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.

 

[pghaffari]

Ok that makes a lot more sense..


So basically I think i figured it out. Ths is what I did:


After opening the switch: I applied 2V to the right hand lop like you said and i get THIS equation:



2 + R1 + i(t) + 1/c integral(0-> infinity) i(t) dt = 0

BUT WE KNOW that i(T) = e^-t/R(C+1)

So we plug that in for i(t) and integrate from 0->inifnity and we end up geting

2 + R1 + i(t) - R(c+1)/c = 0

BUT because this is AFTER we open switch, we know that i(t) has to EQUAL 0, because i(0-) = 0.

SO equation becomes:

2 + R1 - R(c+1)/c = 0


We can solve the top equation and we end up getting:
2C = R


Now we have a R-C relationship. We plug thisback into our old equation:

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

Plug both of them in:

We end up getting a quadratic equation and I end up finding C and R as:


C = 1/10 (5 +- sqrt(26)

and since R = 2C

R = 1/ (5 +- sqrt(26))

But which do we take? the Plus orthe Minus? I am assumnig Positive value because we want it to be postive/higher?

Not sure if this is correct, but it feels right.

Let me know!
THX FOR UR HELP AGAIN@!

 

[jrive]

As you have now figured out with the help Rude Man, after the switch has been closed for a long time, there are 2V stored in the capacitor. As soon as the switch is opened up, the capacitor discharges through R and R1. With this you have all the information you need to solve for R.

You had obtained the equation for the current through the circuit earlier: i(t)=K*e^(-t/RC), where R is R+R1. Knowing the initial voltage across the cap, and the current flowing through te circuit at t=0+, you can determine K, and therefore the value of R.

With the second piece of information given to you in the problem, you can now solve for C. You are given the resistance of R1 and you have the equation of the curren tflowing through it as function of time. Since V_acrossR1=I*R1, V_acrossR1(t)=K*e^(-t/RC)*R1. you can now solve this equation for t=2usecs.

hope this helps...

 

[pghaffari]

my value for i(t) is actually K*e^(-t/(R1(C+1)) where K = 1.

Not sure how you got C(R1+R) under the divider?

ANyways, can you look at my previous post and see if I did it correctly. thanks

 

[jrive]

K has to have some physical meaning...in this case the current at t=0+.
The only way the circuit can produce the 1A at t=0, given that the capacitor has been charged up to 2V, is to divide that voltage by the total resistance in the circuit (after the switch is open) through which the capacitor discharges.

so, K=Vcap_{at t=0}/(R1+R). You're right that K must equal 1...but the 1 has to have physical meaning -- it is not just a unitless number -- it is 1amp.

By the way, your equation should be : i(t)=K*e^-t/((R1+R)C), for t>0...since the capacitor discharges through both resistors.

Keep at it...it will come to you...

sorry if I'm unclear or too cryptic, I'm struggling a little bit between providing responses that help you think through the problem so you can understand what's going on and figure it out, and just "giving it away" in trying to explain the solution.

 

[pghaffari]

You are right, it should be C(R+R1), i had a typo. But also we are given that R = 1 ohm, so I just replaced it in the equation to get C(R1+1).

So now that we have i(t),
we need another equation to relate R and C after switch is open.

We know that before switch is open, 2V is stored in the capacitor and i(0-) is 0.

So rudeman told me that we can basically connect the 2V to the right circut @ t =0 to get a new equation.

So an equation would be:

2 + R1 + i(t) + 1/c integral (0 to infinity) of i(t)

and the i(t) on the left is essentially 0 because i(0-)= 0 = i(0+)

is this correct?

then we solve 2 equations 2 unknowns for R,C ?

 

[jrive]

The problem stated that the switch had been previously closed for a while (at rest). So, the cap has been charged to 2V (the same as the source), and there is no more current flowing through the cap.

Then, the switch was opened. The circuit became a "source free" circuit analysis problem, and if you recall, you wrote the equation for i(t) flowing through the circuit at this time using KVL ("the sum of all the voltage drops equals the sum of all the voltage rises in the loop"). Ie.

1/c∫idt +v(0)-i(R+R1)=0.

(notice how all the terms in the equation above a in units of volts, term 1= volts across capacitor, v(0)=initial voltage across capacitor, and term 3=voltage drop across resistors --always good to verify units are consistent in any equation).

You then solved this differential equation which gave you the result in the form of i(t)=K*e^-t/((R1+R)C).

At this point, you have an equation that describes the flow of current in the circuit from the time the switch is open to eternity(while the switch remains open) This is all you need.


From the initial conditions, you can solve for R as I stated in a previous post, since you know the voltage stored in the cap (and the voltage in a capacitor cannot change instantaneously) and you know the value of R1, and the measured current in the loop, 1 amp.

Then, from i(t)=K*e^-t/((R1+R)C)
(you know the value of K (1amp),and now the value of R as computed from the initial conditions), you can solve or C at t=2usecs.

