Circuit with resistor, switch and capacitor

[johann1301h]

I think I meant;

I₁(t) = V_s/(R₁ + R₂)(1-e^-t)

 

[gneill]

Yes, only there's one thing missing. The exponent needs a time constant.
$$e^{-\frac{t}{\tau}}$$

This is great, but I would never have guessed that the solution would be in this form, what kind of argumentation can be given for this?

Yes, it's the solution for the differential equation. As I mentioned before, ALL values that vary in an RC circuit (or an RL circuit) take the form of exponential growth or decay. You should be able to find the differential equation solved for the trivial RC circuit in your text, or failing that, just look up "RC circuit" on the web.

So about the time constant $\tau$. If you remove the capacitor from the circuit and suppress the voltage supply, what net resistance does the capacitor "see" at its terminal connection points?

 

[johann1301h]

Ok, i have a lot to digest here. One question though, are we not assuming that V_s is constant ?

 

[gneill]

Ok, i have a lot to digest here. One question though, are we not assuming that V_s is constant ?

Yes. It's referred to as being a battery in the problem statement.

 

[johann1301h]

Oh yes! Battery = constant V_s, I get it!

So about the time constant τ. If you remove the capacitor from the circuit and suppress the voltage supply, what net resistance does the capacitor "see" at its terminal connection points?

Do you mean to "remove" the capacitor when it is fully charged? And what du you mean by suppressing the voltage ?

 

[gneill]

Oh yes! Battery = constant V_s, 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.

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?

 

[johann1301h]

Or circuit analysis by superposition?

If you mean by using complex numbers, briefly.

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 R_eq = R₁

And therefore (perhaps) τ = R₁?

 

[johann1301h]

Not only from your sketch though, I get what you said here

what net resistance does the capacitor "see" at its terminal connection points?

 

[gneill]

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 R_eq = 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 $\tau = R_{eq} C$

 

[johann1301h]

Yes, I see, the current would go through both R₁ and R₂...

 

[gneill]

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)?

 

[johann1301h]

Using Kirchhoffs again

V_c = IR₁

and

V_c = IR₂

gives

2V_c = IR₁ + IR₂ =I(R₁ + R₂)

V_c = IR₁ + IR₂ =I(R₁ + R₂)/2

So it "sees" the average of R₁ and R₂?

 

[gneill]

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?

 

[johann1301h]

R_eq = R₁R₂/(R₁ + R₂)

 

[johann1301h]

τ = C*R1R2/(R1 + R2) ?

 

[gneill]

R_eq = 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.

 

[johann1301h]

I see, I will try to find I₂ as well!

 

[gneill]

So, can you now do the same thing for the capacitor voltage? You know it starts at zero and goes up to $V_s \frac{R_1}{R_1 + R_2}$.

 

[gneill]

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.

 

[gneill]

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!

 

[johann1301h]

https://www.uio.no/studier/emner/matnat/fys/FYS1120/h18/index.html

https://www.uio.no/studier/emner/matnat/fys/FYS1120/index-eng.html

 

[johann1301h]

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...

 

[johann1301h]

They call it pioneering, I call it a mistake.

 

[johann1301h]

l₂(t) = V_sR₁/(R₁ + R₂)(1 - e^-t/τ) + V_s/R₂e^-t/τ

 

[johann1301h]

When I find V_c(t), I just use that

I_c = C(dV_c/dt) ?

I just find the derivative of V_c and multiply by C ?

 

[gneill]

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.

 

[gneill]

l₂(t) = V_sR₁/(R₁ + R₂)(1 - e^-t/τ) + V_s/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?

 

[johann1301h]

Yes, I used the wrong value, I think I meant;

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

But this is also not right given that;

And I'd expect a constant offset for the second term, not a decaying exponential.

I can always rewrite it though;

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

l2(t) = V_s/(R₁ + R₂) + (V_s/R₂ - V_s/(R₁ + R₂))e^-t/τ

 

[gneill]

Yes, that looks good.

 

[johann1301h]

And for V_s;

V_s(t) = V_sR₁/(R₁ + R₂)(1 - e^-t/τ)

and therefore

I_c(t) = CV_c'(t) = -C (V_sR₁/(R₁ + R₂))e^-t/τ(-1/τ)

I_c(t) = (C/τ) (V_sR₁/(R₁ + R₂)e^-t/τ)

I_c(t) = (1/R_eq) (V_sR₁/(R₁ + R₂)e^-t/τ)

I_c(t) = (1/ R₂ ) (V_se^-t/τ)

I_c(t) = (V_s/ R₂ )e^-t/τ