Circuit with resistor, switch and capacitor

In summary: Yes, and how does it make the capacitor appear to the surrounding network at that instant of time?The capacitor appears to the surrounding network as an open circuit at t=0.
  • #36
johann1301h said:
Oh yes! Battery = constant Vs, I get it!
Do you mean to "remove" the capacitor when it is fully charged? And what du you mean by suppressing the voltage ?
It doesn't matter when you remove it. With the battery suppressed, the network will be dead anyways.

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By "suppressed" I mean replaced with a short circuit. If it were a current source, you'd replace it with an open circuit. Have you not been introduced to Thevenin equivalents yet? Or circuit analysis by superposition?
 

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  • #37
gneill said:
Or circuit analysis by superposition?
If you mean by using complex numbers, briefly.

gneill said:
Have you not been introduced to Thevenin equivalents yet?
No, my university (University of Oslo) has a tendency to skim the surface of each subject. Its really sad!

From your sketch I can guess that Req = R1

And therefore (perhaps) τ = R1?
 
  • #38
Not only from your sketch though, I get what you said here

gneill said:
what net resistance does the capacitor "see" at its terminal connection points?
 
  • #39
johann1301h said:
If you mean by using complex numbers, briefly.
No, the superposition method of circuit analysis is where you suppress all the sources but one, analyze the resulting circuit to find the effect of the one operating source. Then do the same for each source in turn. When you're done, add up all the individual contributions. The method relies on the fact that circuits with only linear elements (like resistors, capacitors, inductors) behave as linear systems so that the contributions of each source add algebraically.
From your sketch I can guess that Req = R1

And therefore (perhaps) τ = R1?

Nope. How are ##R_1## and ##R_2## connected?

And the time constant will have units of time. So ##\tau = R_{eq} C##
 
  • #40
Yes, I see, the current would go through both R1 and R2...
 
  • #41
johann1301h said:
Yes, I see, the current would go through both R1 and R2...
So how are they connected? In series or in parallel? Does the same current go through both (series connection) or do they take different currents (parallel connection)?
 
  • #42
Using Kirchhoffs again

Vc = IR1

and

Vc = IR2

gives

2Vc = IR1 + IR2 =I(R1 + R2)

Vc = IR1 + IR2 =I(R1 + R2)/2

So it "sees" the average of R1 and R2?
 
  • #43
No, you've made the assumption that the currents through both resistors would be the same ##I##.

How do you "add" resistors in parallel? In series? Have you done any circuit simplification exorcises where you have to combine and reduce the number of resistors?
 
  • #44
Req = R1R2/(R1 + R2)
 
  • #45
τ = C*R1R2/(R1 + R2) ?
 
  • #46
johann1301h said:
Req = R1R2/(R1 + R2)
Yes! That's the resistance that the capacitor "sees" when it's in-circuit.

So the time constant for this circuit is ##\tau = R_{eq} C##. It will apply to every exponential function involved in this circuit.
 
  • #47
I see, I will try to find I2 as well!
 
  • #48
So, can you now do the same thing for the capacitor voltage? You know it starts at zero and goes up to ##V_s \frac{R_1}{R_1 + R_2}##.
 
  • #49
johann1301h said:
I see, I will try to find I2 as well!
Excellent. It would be great if you could do that. It'll be a bit more tricky (but not much!) since it doesn't start or end at zero values, so there will be a constant offset to add.
 
  • #50
johann1301h said:
No, my university (University of Oslo) has a tendency to skim the surface of each subject. Its really sad!
I am curious, if and if you don't mind and have the time, could you tell me what program you are in and in what course in particular this question was posed? The reason I ask is that it seems that you've been pushed into the deep end of the circuit pool without being given the right background information to survive! :nb)
 
  • #52
The thing is this; they have started a project called CSE (Cumputers in Science Education) which basicly means we don't really end up solving differential equations without using a computer all the time, f. ex forward Euler method. We start programming the first year, so a lot(to much) of our focus goes into programming...
 
  • #53
They call it pioneering, I call it a mistake.
 
  • #54
l2(t) = VsR1/(R1 + R2)(1 - e-t/τ) + Vs/R2e-t/τ
 
  • #55
When I find Vc(t), I just use that

Ic = C(dVc/dt) ?

I just find the derivative of Vc and multiply by C ?
 
  • #56
Thank you for that information.

I took part in a similar initiative when I was in university (back when dinosaurs ruled the earth!). The program was for a Bachelor of Computer Science, Electronic Systems. It combined computer science core courses, electrical engineering core courses, and control systems core courses. Nearly no electives at all, so lots of brain-sweat and no time to party. As it turned out, I was the only student to graduate on-time in that first class (*blush*).

To give the school credit, they did go into all the gory details and didn't leave anything out along the way. Lot's of work, but all the groundwork was laid in a smooth progression. So they obviously planned the initiative very well, even if was perhaps a tad "aggressive" given the on-time graduating rate for the first pass at it.
 
