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Charge on capacitor

  • Thread starter huh
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huh
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1. Homework Statement
In the circuit shown in the figure, the applied potential is V_ab = 18.0V . For C_1= 3.15(microFarads) , C_2 = 5.50microF , and C_3 = 6.85microF ,

Find the charge q_1 on the capacitor C_1 (and the same for q2 on c2 and q3 on c3)

Picture is attached...


2. Homework Equations
For capacitors in series: V_total= V_1+V_2+V_3...
= (Q/C)_1+ (Q/C)_2+(Q/C)_3
C_total= Q/V_total 1/C_total= (1/C)_1+(1/C)_2

In parallel: Q= Q_1+Q_2+Q_3=(CV)_1+(CV)_2+(CV)_3

C_total= C_1+C_2+C_3

3. The Attempt at a Solution

I tried to use Q=CV and got 5.67E-5 or 56.7(w/o changing from microF)...wrong

Then, I tried using the series equation to calculate capacitance, but it didn't work.

I'm confusing myself on how to calculate the stupid charge...
1. Homework Statement



2. Homework Equations



3. The Attempt at a Solution
 

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Answers and Replies

957
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maybe you're jumping the gun, first what are the voltages across the block, (Vad), and C3 (Vdb)?
 
Tom Mattson
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3. The Attempt at a Solution

I tried to use Q=CV and got 5.67E-5 or 56.7(w/o changing from microF)...wrong
It looks like you just multiplied [itex]C_1[/itex] by 18V. But the voltage across [itex]C_1[/itex] is not 18V.

Start by finding the equivalent capacitance. That will allow you to find the total charge on all of the capacitors.
 
huh
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Would the equivalent capacitance for the whole thing be 6.85E-6

I calculate C1 and C2 together in parallel and then took that answer and did it again with C3.
 
Tom Mattson
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That's not what I got. Can you show your steps so we can see what you're doing?
 
huh
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(1/3.15e-6)+(1/5.5e-6)=4.99278e5

(1/4.99278)= 2.00289e-6

2.00289e-6+ (1/6.85e-6)= 1.45985e5

(1/1.45985e5)= 6.85e-6
 
Tom Mattson
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Your mistake is line 1. [itex]C_1[/itex] and [itex]C_2[/itex] are in parallel, but you treated them as though they were in series.
 
huh
15
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I see what you mean in calculating it as a series. Whoops :)
So would you just add the values for capacitance together?

(3.15e-6)+(5.5e-6)=8.65e-6

8.65e-6+6.85e-6= 1.55e-5
 
Tom Mattson
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I see what you mean in calculating it as a series. Whoops :)
So would you just add the values for capacitance together?

(3.15e-6)+(5.5e-6)=8.65e-6
Correct.

8.65e-6+6.85e-6= 1.55e-5
Incorrect. The combination of [itex]C_1[/itex] and [itex]C_2[/itex] is in series with [itex]C_3[/itex]. You had that right the first time (although the number you carried over from line 1 was wrong).
 
huh
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It's in series? It looks like it is in the same position as the first two are...

OK so how do I get it for each individual charge (and capacitance)?
 
957
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Capacitors for the purposes of lumping are exactly opposite to resistors: when in parallel they add numerically, and when in parallel, they add as resistors in series. If you think about the latter, it makes sense as one is adding more surface area while keeping the dialectric thicknes constant, ie just making the plates bigger. Now when adding in series, we can't have charge differences under conditions of equilibrium. But those same charges, Q, will lead to different voltages with different capacitances in series. So voltage division around a closed loop will lead to the capacitors sharing voltage unequally. This help at all?
 
huh
15
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umm... I'll try reading this again tomorrow... I understand part of what you are saying, but it's not "sticking" in my head right now.
Thanks though...
 
Tom Mattson
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It's in series? It looks like it is in the same position as the first two are...
They certainly aren't in the same position. Two circuit elements are in series if they have the same current running through them. They are in parallel if they have the same voltage across them. [itex]C_1[/itex] and [itex]C_2[/itex] are obviously in parallel (use KVL to see that), and the combination [itex]C_1[/itex] and [itex]C_2[/itex] are obviously in series with [itex]C_3[/itex]. All of the current that flows through the parallel combination, also flows through [itex]C_3[/itex].

OK so how do I get it for each individual charge (and capacitance)?
Start by analyzing the circuit in which [itex]C_1[/itex] and [itex]C_2[/itex] have been reduced to an equivalent capacitance (call it [itex]C_{12}[/itex]). Since that equivalent capacitor is in series with [itex]C_3[/itex], the two capacitors have to have the same charge. So you should be able to write down a single equation with one unknown: the charge on [itex]C_3[/itex].
 

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