RC circuit with current source

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  • #1
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Homework Statement


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Homework Equations




The Attempt at a Solution


so when t=0, the capacitor is a closed circuit.

The Req of the circuit is (20k)(30k)/(20k+30k)= 12000 ohms.

U = I Req = (7.5x10-3A) (12000 ohms) = 90 Volts

so there is a voltage drop of 90 volts across the 20k resistor (R1), when t=0.



RC= Req C = (12000 ohms) (0.1 x10-6F) = 0.0012 sec
∴ UR1= Uo (e-t/RC) = UR1= 90V (e-t/0.0012)


Ok, so then the current flowing through the capacitor would be the same current flowing through the 30k resistor when t=0.

Using current division, 7.5x10-3A [20 / (20+30)] = 0.003 A

iC1 = io (e-t/RC) = (0.003A)(e-t/0.0012)

This is my final answer, I've erased my initial attempt as it was wrong. Could someone verify if this is correct?
 
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Answers and Replies

  • #2
SammyS
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Homework Statement


FUHimw2.jpg


Homework Equations




The Attempt at a Solution


So when the switch is closed, it becomes a dc circuit, and under dc conditions, we replace each capacitor with an open circuit:
View attachment 101015

Now all the current from the source is going through the 20k ohm resistor, so the voltage drop there is

I R = (0.0075 A) (20,000 ohm) = 150 V

since the capacitor is in parallel with the 20k ohm resistor, the voltage drop on it also is 150 V

Therefore, Ucapacitor = ε [ 1 - e-t/RC]
but ε = I R = (0.0075 A) (20,000 ohm) = 150 V
and RC = (20,000Ω) (0.1x10-6F) = 0.002

∴ Ucapacitor = (150) [ 1- e-t/0.002]
The capacitor is initially uncharged. The voltage across it is zero at t=0. That's like a short circuit, not an open circuit.
 
  • #3
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The capacitor is initially uncharged. The voltage across it is zero at t=0. That's like a short circuit, not an open circuit.
If it is a short circuit, then my drawing is wrong? :(

So a capacitor only behaves as a open circuit when it is fully charged?
 
  • #4
SammyS
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If it is a short circuit, then my drawing is wrong? :(

So a capacitor only behaves as a open circuit when it is fully charged?
Yes. This is when current ceases to flow through the capacitor, like the case for an open circuit.
 
  • #5
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Yes. This is when current ceases to flow through the capacitor, like the case for an open circuit.
Ok, so then the current flowing through the capacitor would be the same current flowing through the 30k resistor when t=0.

Using current division, 7.5x10-3A ( 20 / (20+30)) = 0.003 A

RC = (30,000)(0.1 x 10-6) = 0.003 seconds

iC1 = io (e-t/RC) = (0.003A)(e-t/0.003)

or is RC not 0.003 because I had to use Req of the circuit before capacitor became fully charged?
 
  • #6
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so when t=0, the capacitor is a closed circuit.

The Req of the circuit is (20k)(30k)/(20k+30k)= 12000 ohms.

U = I Req = (7.5x10-3A) (12000 ohms) = 90 Volts

so there is a voltage drop of 90 volts across the 20k resistor (R1), when t=0.



RC= Req C = (12000 ohms) (0.1 x10-6F) = 0.0012 sec
∴ UR1= Uo (e-t/RC) = UR1= 90V (e-t/0.0012)

assuming this is correct, my answer to the previous post should be:
iC1 = io (e-t/RC) = (0.003A)(e-t/0.0012)
 
  • #7
cnh1995
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UR1= Uo (e-t/RC) = UR1= 90V (e-t/0.0012)
This means voltage across R1 is exponentially decreasing. Is it? Think about the steady state of this circuit. Which quantities become zero in the steady state here?
 
  • #8
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This means voltage across R1 is exponentially decreasing. Is it? Think about the steady state of this circuit. Which quantities become zero in the steady state here?
so the capacitor has increasing voltage across it over time, and decreasing current across it, so when current decreases, R2 will have less current passing through it... and since we have a current source, we have a constant flow of current flowing through it, therefore all the current (or most) overtime will have to flow through R1 once the capacitor is fully charged (where it becomes open circuit)

applying ohms law, the voltage drop across R1 should be increasing.

Yeah... I'm having a harder time understanding/remembering the equations than understanding the concepts. :H

The voltage drop across R1 should be (7.5 mA) (20k ohms) = 150 V.
but i'm having trouble obtaining an equation that equates to 150V when t → ∞
 
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  • #9
cnh1995
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so the capacitor has increasing voltage across it over time, and decreasing current across it, so when current decreases, R2 will have less current... and since we have a current source, we have a constant flow of current, therefore all the current (or most) overtime will flow through R1...

applying ohms law, the voltage drop across R1 should be increasing.
Right. Have you studied source transformation? It would be easier.
 
