How Does Opening a Switch Affect a Circuit with a Capacitor and Resistors?

[pghaffari]

Homework Statement

http://imgur.com/VsDrQ,dJ2iv

Homework Equations

Current in loop 1: i_1 going counter-clockwise
Current in loop 2: i(t) going counter-clockwise

Before opening switch: we know that 2 loops exist, left and right. also the current is constant because it says the circuit was at rest for a long time.

After we open the switch, we know that the left loop is shorted out, and we somehow use (a) and (b) and plug it into our equations and solve?


We also know that i(0-) = i(0+)

The Attempt at a Solution

Before opening the switch:

Loop 1 (left) : -v_initial - (i_1 - i(t)) R = 0 ---> v_initial = -R i_1 + R i(t)

Loop 2 (right): i(t) = - R i(t) + r i_1 - 1/c integral i(t)dt

After switch is open:

i(0+) = 1
v(2microseconds) = 1/e


...

I don't understand how to proceed from here? What exactly do I have to do.. I'm stuck. Please let me know as soon as possible. Thanks.

 

[rude man]

Why do you have two current loops?

 

[pghaffari]

Before switch is open, it's been closed for a long time and that means there are 2 loops; one on the left and one on the right. And is current not going into the left loop?

 

[rude man]

How can there be a loop when the switch is open?

 

[pghaffari]

It says switch is closed beforehand, I'm writing the equations for when it's closed..

I don't know that's why I am asking. What am I supposed to do here?

 

[rude man]

It says switch is closed beforehand, I'm writing the equations for when it's closed..

I don't know that's why I am asking. What am I supposed to do here?

You're only interested in what happens after the switch is opened. The switch having been previously closed (for infinite time, in fact) determines your initial condition(s) when the switch is opened.

So, what is/are the initial condition(s) just before, i.e. when, the switch is opened?

 

[pghaffari]

We know that current is a constant value, and that current before opening switch must equal current after opening the switch.

 

[rude man]

Right. So how about writing an expression for that current? What role if any does C play in determining it?

 

[pghaffari]

Because current through a capacitor is C dv/dt , we know that current is constant before switch is open which means current through capacitance is 0 before AND after opening switch?

Thus, we can just say that ( -i(t) / 1 - i(t) R = 0 ) using KVL <-- but isn't voltage through capacitor 1/C integral i dt? So shouldn't we include that?

So we get: i(t) = i(t)*R .. so if we divide by i(t) we know R = 1 ??

But if we included the Capacitor equation in KVL, then it should be:

i(t) = - R i(t) + r i_1 - 1/c integral i(t)dt

not sure which is right..

How do we use part (a) and (b) that they gave us?

 

[pghaffari]

If I actually use KVL on the right loop I get this:

-1/c integral i(t)dt - v(t) - i(t)R = 0

Right?

 

[rude man]

If I actually use KVL on the right loop I get this:

-1/c integral i(t)dt - v(t) - i(t)R = 0

Right?

Right, but you need your equation to have one dependent variable, not two. What can you write in place of v(t)?

Get back to you in about 8-9 hrs.

 

[pghaffari]

-i(t).

So our equation becomes:


-1/c integral i(t) dt - i(t) - i(t) R = 0

So do we jjust plug in i(0+) = 1A?

so -1/c t - 1 - R = 0 ?

How do I use the voltage equation? (v 2us) = 1/e ?

 

[rude man]

-i(t).

So our equation becomes:


-1/c integral i(t) dt - i(t) - i(t) R = 0

How do I use the voltage equation? (v 2us) = 1/e ?

First, I ask you to label the 1 ohm resistor as R₁. That way you can keep track of your units term-by-term. That's a powerful way of catching math errors.

So now your equation is
-(1/C)∫i(t)dt - R₁i(t) -Ri(t) = 0.

This is an integral equation. What do you think should be done with this situation? It's entirely a math problem at this point.

 

[jrive]

Are you familiar with the Laplace transform? This tool enables you to transform your differential equation into an algebraic one, simplifying the math.
You can then use tables to convert from the Laplace domain back to time domain to get your answer.

If not, you can still solve the differential equation. Remove the integral by differentiating the equation and solve for I. Remember then to apply the initial conditions to determine the I when the switch is first opened up.

 

[pghaffari]

First, I ask you to label the 1 ohm resistor as R₁. That way you can keep track of your units term-by-term. That's a powerful way of catching math errors.

So now your equation is
-(1/C)∫i(t)dt - R₁i(t) -Ri(t) = 0.

This is an integral equation. What do you think should be done with this situation? It's entirely a math problem at this point.

Okay so now we solve for i(t) using first-order differentials.

