RC Circuit with Current Source: Is This Solution Correct?

[goonking]

Homework Statement

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
∴ U_R1= U_o (e^-t/RC) = U_R1= 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

i_C1 = i_o (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?

 

[SammyS]

Homework Statement

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:
https://www.physicsforums.com/attachments/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, U_capacitor = 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

∴ U_capacitor = (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.

 

[goonking]

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?

 

[SammyS]

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.

 

[goonking]

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

i_C1 = i_o (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?

 

[goonking]

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
∴ U_R1= U_o (e^-t/RC) = U_R1= 90V (e^-t/0.0012)

assuming this is correct, my answer to the previous post should be:
i_C1 = i_o (e^-t/RC) = (0.003A)(e^-t/0.0012)

 

[cnh1995]

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?

 

[goonking]

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.

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 → ∞

 

[cnh1995]

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.

 

[goonking]

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.

 

[cnh1995]

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.

 

[cnh1995]

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 i_c(t). You can use it in the original circuit to find i1(t) and U1(t).

 

[goonking]

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 i_c(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.

 

[cnh1995]

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?

 

[goonking]

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 i_c(t). You can use it in the original circuit to find i1(t) and U1(t).

U_R1(t) = [I_s - I_C1(t)]/ 20kΩ

Plugging in I_C1(t)] from what I found before:U_R1(t) = [I_s - io (e^-t/RC) ]/ 20kΩ = [I_s - (0.003A) (e^-t/0.0012) ]/ 20kΩ

 

[goonking]

Ok. How about Thevenin's theorem? Have you covered it yet?

Nope. we are low on options haha, sorry.

 

[cnh1995]

UR1(t) = [Is - IC1(t)]/ 20kΩ

This is the right expression.
Edit: Oh dear! How did I miss that.. V=IR and not I/R...

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

The time constant is incorrect.

 

[goonking]

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)

 

[goonking]

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

i_C1 = io (e^-t/RC) = (0.003A)(e^-t/0.003)

 

[cnh1995]

No. The time constant considers both R1 and R2. How are they connected when you open the current source?

 

[goonking]

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.

 

[cnh1995]

the capacitor, and both resistors would be in series.

Right! So what would be your time constant?

 

[goonking]

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.

 

[cnh1995]

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.

 

[cnh1995]

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!

 

[cnh1995]

No. That's right!

So what equations did you get for U1(t) and Ic(t)?

 

[goonking]

So what equations did you get for U1(t) and Ic(t)?

iC1(t) = io (e^-t/RC) = (0.003A)(e^-t/0.003)U1(t) = [Is - iC1(t)] 20k

 

[cnh1995]

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

Req=50k as discussed earlier..

 

[goonking]

Req=50k as discussed earlier..

when the capacitor is short, the Resistors are in parallel so the Req is 20x30 / 50 = 12k ohms.
when the capacitor is fully charged, and is open, only R1 has current flowing through it, so Req is just 20k ohms.
I'm still having trouble understanding if the R in RC equal to the Req when t=0 or when the capacitor is fully charged.
The only way Req would be 50k if is the 2 resistors are in series, and that is only possible when the current source is open, but the question doesn't state the current source is disconnected after the capacitor is fully charged.Unless I'm misunderstanding something, I'll reread the past posts again.

 

[cnh1995]

when the capacitor is short, the Resistors are in parallel so the Req is 20x30 / 50 = 12k ohms.
when the capacitor is fully charged, and is open, only R1 has current flowing through it, so Req is just 20k ohms.
I'm still having trouble understanding if the R in RC equal to the Req when t=0 or when the capacitor is fully charged.
The only way Req would be 50k if is the 2 resistors are in series, and that is only possible when the current source is open, but the question doesn't state the current source is disconnected after the capacitor is fully charged.Unless I'm misunderstanding something, I'll reread the past posts again.

Ok. It would be clearer once you try it using source transformation or Thevenin's theorem, after you cover them in the class.

 

[goonking]

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.

ok, having done some reading on source transformation, i think the part that probably confused me a bit more than it should have, was the source was a current source and not a voltage source.

So yes, after source transformation, it would be a 150 volt battery in series with 20k and 30k resistors, and the capacitor.

then Req would be 20k+30k

 

[cnh1995]

except at the end, you mention to open the current source, does that mean the current source becomes open right after the capacitor is fully charged?

To calculate the equivalent resistance, the current source is open circuited. This equivalent resistance will be the Thevenin resistance viewed from the capacitor terminals. If there were a voltage source instead of the current source, it would be shorted.

 

[goonking]

To calculate the equivalent resistance, the current source is open circuited. Equivalent resistance will be the Thevenin resistance viewed from the capacitor terminals. If there were a voltage source instead of the current source, it would be shorted.

very interesting, so whenever there is a current source, and we need to find Req, we need to open the current source.

and when we have a voltage source, and we need to find Req, we need to short the volt source.

What is the reason for this? I didn't learn thevenin's theorem yet.

 

[cnh1995]

very interesting, so whenever there is a current source, and we need to find Req, we need to open the current source.

and when we have a voltage source, and we need to find Req, we need to short the volt source?

Yes, provided that Req is Thevenin resistance. It will be clearer to you when you'll learn network theorems such as Thevenin, Norton and Superposition.

 

[cnh1995]

What is the reason for this? I didn't learn thevenin's theorem yet.

To calculate equivalent resistance between a and b, you need to consider a source-free network. In fact, Req between a and b is that resistance which is seen by a voltage (or current) source connected between a and b. That voltage source is called Thevenin voltage. It is computed using KVL and the existing sources in the network. Once you get Vth, all the existing sources are removed because you are now viewing the circuit from the terminals a and b and you have found the resultant voltage between a and b due to all the sources present in the network. So, to remove a voltage source, it is shorted, meaning it won't supply any voltage and opening the current source means it won't supply any current.