Help on First order Diff Eq problem

In summary: What do you think should be done with this situation? It's entirely a math problem at this point.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.
  • #1
pghaffari
35
0

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.
 
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  • #3
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?
 
  • #4
How can there be a loop when the switch is open?
 
  • #5
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?
 
  • #6
pghaffari said:
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?
 
  • #7
We know that current is a constant value, and that current before opening switch must equal current after opening the switch.
 
  • #8
Right. So how about writing an expression for that current? What role if any does C play in determining it?
 
  • #9
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?
 
  • #10
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?
 
  • #11
pghaffari said:
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.
 
  • #12
-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 ?
 
  • #13
pghaffari said:
-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.
 
Last edited:
  • #14
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.
 
  • #15
rude man said:
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 ??
 
  • #16
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?
 
Last edited:
  • #17
I thought R_1 is just 1 Ohm here?

so that V(t) = i(t)
 
  • #18
rude man said:
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.
 
  • #19
rude man said:
You're right, sorry. I meant R.

Voltage is somehow still in the Resistor?
 
  • #20
No, a resistor can't store voltage. What componernt can?
 
  • #21
The capacitor
 
  • #22
Right! Now, when the switch was closed for a long time, what voltage do you think might be stored on C?
 
  • #23
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?
 
  • #24
Wait wouldn't it just be 1/c integral i dt
 
  • #25
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.
 
  • #26
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
 
  • #27
pghaffari said:
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?
 
  • #28
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?
 
  • #29
rude man said:
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?
 
  • #30
pghaffari said:
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.
 
  • #31
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@!
 
  • #32
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 VacrossR1=I*R1, VacrossR1(t)=K*e(-t/RC)*R1. you can now solve this equation for t=2usecs.

hope this helps...
 
  • #33
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
 
  • #34
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=Vcapat 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.
 
  • #35
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 ?
 
<h2>1. What is a first-order differential equation?</h2><p>A first-order differential equation is an equation that relates a function to its derivative. It can be written in the form dy/dx = f(x), where y is the dependent variable and x is the independent variable.</p><h2>2. How do I solve a first-order differential equation?</h2><p>To solve a first-order differential equation, you can use various methods such as separation of variables, integrating factors, or substitution. The specific method to use depends on the form of the equation and its initial conditions.</p><h2>3. What are the initial conditions of a differential equation?</h2><p>The initial conditions of a differential equation refer to the values of the dependent variable and its derivatives at a specific point. These conditions are necessary to find a particular solution to the equation.</p><h2>4. Can I use software to solve a first-order differential equation?</h2><p>Yes, there are many software programs and online tools available that can solve first-order differential equations. However, it is important to have a basic understanding of the concepts and methods involved in solving these equations.</p><h2>5. What are some real-life applications of first-order differential equations?</h2><p>First-order differential equations are used to model various physical phenomena, such as population growth, radioactive decay, and chemical reactions. They are also commonly used in engineering, economics, and other fields to analyze and predict behavior of systems over time.</p>

1. What is a first-order differential equation?

A first-order differential equation is an equation that relates a function to its derivative. It can be written in the form dy/dx = f(x), where y is the dependent variable and x is the independent variable.

2. How do I solve a first-order differential equation?

To solve a first-order differential equation, you can use various methods such as separation of variables, integrating factors, or substitution. The specific method to use depends on the form of the equation and its initial conditions.

3. What are the initial conditions of a differential equation?

The initial conditions of a differential equation refer to the values of the dependent variable and its derivatives at a specific point. These conditions are necessary to find a particular solution to the equation.

4. Can I use software to solve a first-order differential equation?

Yes, there are many software programs and online tools available that can solve first-order differential equations. However, it is important to have a basic understanding of the concepts and methods involved in solving these equations.

5. What are some real-life applications of first-order differential equations?

First-order differential equations are used to model various physical phenomena, such as population growth, radioactive decay, and chemical reactions. They are also commonly used in engineering, economics, and other fields to analyze and predict behavior of systems over time.

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