# Homework Help: Capacitance in circuit

1. Apr 13, 2008

### ~christina~

1. The problem statement, all variables and given/known data
Assume that $$C_1= 10.0 \mu F$$, $$C_2= 5.00\mu F$$, and$$C_3= 4.00\mu F$$. A potential difference of $$V$$= 100V is applied across this network of capacitors.

1st picture is original (after that is when I simplified the circuit or tried to anyhow)

a) find the equivalent capacitance of this circuit
b) find the potential difference across capacitor 2
c) what is the charge on this capacitor?
d) how much energy is stored by this capacitor?

http://img397.imageshack.us/img397/4125/21107020ai0.th.jpg [Broken]

2. Relevant equations
$$C_{eq}= C_1+C_2+C_3...$$
$$\frac{1} {C_{eq}} = \frac{1} {C_1} + \frac{1} {C_2} + ...$$

3. The attempt at a solution

a) find the equivalent capacitance of this circuit

well I broke down the circuit into series and parralel sort of. (in the picture)

not sure if I arranged it correctly first of all. (by breaking it down in to series and parallel) because I only came up with one step for breakin it down which seems incorrect

would I calculate it at the begining as if it was parrallel or series??

Thanks

Last edited by a moderator: May 3, 2017
2. Apr 13, 2008

### Nabeshin

Your breakdown of the capacitors looks fine to me. What seems to be the problem here? Not sure what you mean when you say, "
would I calculate it at the begining as if it was parrallel or series??"... perhaps some elaboration on that point would help clear up your question?

3. Apr 13, 2008

### ~christina~

okay. (I was just confused as to what was series and what was parallel)
but I think I get it now.

a) find the equivalent capacitance of the circuit

C1 and C2 are in series so:
$$1/C_{12}= 1/C_1 + 1/C_2$$
$$1/C_{12}= 1/10\mu F + 1/5.00\mu F$$
$$C_{12}= 3.33\mu F$$

$$C_{eq}= C_{12} + C_{3}$$
$$C_{eq}= 3.33\mu F + 4.00\mu F$$
$$C_{eq}= 7.33\mu F$$

I think this is fine.

b) find the potential difference across capacitor 2

not sure how to find this though.

Thanks

4. Apr 13, 2008

### Nabeshin

Well, a total of 100V is applied to the circuit, so you should be able to use the knowledge that voltage is the same in parallel and the sum of the voltage drops in series is equal to the total voltage drop to break it down more... After that, the equation Q=CV should solve most of your woes.

5. Apr 13, 2008

### kamerling

You also need that capacitors in series will have the same charge. (if they started out with the same charge before connecting the voltage source)

6. Apr 13, 2008

### ~christina~

a) find the equivalent capacitance of the circuit

okay I still am not sure how to find it but only on one part I think

Since C1 and C2 are in series then:

$$\Delta V_{total}= \frac{Q} {C_{eq}}$$

but if this is so then

would it be this?

$$V_{tot}= \frac{1} {C_1} + \frac{1} {C_2}$$

but if so then how would I find the individual V across 2?

I know that $$\Delta V_{tot}= \Delta V_1 + \Delta V_2$$

Last edited: Apr 13, 2008
7. Apr 13, 2008

### ~christina~

am I correct in my thinking? anyone?

8. Apr 13, 2008

### Nabeshin

Your thinking is correct, but what I left out is what kamerling so kindly said. Charge on series capacitors is also the same. Therefore C is the same in that first equation you have. Try to use a ratio to get the exact voltage drop :)

9. Apr 13, 2008

### ~christina~

not really getting what your saying but..lets see:

since charge on series capacitors is the same then

Q1=Q2

Do I do this?

$$\Delta V_{total}= \frac{Q} {C_{eq}}$$

solve for Q

$$Q= \Delta V_{total} C_{eq}$$

then I would plug that into

$$Q/C_{eq}= Q_1/C_1 + Q_2/C_2$$

but I don't know C1
and solve for $$Q_2/C_2$$ ??

10. Apr 13, 2008

### Nabeshin

Hrm.. I'm not quite sure what you're doing to be perfectly honest, lol. What I meant by my ratio comment was this:
$$V_{1}$$+$$V_{2}$$=$$V_{tot}$$
Because voltage drop in parallel will be equal.

We also know $$Q=CV$$ for individual capacitors, and $$Q_{1}=Q_{2}$$

So $$Q_{1}=C_{1}V_{1}$$ and, because of the above $$Q_{1}=C_{2}V_{2}$$

I'm going to leave the rest to you, but hopefully you can see the ratio here, and use the first equation provided. Hope this helps to clarify

11. Apr 13, 2008

### ~christina~

is the ratio because
$$Q_{1}=C_{1}V_{1}$$ &
$$Q_{1}=C_{2}V_{2}$$

then plugging into the equation
$$V_1 + V_2= V_{total}$$

I willl get

$$\frac{Q_1} {C_1} + V_2= V_{total}$$

but I still don't understand how If I don't have Q then how will I find the V2 ?

so I think that V total will be 100V

thus the equation would be

$$\frac{Q_1} {10.00\mu F} + V_2= 100V$$

but would Q1 be Vtotal(Ceq)= Q1 ??

so Q1= 100V(7.33x10^-6 F) = 7.33x10^-4 C ?

Last edited: Apr 13, 2008
12. Apr 13, 2008

### Nabeshin

Yes this is correct. Set the first equations equal to eachother, and you can solve for $$V_{1}$$ in terms of $$V_{2}$$! Then use the other equation, and your only unknown is $$V_{2}$$.

13. Apr 13, 2008

### ~christina~

oh...

$$C_1V_1= C_2V_2$$

$$V_1= \frac{C_2V_2} {C_1}$$

then
$$V_1 + V_2= V_{total}$$

$$\frac{C_2V_2} {C_1} + V2 = V_{total}$$

d)find the total energy is stored by the capacitor

hm..

$$U= \frac{Q^2} {2C}= \frac{1} {2} Q\Delta V= \frac{1} {2} C( \Delta V )^2$$

Last edited: Apr 13, 2008
14. Apr 13, 2008

### Nabeshin

Yep. :)

15. Apr 14, 2008

### ~christina~

which of the equations should I use??

$$U= \frac{Q^2} {2C}= \frac{1} {2} Q\Delta V= \frac{1} {2} C( \Delta V )^2$$

and is Delta V, for the whole circuit? or do I calculate it in parts and etc. (Q, C)

16. Apr 14, 2008

### ~christina~

I'm still not sure as to what number to use for the total energy stored in the capacitor.

$$U= \frac{Q^2} {2C}= \frac{1} {2} Q\Delta V= \frac{1} {2} C( \Delta V )^2$$

what Q do I use? and what C do I use? or what change in potential do I use for the equation?

C would be Ceq right but as for the rest..not so sure.

Thanks