We end up getting this:

i(t) = Ke^ (-t/ (C(R+1))

Now, we can use CONDITION a) that i(0+) = 1A to find K, which we find out K = 1.

So our equation is now:

i(t) = e^ (-t/(C(R+1))

Now we can use our second condition v(2us) = 1/e to get a RC relation.

v(t) = i(t)*1 <-- due to the voltage drop across the resistor

so v(2us) = i(2us) = 1/e

Plugging it in:


e^-(2*10^-6)/C(R+1) = 1/e

Take natural log of both sides, we end up getting:

C = ( 2*10^-6 ) / ( R + 1 )

But now.. I still have 2 variables with 1 equation. How do I solve for C and R ??

 

[rude man]

EDIT: oops! You're right!

OK, you know that i(0+) = 1 amp. So what does that tell you about R₁?

Another way of putting it - what causes i(t) to flow once the switch is open and the 2V source is out of the picture?

 

[pghaffari]

I thought R_1 is just 1 Ohm here?

so that V(t) = i(t)

 

[rude man]

EDIT: oops! You're right!

OK, you know that i(0+) = 1 amp. So what does that tell you about R₁?

Another way of putting it - what causes i(t) to flow once the switch is open and the 2V source is out of the picture?

You're right, sorry. I meant R.

 

[pghaffari]

You're right, sorry. I meant R.

Voltage is somehow still in the Resistor?

 

[rude man]

No, a resistor can't store voltage. What componernt can?

 

[pghaffari]

The capacitor

 

[rude man]

Right! Now, when the switch was closed for a long time, what voltage do you think might be stored on C?

 

[pghaffari]

entire 2 Volts? ..what about the v(t) ? Is that just the voltage drop across the resistor? SO it doesn't matter to the Capacitor?

 

[pghaffari]

Wait wouldn't it just be 1/c integral i dt

 

[rude man]

Yes. After a long time, C is fully charged and i = 0, so no drop across R₁.

So now we open the switch, and what do you think i(0+) has to be?

I have to take a break, will be back in 3-4 hrs. Think some more about what the circuit is really doing once the switch is opened, what the source of i(t) is, what i(0+) has to be. The result will be your value for R.

 

[pghaffari]

Source of i(t) is the Capacitor.. where Vc = 1/c integral (i dt)

We set that equal to 2 because that's how much voltage is stored in the capacitor at before switch is open for a very long time. If we do that we get ::

2 = 1/c integral idt

2c = integral idt

take derivative of both sides:

i(t) = 0.

This makes sense I think because I(0-)has to be I(0+) = 0 .. ?

butthen how would i find R? R = V/I .. if I = 0, then I can't solve for R,

Not sure about this :| Ill think some more> Let me know if any of this makes sense when you get back.

Thank you

 

[rude man]

Source of i(t) is the Capacitor.. where Vc = 1/c integral (i dt)

We set that equal to 2 because that's how much voltage is stored in the capacitor at before switch is open for a very long time. If we do that we get ::

2 = 1/c integral idt

2c = integral idt

take derivative of both sides:

i(t) = 0.

Thank you

∫i(t)dt from -∞ to 0 is a constant, it's a definite integral and its time-derivative is thus zero. You can't come up with the voltage on C at t=0+ that way.

Instead, just look at the circuit and what you have after the switch has been on a long time: +2V sitting to the left of R₁, charging up C to what value?

 

[pghaffari]

Charges up to 2V? So C = 2? How many Farads would that be ?

IF C = 2, Then we can just plug it into the euqationC = (-2*10^-6)/(R+1) and solve for R?

Can you explain to me why entire 2V is stored into the Capacitor?

 

[pghaffari]

Yes. After a long time, C is fully charged and i = 0, so no drop across R₁.

So now we open the switch, and what do you think i(0+) has to be?

I have to take a break, will be back in 3-4 hrs. Think some more about what the circuit is really doing once the switch is opened, what the source of i(t) is, what i(0+) has to be. The result will be your value for R.

Source of i(t) after switch is open: The Conductor
if i(0-) = 0, then i(0+) = 0

Right?

 

[rude man]

Charges up to 2V? So C = 2? How many Farads would that be ?

IF C = 2, Then we can just plug it into the euqation


C = (-2*10^-6)/(R+1) and solve for R? NO.

Can you explain to me why entire 2V is stored into the Capacitor?

Why would C=2 just because its voltage is 2V?

Take the 2V source, connect via R1 to an uncharged C, and write the equation for i(t). Tne voltage on C will be (1/C)∫i(t)dt integrated from t= 0 to t = ∞. That'll give you your initial voltage on C for your particular problem, where the switch is opened at t = 0.

Have to go to bed. Will probably not get much of a chance to help further until tomorrow late afternoon.