  • #10
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Right. Have you studied source transformation? It would be easier.
No, I'm in an intro class, and we haven't even (or just started) AC circuits.
 
  • #11
cnh1995
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No, I'm in an intro class, and we haven't even (or just started) AC circuits.
Ok but source transformation means converting current source into voltage source.
You can replace a current source of magnitude I with parallel resistance R with a voltage source of magnitude I*R with the same resistance R in series. This way, you'll end up with a simple RC circuit with two resistors in series with a capacitor and a voltage source.
images.png
 
  • #12
cnh1995
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The voltage drop across R1 should be (7.5 mA) (20k ohms) = 150 V.
but i'm having trouble obtaining an equation that equates to 150V when t → ∞
Initial current through the capacitor will be 30mA and that through R1 is 20mA, as you calculated earlier. Now, voltage across R2 decays to 0 exponentially and voltage across R1 increases to 150V exponentially. Also, capacitor voltage increases exponentially to 150V. Using the source transformation, you'll get the expression for ic(t). You can use it in the original circuit to find i1(t) and U1(t).
 
  • #13
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Initial current through the capacitor will be 30mA and that through R1 is 20mA, as you calculated earlier. Now, voltage across R2 decays to 0 exponentially and voltage across R1 increases to 150V exponentially. Also, capacitor voltage increases exponentially to 150V. Using the source transformation, you'll get the expression for ic(t). You can use it in the original circuit to find i1(t) and U1(t).
I'm trying to answer it without using source transformation, I know it'd be easier but we haven't covered it yet in class.
 
  • #14
cnh1995
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I'm trying to answer it without using source transformation, I know it'd be easier but we haven't covered it yet in class.
Ok. How about Thevenin's theorem? Have you covered it yet?
 
  • #15
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Initial current through the capacitor will be 30mA and that through R1 is 20mA, as you calculated earlier. Now, voltage across R2 decays to 0 exponentially and voltage across R1 increases to 150V exponentially. Also, capacitor voltage increases exponentially to 150V. Using the source transformation, you'll get the expression for ic(t). You can use it in the original circuit to find i1(t) and U1(t).
UR1(t) = [Is - IC1(t)]/ 20kΩ

Plugging in IC1(t)] from what I found before:


UR1(t) = [Is - io (e-t/RC) ]/ 20kΩ = [Is - (0.003A) (e-t/0.0012) ]/ 20kΩ
 
  • #16
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Ok. How about Thevenin's theorem? Have you covered it yet?
Nope. we are low on options haha, sorry.
 
  • #17
cnh1995
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UR1(t) = [Is - IC1(t)]/ 20kΩ
This is the right expression.
Edit: Oh dear! How did I miss that..:-p V=IR and not I/R...
iC1 = io (e-t/RC) = (0.003A)(e-t/0.0012)
The time constant is incorrect.
 
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  • #18
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This is the right expression.

The time constant is incorrect.
I didn't derive this equation, I just thought for a bit and this equation made sense.
How would one derive this?


I wasn't sure if I had to use the resistors of R2 (the resistor it was in series with, so they have the same current) or Req of the circuit when the capacitor was short.

would I be wrong if I said Time constant for a circuit = (Req) (Ceq)? (assuming there were more than 1 capacitor)
 
  • #19
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RC = (30,000)(0.1 x 10-6) = 0.003 seconds

iC1 = io (e-t/RC) = (0.003A)(e-t/0.003)
 
  • #20
cnh1995
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No. The time constant considers both R1 and R2. How are they connected when you open the current source?
 
  • #21
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No. The time constant considers both R1 and R2. How are they connected when you open the current source?
the capacitor, and both resistors would be in series.
 
  • #22
cnh1995
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the capacitor, and both resistors would be in series.
Right! So what would be your time constant?
 
  • #23
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Right! So what would be your time constant?
it would be the Req of the two resistors.

but in this case, the R in time constant = RC is 30,000 ohms? because it is in series with the capacitor, thus having the same current as the current flowing through the capacitor.
 
  • #24
cnh1995
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it would be the Req of the two resistors.

but in this case, the R in time constant = RC is 30,000 ohms? because it is in series with the capacitor, thus having the same current as the current flowing through the capacitor.
But the current (and rate of change of current)through the capacitor also depends on R1. Change R1 and verify. Current division changes with R1. Voltage across the capacitor also depends on R1. Also, using source transformation, the circuit will look like this:
150V source in series with 20k and 30k and the capacitor. This means R1 and R2 are in series when it comes to time constant calculation. Open the current source and short the voltage source and see how the elements are connected.
 
  • #25
cnh1995
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would I be wrong if I said Time constant for a circuit = (Req) (Ceq)? (assuming there were more than 1 capacitor)
No. That's right!
 
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