# Circuit with three capacitors

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1. Feb 18, 2016

### vizakenjack

1. The problem statement, all variables and given/known data
A potential difference of V = 38.0 V is applied across a circuit with capacitances C1 = 4F, C2 = 5F, and C3= 2F, as shown in the figure.
1. What is the magnitude and sign of q3l, the charge on the left plate of C3 (marked by point A)?
2. What is the electric potential difference, V3, across C3?
3. What is the magnitude and sign of the charge q2r, on the right plate of C2 (marked by point B)?

2. Relevant equations

Q = CV
Useful concepts:

Charge for capacitors in series is the same.

Current for resistors/capacitors in series is the same.

Voltage for resistors/capacitors in parallel is the same.

3. Solution:

1. C1+C2=C12 = 9F

Cequivalent = (C12 * C3)/C12 + C3) = ~1.6F

q3l = Ceq * ∆V = 60.8V

2.

V3 = Q3/C3 = 60.8V = 30.4 V

3.

q2r = C2/(C1 +C2 ) * Q3 = 34.05 C

Already have the solutions, just need explanations.
C1 and C2 are in parallel, therefore should have the same Voltage (potential difference).

Third one is the one I'm having trouble with.

It's so similar to the "resistance of interest equation", like in here, when finding V_1.

It's so frustrating, I know all the formulas, Q = CV, capacitance in series just add up (e.g. C12 + C3).

But no one ever tells how to find different voltages / charges on different elements on the circuit.

Like, in this example, would you still be able to use the "resistance of interest" formula:

R1/(R1+R2) * Vs (supplied voltage)
if additionally further down the circuit there were a bunch of other resistors connected in series or neither in series nor in parallel?
Would you have to account for them when finding Voltage (V1) across R1?

-----

1. The voltage across C1 and C2 is the same, correct? But their sum doesn't equal 38V, nor does it mean that voltage across C1 is 38V, correct?

But whatever the voltage across C1 is, it's of the same magnitude in C2, despite the fact that capacitors have different capacitance?

2. Why is the equivalent capacitance and the given voltage is used to calculate the charge on C3 instead of Q3 = C3 * V3?

I know the formula Q3 = C3 * V3 is valid, but how come the equivalent capacitance * given voltage also yields V3?

Because at the left plate of C3, the current have traveled far enough to have passed all capacitors (hence the equivalent capacitance), and at that point voltage must therefore be the same as it was supplied initially? So you treat the circuit at that point as if it has one capacitor (Ceq)?

Or because in the end, the same voltage must return to the negative end of Voltage supply...? And because of the KVL loop rule, that voltages must add to 0...?

But I thought Voltage either rises or drops across a capacitor. Once current passes either through a capacitor or a resistor, once it leaves an element, it will have different Voltage... uhm?

3. How does one figure the Voltage across C2? Like, how do you know how much of the supplied voltage would go to both V1 and V2? Is it even correct to say: "go to"? I mean, Voltage has a relationship with current. And the more current goes to a capacitor, the higher the voltage will be, right?

Last edited: Feb 18, 2016
2. Feb 18, 2016

### cnh1995

Unit of charge is coulomb, not volt.
This will give you the total charge on the equivalent capacitance. Since parallel combination of C2C1 is in series with C3, charge on equivalent capacitance will also be equal to that on C3.

3. Feb 18, 2016

### vizakenjack

it does give you the charge on the left plate of the Capacitor3 (meaning, it is the correct answer).

4. Feb 18, 2016

### cnh1995

Right. I just edited my post. I myself had drawn the diagram wrong, showing C2 in place of C3. Apologies..

Last edited: Feb 18, 2016
5. Feb 18, 2016

### cnh1995

If there are any additional resistors in series, parallel or in any other combination, you can reduce the network to a simple series network like the one above using simplification techniques such as series-parallel reduction, star-delta conversion, symmetry etc.and use the above formula. Voltage division rule for capacitors is not same as it is for resistors.

6. Feb 18, 2016

### vizakenjack

Indeed, but I already know it.