  • #57
johann1301h said:
l2(t) = VsR1/(R1 + R2)(1 - e-t/τ) + Vs/R2e-t/τ
Something's gone amiss. The units in the first term do not work out to current (amps). And I'd expect a constant offset for the second term, not a decaying exponential. But it's really close to what I'd expect.

Can you show some details of how you arrived at your result?
 
  • #58
Yes, I used the wrong value, I think I meant;

l2(t) = Vs/(R1 + R2)(1 - e-t/τ) + Vs/R2e-t/τ

But this is also not right given that;
gneill said:
And I'd expect a constant offset for the second term, not a decaying exponential.

I can always rewrite it though;

l2(t) = Vs/(R1 + R2) - Vs/(R1 + R2)(e-t/τ) + Vs/R2e-t/τ

l2(t) = Vs/(R1 + R2) + (Vs/R2 - Vs/(R1 + R2))e-t/τ
 
  • #59
Yes, that looks good.
 
  • #60
And for Vs;

Vs(t) = VsR1/(R1 + R2)(1 - e-t/τ)

and therefore

Ic(t) = CVc'(t) = -C (VsR1/(R1 + R2))e-t/τ(-1/τ)

Ic(t) = (C/τ) (VsR1/(R1 + R2)e-t/τ)

Ic(t) = (1/Req) (VsR1/(R1 + R2)e-t/τ)

Ic(t) = (1/ R2 ) (Vse-t/τ)

Ic(t) = (Vs/ R2 )e-t/τ
 
  • #61
Yup. Looks good.
 
  • #62
This is great!

But you mentioned in post #13
gneill said:
That is the "starting from the basics" approach, yes.

The thing is this, in the next task I'm asked to verify the result using a numerical method; ie. using a differential equation! (forward Euler for instance)

But I/we have solved it, by simply assuming the "form" of the function; I guess I am puzzled by the "starting from the basics" approach" you mentioned...

How does one go about finding the differential equation?
 
  • #63
johann1301h said:
How does one go about finding the differential equation?
One writes the circuit equations using the differential or integral forms for the capacitor voltage or current, then solve the resulting differential equation. So for example, for the simple case of a charged capacitor discharging through a resistor, writing KCL:

##C \frac{dV}{dt} +\frac{V}{R} = 0##

##\frac{dV}{dt} = -\frac{V}{R C}##

##\frac{dV}{V} = -\frac{dt}{R C} ##

and so on
 
<h2>1. What is the purpose of a resistor in a circuit?</h2><p>A resistor is used to limit the flow of electric current in a circuit. It helps to control the amount of current that passes through a circuit and prevents damage to other components.</p><h2>2. How does a switch affect the flow of electricity in a circuit?</h2><p>A switch is used to open or close a circuit, which can either allow or prevent the flow of electricity. When the switch is closed, it completes the circuit and allows electricity to flow. When the switch is open, it breaks the circuit and stops the flow of electricity.</p><h2>3. What is the role of a capacitor in a circuit?</h2><p>A capacitor stores electrical energy in the form of an electric field. It can release this energy when needed, which can be useful for smoothing out voltage fluctuations in a circuit or providing a temporary power source.</p><h2>4. How does the presence of a capacitor affect the behavior of a circuit?</h2><p>The presence of a capacitor can affect the behavior of a circuit in several ways. It can cause a delay in the flow of electrical current, filter out certain frequencies, and store and release energy. It can also help to stabilize the voltage in a circuit.</p><h2>5. What happens when a resistor, switch, and capacitor are combined in a circuit?</h2><p>When a resistor, switch, and capacitor are combined in a circuit, they can work together to control the flow of electricity and store and release energy. The resistor limits the current, the switch controls the flow, and the capacitor stores energy. This combination can be useful in various electronic applications.</p>

1. What is the purpose of a resistor in a circuit?

A resistor is used to limit the flow of electric current in a circuit. It helps to control the amount of current that passes through a circuit and prevents damage to other components.

2. How does a switch affect the flow of electricity in a circuit?

A switch is used to open or close a circuit, which can either allow or prevent the flow of electricity. When the switch is closed, it completes the circuit and allows electricity to flow. When the switch is open, it breaks the circuit and stops the flow of electricity.

3. What is the role of a capacitor in a circuit?

A capacitor stores electrical energy in the form of an electric field. It can release this energy when needed, which can be useful for smoothing out voltage fluctuations in a circuit or providing a temporary power source.

4. How does the presence of a capacitor affect the behavior of a circuit?

The presence of a capacitor can affect the behavior of a circuit in several ways. It can cause a delay in the flow of electrical current, filter out certain frequencies, and store and release energy. It can also help to stabilize the voltage in a circuit.

5. What happens when a resistor, switch, and capacitor are combined in a circuit?

When a resistor, switch, and capacitor are combined in a circuit, they can work together to control the flow of electricity and store and release energy. The resistor limits the current, the switch controls the flow, and the capacitor stores energy. This combination can be useful in various electronic applications.

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