm not tracking at all what you are doing.

When you have two series capacitors, both with value C, what is the total capacitance?
 
  • #9
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I'm sorry, I made a huge mathematical mistake from the beginning, I edited it so it should make better sense now.

when I say the capacitors in the series, I'm talking about the one labled C2 in my picture and the one right below it.

Then the one that is parallel to those 2 would be the one I combined next and followed by combining the last capacitor on the bottom right of my picture.
 
  • #10
berkeman
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Capacitance across the series is C2*C/C+C or C2*C/2 which I assume I can make into just C^2/2Csince all the C's are equal
This still doesn't make sense to me...

Capacitance across all three would be 3C/2
But that is correct.

Then if I make it all into just 1 capactior it would be 3C/5
Also correct, but I don't think you need the total capacitance including the bottom cap. You need to figure out how the charge from the 10V bias distributes between the bottom capacitor and the top 3 capacitors, in order to figure out what the voltage is at the top node of the bottom capacitor.

The current across it would be q=CV so q=10(3*10)/(5) since it's microfarods it would really be 10V*.000006F so .00006amps or 6.x10^-5 amps is the current the the bottom right of the circut basically I think.
q is not a "current", it is the electric charge.

When you have two capacitances in series with an overall voltage across the series combination, how does the voltage divide between them? Use Q=CV to figure out how the voltage divides between the top combine 3-caps and the one bottom cap. Then use that voltage to work your way back to the voltage across C2.
 
  • #11
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I was distinguishing between capacitor C2 and C, but since they are equal it would just be C&C. Does that clear it up? If not I did this C*C/C+C=C/2

current, charge, voltage...I still can't get the differenct between them and it's really hard becuase I can't picture the difference in my head.

ok so in a series the charge is the same so if the voltage across all 4 is 60microC then the charge of the 3 combined has to be 60 as well. now its 60mC=V*3C/2 so the V=4

Then voltage is the same across parallel capacitors so I can disreguard the lone capacitor on the right. so now q=V*C/2 to find the charge across the 2 capacitors and since they're in a series it will be the same for both. q=4*10/2 q of C2= 20mincroC or 2.x10^-5C and I think that is actually the correct answer. Please tell me I didn't break and laws in figureing that out because I'm good about that. If it looks good, thank you so much for your help.
 
  • #12
berkeman
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I was distinguishing between capacitor C2 and C, but since they are equal it would just be C&C. Does that clear it up? If not I did this C*C/C+C=C/2
I think you have a typo there. Before you correctly got 3C/2 for the combination of the top 3 capacitors...

BTW, I have trouble parsing your equation because I'm used to seeing it written differently:

[tex]C_{tot} = C + \frac{1}{\frac{1}{C} + \frac{1}{C}} = C + \frac{C}{2} = \frac{3C}{2}[/tex]

EDIT -- Oh, by adding parens I think I see what you did: (C*C)/(C+C)=C/2

current, charge, voltage...I still can't get the differenct between them and it's really hard becuase I can't picture the difference in my head.
Charge is just the number of electrons multiplied by the charge on a single electron. The units of charge are Coulombs.

Voltage is the potential difference between two places. It can be the voltage between the plates of a capacitor, or the voltage drop across a resistor from a current flowing. Units of voltage are volts.

Current is the flow of charge (usually electrons). A voltage difference between two points generates an electric field, which generates a force on charges and causes them to flow. The voltage can be across a resistor's terminals, forcing a current to flow through the resistor. Or you can have a voltage source like your 10V source in this problem, and that forces electrons to flow onto the plates of the capacitors, charging them up to a total of 10V across the capacitor network.

ok so in a series the charge is the same so if the voltage across all 4 is 60microC then the charge of the 3 combined has to be 60 as well. now its 60mC=V*3C/2 so the V=4
I don't know what you mean by "in series the charge is the same". Not in general. Do you mean because they have the same capacitance in this problem? In that case it would be correct. You appear to be mixing mC and uC in your units, however.

If the total capacitance across the 10V is 3C/5, then Q = CV = 3uF/5 * 10V = 6uC total.

Then voltage is the same across parallel capacitors so I can disreguard the lone capacitor on the right. so now q=V*C/2 to find the charge across the 2 capacitors and since they're in a series it will be the same for both. q=4*10/2 q of C2= 20mincroC or 2.x10^-5C and I think that is actually the correct answer. Please tell me I didn't break and laws in figureing that out because I'm good about that. If it looks good, thank you so much for your help.
I need to look at this last part a bit more...
 
  • #13
berkeman
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One of the things that you can use to help solve this problem is the voltage division rule for capacitor circuits:

http://www.aikenamps.com/VoltageDividerRule.htm [Broken]

It is kind of opposite of the voltage divider equation for resistors...

BTW, capacitive dividers wouldn't pass a DC voltage, but problems like this assume that you turned on the power supply at some time, and that caused currents to flow to charge up the caps to their final DC values. Note that the same current I = dQ/dt flows through two series capacitors in a voltage divider, so the same change in charge would have to happen on each capacitor in the series string. That's why you get different voltages across the two caps in a series voltage divider -- because the same charge is being stored on each of the caps, but the different capacitance values Cx result in different cap voltages Vx.
 
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  • #14
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What I meant was in a series, the Q's in Q=CV are the same for each capacitor in that series and in a parallel series the V's in Q=CV are the same.

Thanks for explaining the differences, why the book couldn't make it that easy I'll never understand
 

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