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Circuit with resistor, switch and capacitor

  • #1

Homework Statement


We have a circuit with a battery, two resistors, a switch and a capacitor. We assume that the battery has a switch so that we can choose whether it should be a voltage source for the circuit.
krets_element.png

a) At time t = 0, we put on a voltage Vs
over the battery in the circuit. What is the current over the resistors and over the capacitor?

b) At time t = 0, we put on a voltage Vs
in the circuit. What is the charge on the capacitor as a function of time? What is the current over R2
as a function of time?

The Attempt at a Solution


I assume that the circle with the V inside is the switch. I guess I have to use
Kirchhoffs first and second law somehow, can someone give me hints? I not sure if understand the different components of the diagram, where is the switch for instance?
 

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Answers and Replies

  • #2
gneill
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Usually a circle with a V inside represents a voltmeter. The switch would be associated with the Vs(t) component (the battery).

What have you learned about the basic RC circuit so far?
 
  • #3
Not much, so there is no resistance in the voltmeter?
 
  • #4
gneill
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Not much, so there is no resistance in the voltmeter?
The voltmeter, unless stated otherwise, is considered to be ideal and does not affect the circuit at all. It effectively has infinite resistance (open circuit).

You'd best go to your class notes and textbook and look at the sections that describes capacitors and the RC circuit. There's a formula or two you'll need.
 
  • #5
Yes, sadly the course I'm taking is horrible, and there really isn't a book available, but I do know that

for the capacitor: I = C*dV/dt
and for the resistor V = RI,

But "open circuit", does that mean that the circuit is equivalent with this image, see picture:
 

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  • #6
gneill
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Yes, sadly the course I'm taking is horrible, and there really isn't a book available, but I do know that

for the capacitor: I = C*dV/dt
and for the resistor V = RI,
Yes, but you should also find expressions for I(t) , Q(t), and V(t) for charging and discharging the basic RC circuit configuration.

If they're not in your notes then a web search will turn up what you need.
But "open circuit", does that mean that the circuit is equivalent with this image, see picture:
Yes.
 
  • #7
I found, using Kirchhoffs laws, the following: but I don't know how to go further
 

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  • #8
How do i find I (current) as function of time?
 
  • #9
gneill
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How do i find I (current) as function of time?
You would have to recognize that for a capacitor,
$$I_c = C \frac{dV_c}{dt}$$
and similarly,
$$V_c = \frac{1}{C} \int{I_c} \, dt$$
then solve the resulting differential equation(s).

It is much faster to recognize that for a DC driven RC circuit, all currents and potential differences will have the form of decaying or growing exponential functions (##A e^{t/\tau}## or ## A(1 - e^{t/\tau}##). Then it's a matter of determining the time constant ##\tau## for the circuit and finding the initial and steady state values for the quantities in question and writing the appropriate equations to fulfill those conditions.

Have you studied how to determine initial and steady state conditions for a circuit with capacitors?
 
  • #10
I think I understand more now, I have to assume that I and V are functions of time, find a/the differential equation and solve it?
 
  • #11
Have you studied how to determine initial and steady state conditions for a circuit with capacitors?
Well, I assume that there is no charge in the capacitor at t=0, is that what you mean?
 
  • #12
gneill
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Well, I assume that there is no charge in the capacitor at t=0, is that what you mean?
Yes, and how does it make the capacitor appear to the surrounding network at that instant of time?

When steady state is achieved, how much current does a capacitor draw or supply in the circuit? How then does the capacitor appear to the surrounding network at that time?

How can you use this information to find the initial and final values of voltages and currents?
 
  • #13
gneill
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I think I understand more now, I have to assume that I and V are functions of time, find a/the differential equation and solve it?
That is the "starting from the basics" approach, yes.

It's much quicker to already know what the solution is going to look like and then simply writing down the function of time. :smile:
 
  • #14
At t = 0 i suppose all charged particles would move into the capacitor, and none would go through R1. And at the steady state it would be the other way around; the capacitor draws zero current (none can pass). And I guess it would appear as if the capacitor would be like an open circuit? Im a on to something?
 
  • #15
gneill
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At t = 0 i suppose all charged particles would move into the capacitor, and none would go through R1. And at the steady state it would be the other way around; the capacitor draws zero current (none can pass). And I guess it would appear as if the capacitor would be like an open circuit? Im a on to something?
You are definitely on to something.

So for the capacitor, can you tell me what the initial and final voltages will be?
 
  • #16
gneill
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Let's re-draw the circuit so that it's in a more familiar layout:
upload_2018-11-1_13-46-10.png
 

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  • #17
Since there are no charged particles in the capacitor at t=0, I guess the voltage Vc would be zero.
 
  • #18
But also, because Vc = Q/C = 0/C = 0
 
  • #19
For the final voltage I guess;

Vs - I2R2 - Vc = 0

Vc = I2R2 - Vs
 
  • #20
But also;

Vs - I2R2 - I1R1 = 0

Vs = I2R2 + I1R1

But because of Kirchhoffs current law, I1 = I2

Which gives

Vs = I2(R2 + R1)

Therefore

Vc = I2R2 - Vs = I2R2 - I2(R2 + R1) = I2R1

Vc = I2R1?
 
  • #21
gneill
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To determine what the final potential across the capacitor will be, remove the capacitor from the circuit (it draw no current anyways). Then look at what remains of the circuit. There will be no time-varying values anywhere in the circuit so you can treat it just as you would any resistor network being driven by a DC source. Determine the potential between the points where the capacitor was removed.
 
  • #22
Determine the potential between the points where the capacitor was removed.
That would be the voltage over R1? Which is I2R1
 
  • #23
gneill
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That would be the voltage over R1? Which is I2R1
Sure, but what is I2? You should be able to find these values using only the given component variable names, no introduced unknown values.
 
  • #24
l2 = Vs/(R1+R2)

Giving

Vc = l2R1 = VsR1/(R1+R2)
 
  • #25
gneill
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Right! Now you have final "values" for the capacitor voltage and the currents i1 and i2.

Do the same for their initial values. To find initial values, take the capacitor voltage to be zero (replace the capacitor with a short circuit) and analyze the circuit.
 

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