Capacitors in Electrical Circuits

  • #1

Homework Statement


In the circuit below, C1 = 5 microfarads, C2 = 10 microfarads and R = 1000 ohms. Initially, the switch is open, C1 is charged to 20 volts, and C2 is uncharged. At time t=0 the switch is closed.
I think I managed to solve the first two but I have troubles with part (c):
(a) Calculate the voltage across C1 at a much later time.
(b) The energy stored in a charged capacitor is given by U = (CV^2)/2. Calculate the energy stored in C1 and C2 before and after closing the switch.
(c) Derive an expression for the power dissipated in R as a function of time for t>0.

Diagram:
|--------- switch -------- R------|
C1...........................................C2 [ignore the dots, they are just used for spacing)
|--------------------------------|


Thank you so much for your time,


Homework Equations


Q=CV
U = (CV^2)/2
P = I^2R


The Attempt at a Solution


a) Q initial=CV=(5)(20)=100 microF
when the switch is closed, the caps are in parallel which means that C total=15 microF
thus V=Q/C=100/15=6.7V

b) Before: U1 = (CV^2)/2=1000microJ
U2 = (CV^2)/2=0J
After: U1 = (CV^2)/2=111microJ
U2 = (CV^2)/2=222microJ

c) I know that total energy lost in resistor is 1000-333=667microJ and I think P = I^2R can be used somehow, but I need some help from here.

Thank you for your help!
 

Answers and Replies

  • #2
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If I understand your diagram correctly, C1 and C2 are not in parallel.
 
  • #3
Sorry about that... here's a better picture of the circuit.
 

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  • #4
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Okay, that's what I thought it was. In order for two elements to be in parallel they have to share the same two nodes. The series combination of R and C2 would be in parallel with C1. Or, perhaps more useful, the series combination of C1 and C2 would be in parallel with R.
 
  • #5
Would that mean that answers for (a) and (b) are incorrect since I considered them to be in parallel?
 
  • #6
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Yes, but I can give you some pointers in reworking.

Let's just solve it intuitively because I have a feeling the formal way will be tedious. C1 starts out with a voltage, which means that a current flows from C1 through R, to C2. Now, C2 is hungry, he has no charge, and so he eats up all the current that C1 sends out. So, consequently, C1 is losing charge, in turn losing potential, while C2 is gaining charge, and gaining potential. So, I ask you, after a long time, what happens in this scenario?
 
  • #7
I would think that eventually C2 will be completely charged and C1 will be empty, ie. the opposite from how it started. Now wouldn't the charge in C2 move back to C1 eventually and the cycle will repeat itself until all energy is converted to heat through the resistor?
 
  • #8
SammyS
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Homework Statement


In the circuit below, C1 = 5 microfarads, C2 = 10 microfarads and R = 1000 ohms. Initially, the switch is open, C1 is charged to 20 volts, and C2 is uncharged. At time t=0 the switch is closed.
I think I managed to solve the first two but I have troubles with part (c):
(a) Calculate the voltage across C1 at a much later time.
(b) The energy stored in a charged capacitor is given by U = (CV^2)/2. Calculate the energy stored in C1 and C2 before and after closing the switch.
(c) Derive an expression for the power dissipated in R as a function of time for t>0.

Diagram:
|--------- switch -------- R------|
C1...........................................C2 [ignore the dots, they are just used for spacing)
|--------------------------------|


Thank you so much for your time,


Homework Equations


Q=CV
U = (CV^2)/2
P = I^2R


The Attempt at a Solution


a) Q initial=CV=(5)(20)=100 microF
when the switch is closed, the caps are in parallel which means that C total=15 microF
thus V=Q/C=100/15=6.7V

b) Before: U1 = (CV^2)/2=1000microJ
U2 = (CV^2)/2=0J
After: U1 = (CV^2)/2=111microJ
U2 = (CV^2)/2=222microJ

c) I know that total energy lost in resistor is 1000-333=667microJ and I think P = I^2R can be used somehow, but I need some help from here.

Thank you for your help!

(a) Technically, the capacitors are not in parallel, (they're in series) but for the purpose of finding the final charge of each of the capacitors you can treat them as if they are in parallel.

For part (b): What are your units for energy? micro__?__

part (c):
Here, you need to realize that the capacitors are in series, an calculate the equivalent capacitance. Ceq = (C1C2)/(C1+C2)

The time constant for the current is RCeq .

I=I0e(-t/(RCeq))
 
  • #9
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Okay, well, you have to also consider that as C1 discharges and releases current that C2 is charging on and then sending out an opposing current from itself in the opposite direction. C1's current gets slightly weaker as C2's current gets slightly stronger. Will C2 ever overpower C1, as you suggested?
 
  • #10
SammyS
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I would think that eventually C2 will be completely charged and C1 will be empty, ie. the opposite from how it started. Now wouldn't the charge in C2 move back to C1 eventually and the cycle will repeat itself until all energy is converted to heat through the resistor?

No. As C1 discharges, C2 charges until both have equal voltage.
 
  • #11
Thanks so much for clearing it out guys! I think I'm finally starting to understand how caps work.
Also, why do we need to calculate the equivalent capacitance?
 
  • #12
1,860
0
You don't need to calculate the equivalent capacitance, but just like equivalent resistance can make things easier so can equivalent capacitance.

So, now that you understand the problem, do you stand by your answers? Have you reworked anything yet? Sorry for not being more direct with help, but it's better for your learning this way. :)
 
  • #13
I understand that the capacitors will have to reach some equilibrium eventually (ie. the long run) so treating them in series would work as SammyS suggested. Regarding part (b), it's just conservation of energy problem so it seems right.
For (c) the expression will have to be some decay function since energy is being lost through the resistor and I=I0e^(-t/(RCeq)) is also what we did in class so it makes sense.
I think I get it know. Is there something I don't see?
Thanks again so much, I really appreciate the help.
 
  • #14
gneill
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I would think that eventually C2 will be completely charged and C1 will be empty, ie. the opposite from how it started. Now wouldn't the charge in C2 move back to C1 eventually and the cycle will repeat itself until all energy is converted to heat through the resistor?

If the circuit components are ideal there won't be any "inertia" in the current to enable C1 to empty; as soon as the potential difference driving the current from C1 to C2 is gone, the current goes to zero. Current inertia in electrical circuits involves inductance.

In the real world, all components have a certain amount of inductance (a perfectly straight piece of wire 1cm long has some inductance!). So in that real world, what you'd be looking at would be a form of LRC circuit - one with inductance, resistance, and capacitance. Fortunately, even in the real world the value of L would be minuscule, and thanks to the presence of R, it would be seriously overdamped. So still no worries about significant oscillations breaking out!
 

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