# Series Parallel Capacitors

Discussion in 'Homework Help' started by paul_alan, Dec 12, 2011.

1. ### paul_alan Thread Starter Member

Nov 5, 2011
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I'm having trouble with calculating a series parallel capacitor circuit. I have a 50Hz 10V AC source. a 2 micro farad (C1) and 3 micro farad (C2) on branch 1; and a 0.8 micro farad capacitor (C3) on branch 2 in parallel to B1. finally there is a 4 micro farad capacitor (C4) at the end of the circuit right by the source. I'm trying to figure out Ct, XCt, IC4 and VC3 and I'm having troule because our instructor is gone and we have a substitute that doesn't know anything. And he's going to be there until the end of the term.

2. ### praondevou AAC Fanatic!

Jul 9, 2011
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Is that what you have?

Where are you stuck? Start by calculating the series/parallel capacitances, then their impedances.

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3. ### paul_alan Thread Starter Member

Nov 5, 2011
43
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Yeah, that's almost exactly what the schematic looks like. I calculated down the total capacitance, and I came up with 2.789 micro farads. And I calculated the total reactance and got 1.142 kilo ohms. I'm having some trouble with voltage drops and current in a series-parallel capacitor circuit.

4. ### paul_alan Thread Starter Member

Nov 5, 2011
43
0
Series parallel capacitor circuit

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5. ### praondevou AAC Fanatic!

Jul 9, 2011
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Ok let's start here. How did you do it? It's not correct.

6. ### thatoneguy AAC Fanatic!

Feb 19, 2009
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Take a glance At This

It's like a 2 dimensional slide rule for impedance/frequency/inductors/capacitors, and gives you a quick rough idea of what the answer roughly should be around.

7. ### paul_alan Thread Starter Member

Nov 5, 2011
43
0
I thought those values were calculated similar to resistors. i guess that wasn't correct

8. ### crutschow Expert

Mar 14, 2008
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5,190
No, it's the reverse of resistors.

Capacitor values in parallel add together, like resistors in series, and capacitors in series are the inverse (reciprocal) of the sum of the inverse values, like resistors in parallel.

Last edited: Dec 13, 2011
9. ### paul_alan Thread Starter Member

Nov 5, 2011
43
0
I know and understand series capacitors and parallel capacitors. what i did for this circuit was product over sum (2μF x 3μF) \ (2μF + 3F), then add the 0.8μF on branch 2 parallel to branch one, and again used the product over sum with that value and the 4μF capacitor at the end of the circuit. and i was told that mly final value (2.789μF) was incorrect.

10. ### praondevou AAC Fanatic!

Jul 9, 2011
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So are you ok to solve the problem now? Do you want to post it?

11. ### thatoneguy AAC Fanatic!

Feb 19, 2009
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Try finding the impedance of all the capacitors at the given frequency.

Then treat it as a series/parallel resistance problem to find the current and voltages.

12. ### paul_alan Thread Starter Member

Nov 5, 2011
43
0
No, because i don't know the right way to get the total capacitance.

13. ### praondevou AAC Fanatic!

Jul 9, 2011
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1. C1 and C2 --> 1/C12 = 1/C1 + 1/C2

2. Cseries in parallel with C3 --> C123 = C12 + C3

3. C4 in series with the rest --> 1/Ctotal = 1/C123 + 1/C4

14. ### thatoneguy AAC Fanatic!

Feb 19, 2009
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Draw the entire circuit with the capacitors in a vertical stack along one side, making sure the connections are all the same.

Capacitors in parallel are just like resistors in series
Capacitors in series are just like resistors in parallel

Here is what I came up with, though there are errors in the below. It should give you an idea of how to calculate it though. The formulas are almost there.

$C_{total}=\frac{1}{\frac{1}{2\mu F + 3\mu F}+0.8uF}+\frac{1}{4\mu F}\\
C_{total}=\frac{1}{1.2 \mu F + 0.8\mu F}+ 0.250 \mu F\\
C_{total}=0.5 \mu F + 0.250 \mu F
C_{total}=0.750 \mu F
C_{total}=750nF
$

$
Freq=50Hz
Z=\frac{1}{2\pi f\cdot C}
Z_{C1}=\frac{1}{2\pi 50\cdot 2\mu F}=1.592k\Omega
Z_{C2}=\frac{1}{2\pi 50\cdot 3\mu F}=1.061k\Omega
Z_{C3}=\frac{1}{2\pi 50\cdot 0.8\mu F}=3.979k\Omega
Z_{C4}=\frac{1}{2\pi 50\cdot 4\mu F}=795.8\Omega
Z_{total}=\frac{1}{\frac{1}{2.653 k\Omega}+\frac{1}{3.979k\Omega}}+795.8\Omega\\
Z_{total}=1.592 k\Omega + 795.8\Omega
Z_{total}=2.388 k\Omega
$

15. ### paul_alan Thread Starter Member

Nov 5, 2011
43
0
This is the problem. I'm getting so frustrated with it because our instructors talk to us like we're electrical engineers. if someone could please dumb it down as much as possible so i can figure it out. i'm getting ready to throw my book through a window. I need to find Ct, which i got 1.333μF. i need XCt, which i got 2.389kΩ. and i need to find the current for C4 and voltage across C3.

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16. ### praondevou AAC Fanatic!

Jul 9, 2011
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That's correct. Current through C4 will be the total current, i.e. total voltage divided by total impedance.
For the voltage on C3 find the impedance of C4. C3 is total voltage minus that.

17. ### thatoneguy AAC Fanatic!

Feb 19, 2009
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You are doing fine.

C1 and C2 are in series, both of which are in parallel with C3. Then that whole set is in series with C4.

Get your impedances and see if it matches what I have above. Then treat it like a resistor voltage divider. (I am assuming you know how do do mesh analysis or node analysis?)

18. ### paul_alan Thread Starter Member

Nov 5, 2011
43
0
I got 4.186mA for total current. I took Vs / XCt (10V / 2389kΩ). one thing i don't get is why isn't the voltage across C3 10V if it is parallel the the source?

19. ### thatoneguy AAC Fanatic!

Feb 19, 2009
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It's not quite in parallel, C4 is down there. The voltage across (C1+C2) should equal the voltage across C3. That voltage subtracted from the source is the voltage across C4.

20. ### paul_alan Thread Starter Member

Nov 5, 2011
43
0
Okay, is the current Vs / XCt of C1 +C2 (3.768mA) added to Vs / XCt of C3 (2.521mA) (6.281mA total). then that current times XC4 (796.178Ω) which gives me a voltage across C4 of 5V?