My work: your thoughts needed!

  • #1
My question is:

A 3.00 uF and a 5.00 uF capacitor are connected in series across a 30.0 V battery. A 7.00 uF capacitor is then connected in parallel across the 3.00 uF capacitor. Determine the voltage across the 7.00 uF capacitor.

Ok,

there are two situations, one with the 7 capacitor and one without.

Ok, so I guess I can find the equivalent capacitances by combining some of the capacitors.

So for situation 1: 1/Cs=1/(3x10^-6F)+1/(5.00x10^-6F), (1/Cs)^-1=Cs

Cs=1.88x10^-6F

Ok now for situation 2: 1/Cs=1/(7.00x10^-6F+3.00x10^-6F)+1/(5.00x10^-6F)
My Cs here is: 8.33x10^-6 F.

Here's where I'm stuck:


I think I can go, CV=CV+CV Or Ctotal x Vtotal= Ctotal x Vtotal +C at 7 x V at 7.

Any ideas?
Thanks.
 

Answers and Replies

  • #2
21
0
NotaPhysicsMan said:
A 3.00 uF and a 5.00 uF capacitor are connected in series across a 30.0 V battery. A 7.00 uF capacitor is then connected in parallel across the 3.00 uF capacitor. Determine the voltage across the 7.00 uF capacitor.
Since it is DC, the charge builds up on the capacitor until the potential across the capacitor is equal to the applied voltage. The voltage drop across the capacitor is Q/C.

The two parallel capacitors are equivalent to 10 uF. The total capacitance is 10/3 uF. And the charges on the 'inside' are at equal potential.

That means the charge on the two parallel capacitors is 1/3 of what it would be if the 30 v was applied only to it. And on the 5 uF capacitor, the charge is 2/3 of what it would be if the 30 v. was applied to it. You take 'er from there now.
 
  • #3
Ok, I get everything up to 1/3 this and 2/3 that. I know how you got the 10/3 uF. The next part is fuzzy though. Where did 1/3 come from?

It seems that your saying 1/3 x 30V=10V is the voltage drop across the 7, but any clarifications?
Thanks.
 
  • #4
Uh yea, *bump*. Anyone want to just explain what Calculex is saying?
 
  • #5
21
0
NotaPhysicsMan said:
Ok, I get everything up to 1/3 this and 2/3 that. I know how you got the 10/3 uF. The next part is fuzzy though. Where did 1/3 come from?

It seems that your saying 1/3 x 30V=10V is the voltage drop across the 7, but any clarifications?
Thanks.
The total charge in the system adds to 0 and the total charge between the capacitors is 0. So the charge Q on the 'outside' plate of the 10 uF capacitor is equal and opposite to the charge Q on the outside plate of the 5 uF capacitor and the charge on the inside plates (which totals 0) distributes itself so there is no potential difference between the two plates. In otherwords, the charge on each of the plates of the two series capacitors (treat the two parallel capacitors as one) is the same.

So:

V1 + V2 = 30

Q/C1 = V1 (C1 = 3 + 7 = 10 uF)

Q/C2 = V2 (C2 = 5 uF)

Capacitors in series will grab enough voltage to hold the charge Q that the other capacitor is holding. Capacitors are trickly little devils and to out-smart them you have to try to think like they do.
 
  • #6
Ok, so what you're saying is:

Q=Q
Since I don't have the value for Q, I will eliminate it.

C1V1=C2V2, also I can sub in V2=30-V1

so C1V1=C2(30-V1)
C1V1=30C2-C2V1
Solve for V1
C1V1+C2V1=30C2
V1(C1+C2)
V1=30C2/(C1+C2)
V1=10Volts

and V2=20Volts I guess since V1+V2=30

I see. Thanks.
 

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