So it's a no then? So as I suspected, additional resistors screw that formula up.

So that formula for Resistance of interest... which btw is weird and I don't know how it was derived, only applies to resistors in series, and if there's no other combination of resistors in the circuit?

7. Feb 18, 2016

### cnh1995

Yes. From basic network theory, elements in parallel share equal voltage. As you know
V=Q/C, Q1/C1=Q2/C2. The charges on the capacitors will be different if the capacitances are different.
As I said in my edited post #2, parallel combination of C1C2 is in series with C3. So, charge on C3 will be same as that on the equivalent capacitance( charge on series capacitors is same).

8. Feb 18, 2016

### cnh1995

Its not weird if you know how its derived..
Voltage across R1,
V1=(Current through R1)*R1
Current through R1=current through whole circuit= Vs/(R1+R2).
Substitute this in the equation for V1 above and see..
And yes, it applies only to resistors in series.

9. Feb 18, 2016

### ehild

The voltages across series elements add up. So the voltage across C3 + the voltage across Ceq is equal to 38 V.
You got the charge on C3 by multiplying 38 V by the total capacitance. What is the voltage V3 across C3? So what is the voltage across Ceq, the parallel combination of C1 and C2?

10. Feb 19, 2016

### vizakenjack

Why does it give me the charge Q3?

Why Q3 = 38V * Ceq? It should give you the overall charge, if you combine all capacitors. But you only need the charge Q3...

11. Feb 19, 2016

### cnh1995

Charge on series capacitors is same as that on the equivalent capacitance.
So, 38V*Ceq=C3*V3

Last edited: Feb 19, 2016
12. Feb 19, 2016

### vizakenjack

That'd be true if you had the charge Q12 (combined charge of C1+C2), then the charge Q3 = Q12

That's not the case here, when you do 38V * C123 (eq) you're finding the charge Q123, not Q12...

13. Feb 19, 2016

### cnh1995

Q3=Q12=Qequivalent capacitance of C3 in series with C12

14. Feb 19, 2016

### cnh1995

Suppose C1 and C2 are in series. Their equivalent capacitance=Ceq=C1C2/C1+C2
Now, charge on C1=charge on C2=charge on Ceq.

15. Feb 19, 2016

### vizakenjack

"equivalent capacitance of C3" doesn't even make sense, what is it??

Ceq = C123.

Q1=Q2
but it can't also be = Qeq, maybe they add up to Qeq, but Q1 < Qeq for sure. How can one capacitor hold the same amount of charge than the two capacitors one of which is the same?

16. Feb 19, 2016

### cnh1995

I meant equivalent capacitance of "C3 in series with parallel equivalent of C1C2"..forgot to add double quotes..
Nope, not for series capacitors!Think about it. Suppose you have two capacitors of 5F in series and you apply 10V across their series combination. Voltage across each capacitor will be 5V, so charge on each capacitor=Q=CV=25C. Their series equivalent capacitance will be 5/2=2.5V and voltage across equivalent capacitance will be 10V. So charge on the equivalent capacitance will be Q=Ceq*V=2.5*10=25C, same as that on the individual capacitors.
When capacitors are in series, charge on individual capacitor is equal to the charge on their equivalent capacitance. Ceq for series capacitors is less than the individual capacitances but voltage across the equivalent capacitance is greater such that,
CindividualVindividual=CequivalentVequivalent=Qindividual=Qequivalent.
Here, Cequivalent<Cindividual but Vequivalent>Vindividual such that their product(i.e. charge) remains the same.
Hope this makes sense now...

Last edited: Feb 19, 2016
17. Feb 19, 2016

### cnh1995

If you remember how series capacitors are charged, you'll see that the charging current and charging time are same for each capacitor. This means the charges on each capacitor MUST be equal.
This is true for capacitors in parallel, not in series. Parallel capacitors have same voltage but charges may be different. Since V=Q/C,
you can see charges on the capacitors are (have to be) proportional to their capacitances in order to maintain the same voltage V across them.

Last edited: Feb 19, 2016