 

[pghaffari]

OK that makes a lot more sense

SO since voltage across capacitor is 2V

We can set this equation for after short is open1/c integral (0 to infinity) of i(t) = 2when i integrate the first one.. I stil have 2 unknow variables, C and R1. how do I get rid of one of them?

i don't know the value of R1

EDIT:
is R1 just 0 since there is NO drop across it?? when C is fullycharged to 2v?

so if i set R1 to 0 in that integration i would get C = 1/2?

 

[rude man]

jrive has summed it up very nicely and I'm handing this over to him.

 

[jrive]

there's nothing to integrate...you have the equation, i(t)=Ke^-t/((R+R1)C).

at t=0, i(t)=K; K=1amp. But, 1 Amp flows because there are 2V stored on the cap and you have a discharge path through the 2 resistors. So,

2V/(R+1ohm)=1, solve for R.

Now,
i(t)=i(t)=1e^-t/((R+R1)C).

the only unknown is C.

You know that the voltage across a resistor is given by V=IR. Since i(t) flows through the resitor R1, mulitply i(t)*R1, equate it to 1/e and solve for C at t=2usec.

 

[pghaffari]

I understand what you did but i still would appreciate it if you could explain afew things to me.

What exactly HAPPENS to this circuit when the switch is open? Also what does it mean to have a discharge path.

So basically I understand this:

Since the circuit has been at rest while switch is closed. Current is constant and the capacitor is fully charged to 2V. (I don't understand why i = 0 , shouldn't it still be a constant value ?)

Then once the switch opens.. what exactly happens to the circuit. Can you explain to me what a circuit behaves like when a switch is open/closed. What causes current to flow after a switch is open (didn't we say current was 0 right before switch was open?)Thank you for everything.

I ended up getting R1 = 1 and C = 1uF

 

[jrive]

When the switch closes, the capacitor charges though the 1ohm resistor. After sometime (as determined bythis equation Vc=Vin(1-e^(-t/(RC))) it is fully charged and since Vc=2V just like the source, current stops flowing through it. Resistor R, though, still conducts the current from the source to ground. However, there is no more current flowing through R1 and C.

When the switch opens, the voltage source in the circuit is now the capacitor, initially at 2V (since that is what it was charged to). So, the capacitor dicharges through the resistors to ground -- that means, current flows from the top terminal of the capacitor, charged to 2V with respect to the bottom terminal, through the resistors to the bottom terminal of the capacitor completing the loop.

As the current flows, the voltage across the capacitor discharges or decays. You can see that relationship when you apply vc=1/c ∫(-i*dt). The negative sign is just convention, since the current is flowing "out" of the capacitor. Remember that you had found i(t) as i(t)=Vinitial/R*e^-t/(RC)), so, this intregral becomes Vc=Vinitial*e^-t/(RC). So, as t gets bigger, the voltage across the capacitor decreases,eventually discharging the cap to 0V (0 V across the capacitor). No more current flows.

If you close the switch again, current begins to flow in two directions. 1 path is through R to ground. The other current path is through the 1ohm resistor and through the capacitor to ground charging the capacitor until the voltage across it matches the source. At that point, the capacitor is fully charged and current stops flowing through the capacitor. just as before, though, current is always flowing through resistor R while the switch is closed, sourced by the 2V voltage source.

 

[pghaffari]

When the switch closes, the capacitor charges though the 1ohm resistor. After sometime (as determined bythis equation Vc=Vin(1-e^(-t/(RC))) it is fully charged and since Vc=2V just like the source, current stops flowing through it. Resistor R, though, still conducts the current from the source to ground. However, there is no more current flowing through R1 and C.

When the switch opens, the voltage source in the circuit is now the capacitor, initially at 2V (since that is what it was charged to). So, the capacitor dicharges through the resistors to ground -- that means, current flows from the top terminal of the capacitor, charged to 2V with respect to the bottom terminal, through the resistors to the bottom terminal of the capacitor completing the loop.

As the current flows, the voltage across the capacitor discharges or decays. You can see that relationship when you apply vc=1/c ∫(-i*dt). The negative sign is just convention, since the current is flowing "out" of the capacitor. Remember that you had found i(t) as i(t)=Vinitial/R*e^-t/(RC)), so, this intregral becomes Vc=Vinitial*e^-t/(RC). So, as t gets bigger, the voltage across the capacitor decreases,eventually discharging the cap to 0V (0 V across the capacitor). No more current flows.

If you close the switch again, current begins to flow in two directions. 1 path is through R to ground. The other current path is through the 1ohm resistor and through the capacitor to ground charging the capacitor until the voltage across it matches the source. At that point, the capacitor is fully charged and current stops flowing through the capacitor. just as before, though, current is always flowing through resistor R while the switch is closed, sourced by the 2V voltage source.

thank you for that and your time! it's clear to me now. (: