RL circuit differential equations

In summary, the task is to write a differential equation for a given circuit with constants E, R1, R2, R3, and L and to determine the current and voltage through the inductor. The conversation discusses the use of Thevenin's theorem to simplify the equations, and the correct expressions for the Thevenin resistance and voltage. The conversation also addresses the initial conditions for the current and the process for solving the differential equations. The conversation also mentions two additional tasks involving current sources and provides the solutions to those tasks.
  • #1
evol_w10lv
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Homework Statement



zidurtja43lr71tzla.png


Task is to write differential equation for this circuit.
E, R1, R2, R3, L are constants.

Homework Equations



Ul = L di/dt

The Attempt at a Solution



I guess, we have to use current method for each contour.
1st contour equation:
E = U1 + U2 + Ul = i*(R1+R2) + L di/dt

Equation arround the loop:
E = i*R1 + i2*(R1+R3)

It means that I got system:
E=i*(R1+R2) + L di/dt +i2*R1
E = i*R1 + i2*(R1+R3)

The problem is i2. I need to get i2, I guess, from second contour using L, R2 and R3. Can someone help quickly? I need this task until tomorrow.
Or I have done something wrong at the begining?
 
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  • #2
Yes, you'll need to write two loop equations. I suggest choosing two of the branch currents and letting the third current be determined by KCL. Thus:

attachment.php?attachmentid=62331&stc=1&d=1380567543.gif


So the current through R1 is i1+i2.

If you've studied Thevenin Equivalents already you can make your life much simpler by replacing the voltage source and resistor network with its Thevenin Equivalent. Then you'll have a circuit consisting of a source, one resistor, and one inductor :wink: Otherwise, continue on the way you're going.
 

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  • #3
Using Thevenin's theorem, I will get this?
ooua8azemn0zetdyibmp.png


We have spoken about Thevenin's and Northon's theorems, but we haven't got any examples.
Actually, I guess, there's no matter, which way I use. I need to solve current and voltage througt inductor, that's why I have to use differential equation and U = L di/dt.

If I use Thevenin's theorem, then it should be:

L di/dt + R(Thevenin)*i = E (Thevenin)

Is it R(Thevenin) = R1 + (R2*R3)/(R2+R3) ?
What is the E (Thevenin)? :shy:
 
  • #4
evol_w10lv said:
Using Thevenin's theorem, I will get this?
ooua8azemn0zetdyibmp.png


We have spoken about Thevenin's and Northon's theorems, but we haven't got any examples.
Actually, I guess, there's no matter, which way I use. I need to solve current and voltage througt inductor, that's why I have to use differential equation and U = L di/dt.

If I use Thevenin's theorem, then it should be:

L di/dt + R(Thevenin)*i = E (Thevenin)
Yup.
Is it R(Thevenin) = R1 + (R2*R3)/(R2+R3) ?
[STRIKE]Yup.[/STRIKE] Correction: That would be the Thevenin resistance as seen from the voltage source (assuming the resistance of the inductor is zero). You want the resistance seen looking into the circuit from the inductor's location. Sorry about that.

What is the E (Thevenin)? :shy:
If you temporarily remove the inductor and measure the open-circuit voltage there, what would you get?
 
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  • #5
Do you mean:
E (Thevenin) = (E*R2)/(R1+R2) ?
 
  • #6
evol_w10lv said:
Do you mean:
E (Thevenin) = (E*R2)/(R1+R2) ?

Not quite. R1 and R3 form the voltage divider. With the inductor out of the circuit, R2 is presented with an open circuit at the "inductor end", so no current flows through it.

It might help to remember that the two branches (the one with R3 and the one with R2) are in parallel, so you are free to exchange their positions. That will make the voltage divider comprising R1 and R3 more obvious.
 
  • #7
Then it has to be:
E (Thevenin) = (E*R3)/(R1+R3)
Or again I have made akward mistake? :D

And after that:
(E*R3)/(R1+R3) = (R1 + (R2*R3)/(R2+R3))*i + L di/dt.
 
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  • #8
evol_w10lv said:
Then it has to be:
E (Thevenin) = (E*R3)/(R1+R3)
Or again I have made akward mistake? :D

And after that:
(E*R3)/(R1+R3) = (R1 + (R2*R3)/(R2+R3))*i + L di/dt.

One small correction. Back in post #4 I Okay'd your Thevenin resistance but misremembered the placement of the resistor labels. It turns out that "shape" of your expression is correct but it actually gives the Thevenin resistance of the circuit as "seen" from the voltage source rather than looking into the circuit from the inductor's location. Fix that up and you're good to go.
 
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  • #9
Yes, of course.
In that case: R(Thevenin) = R2 + (R1*R3)/(R1+R3)

And equation:
(E*R3)/(R1+R3) = (R2 + (R1*R3)/(R1+R3))*i + L di/dt.

And I've got another question. If there is no R3 before comutation proces (t<0), then I(0) = E/(R1+R2)?
 
  • #10
evol_w10lv said:
Yes, of course.
In that case: R(Thevenin) = R2 + (R1*R3)/(R1+R3)

And equation:
(E*R3)/(R1+R3) = (R2 + (R1*R3)/(R1+R3))*i + L di/dt.

And I've got another question. If there is no R3 before comutation proces (t<0), then I(0) = E/(R1+R2)?

How does an inductor behave when a sudden change of potential happens? If you suppose that the current for t < 0 is zero, when the voltage source is suddenly turned on what's the current through an inductor at time t = 0+ ?
 
  • #11
Hmm.. I guess, then initial condition (current through inductor) I = 0, when (t=0). Am I wright?
And, when t>0, then current will be solution of equation, which we wrote:
((E*R3)/(R1+R3) = (R2 + (R1*R3)/(R1+R3))*i + L di/dt.).
 
  • #12
evol_w10lv said:
Hmm.. I guess, then initial condition (current through inductor) I = 0, when (t=0). Am I wright?
Yes that's right.

And, when t>0, then current will be solution of equation, which we wrote:
((E*R3)/(R1+R3) = (R2 + (R1*R3)/(R1+R3))*i + L di/dt.).
Sure.
 
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  • #13
And also..
To get voltage through inductor (t>0), I have to integrate result of current?
And, when I have voltage, to get current, I have to derive result of voltage?
 
  • #14
evol_w10lv said:
And also..
To get voltage through inductor (t>0), I have to integrate result of current?
And, when I have voltage, to get current, I have to derive result of voltage?
I'm not sure what you're getting at here, but once you've solved the differential equation for i(t) you can write the voltage across the resistor using Ohm's law, then the voltage across the inductor using KVL.
 
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  • #15
Never mind.
Actually this task is solved. Thanks for helping me. :)
 
  • #16
Hei!
Again it's me. Can you chek these two tasks?
1.
kz3cd9z9vy5mw8qgl.png

System of:
L*di_L/dt + R1*(I_L - I_2) = V
R1*(I_2 - I_L) + (R_2+R_3)*I_2=0

from second equation: I_2 = (R1*I_L)/(R1+R2+R3)
then:
L*di_L/dt + R1*(I_L - (R1*I_L)/(R1+R2+R3)) = E

And if I use Thevenin method, then:
R(Thevenin) = ((R2+R3)*R1)/(R1+R2+R3)
E(Thevenin) = V

L*di_L/dt + (((R2+R3)*R1)/(R1+R2+R3))*I_L = V

2.
xb8euajesuqx7oy3e56l.png

R(Thevenin) = R3 * ((R1*R2)/(R1+R2))
E(Thevenin) = (V*R2)/(R2+R1)

L*di_L/dt + (R3 * ((R1*R2)/(R1+R2)))*I_L = (V*R2)/(R2+R1)

And what do you suggest, when current source given?
Hope that you'll chek the tasks. I guess, it's boring for you so easy tasks. I tried to solve by myself, but I can't chek whether they are correct. :)
 
  • #17
It is preferred that a new thread be started for each new question rather than tacking onto an old thread. Please keep that in mind for the future.

The first one looks fine. The second one has a problem with R(Thevenin); R3 is in series with the parallel combination of R1 and R2.
 
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  • #18
Ohh.. of course about second one.. typing mistake.
OK, next time I'll make new topic. Just thought.. almost same circuit and same task.
 
  • #19
evol_w10lv said:
Ohh.. of course about second one.. typing mistake.
OK, next time I'll make new topic. Just thought.. almost same circuit and same task.

No worries. Looks like you've got the hang of these problems. Well done.
 

1. What is an RL circuit?

An RL circuit is an electrical circuit that contains a resistor (R) and an inductor (L). These circuits are commonly used to control the flow of current in a circuit, and they can be found in many electronic devices such as radios, televisions, and computers.

2. What is a differential equation?

A differential equation is a mathematical equation that describes how a variable changes over time. In the context of an RL circuit, differential equations are used to model the relationship between the current (I) and voltage (V) in the circuit.

3. How are differential equations used in RL circuits?

Differential equations are used to describe the behavior of an RL circuit by relating the rate of change of current (dI/dt) to the voltage (V) and inductance (L) in the circuit. These equations can be solved to determine the current in the circuit at any given time.

4. What is the time constant of an RL circuit?

The time constant of an RL circuit is a measure of how quickly the current in the circuit changes. It is calculated by dividing the inductance (L) by the resistance (R) in the circuit. A larger time constant indicates a slower change in current, while a smaller time constant indicates a faster change.

5. How can I solve RL circuit differential equations?

There are several methods for solving RL circuit differential equations, including analytical methods such as substitution and separation of variables, as well as numerical methods such as Euler's method or Runge-Kutta methods. The method used will depend on the complexity of the circuit and the desired level of accuracy.

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