Me Understand This Before The Test 2morrow: Current In Capacitor Circuit

In summary, a student was struggling to understand a problem given by their teacher. They had difficulties with step c from point 4 onwards. After receiving help, they understood steps 4 & 5, but not the remaining steps. The teacher had isolated I1 and I2 in previous steps and then plugged them into the differential equation of I2 to solve it. A rule in math was used to guess the form of the solution, and then the initial condition was used to determine the constants. The student eventually understood the problem with the help of an expert summarizer.
  • #1
Lisa...
189
0
Hey!

My teacher gave me the following explanation to the following problem, but I've lost him in part c from point 4 and on. PLEASE help me to understand what he's doing, cause I really don't get the last part of question c.

http://img230.imageshack.us/img230/7448/problem6ak.th.gif [Broken]
 
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  • #2
Okay I've figured out 4 & 5 & 6 just now, but from 7 and on I still don't get anything of the explanation...
 
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  • #3
Lisa... said:
Okay I've figured out 4 & 5 just now, but from 6 and on I still don't get anything of the explanation...


In step 5, he simply isolate I1 interms of I2, right?
In step 4 he had isolated the derivative of I2 in terms of I1 and I2.

He then plugs the expression for I1 of step 5 in the differential equation of I2, giving an equation for I2 in terms of I2 alone and all the constants. He then solves the differential equation.

Pat
 
  • #4
But how does he solve the differential equation? I don't get that part (haven't had it yet in math courses)
 
  • #5
Lisa... said:
But how does he solve the differential equation? I don't get that part (haven't had it yet in math courses)

So you are ok about how he gets to the equation?

Ok. An equation of the form [itex] {di \over dt} = A + B \,i(t)[/itex] has a solution of the form [itex] i(t) = a + b e^{ c t} [/itex] where a,b,c,A,B are all constants. This is why he tries a solution of the form given. Just plug this "guessed solution" in the differential equation and you get the result he gives.

I am not sure if your question is i) How did he guessed the form of the solution or b) what did he do after that? (answer: he just plugged it in the diff equation for the current)
 
  • #6
Yeah I understand how he got to the equation, thanks for helping btw...

I guess the suggestion is just a rule from maths, but I plugged it in the equation and I can't seem to solve it:

I have:

[tex]bc e^{ c t}[/tex] = [tex]\frac{E}{R_1 R_2 C} - a \frac{R_1 + R_2}{R_1 R_2 C}- b e^{ c t}\frac{R_1 + R_2}{R_1 R_2 C} [/tex]

and don't know what to do next...
 
  • #7
I've tried this:

[tex]a \frac{R_1 + R_2}{R_1 R_2 C}[/tex] = [tex]\frac{E}{R_1 R_2 C} - b e^{ c t}\frac{R_1 + R_2}{R_1 R_2 C} - bc e^{ c t} [/tex]

Then I've multiplied by [tex]R_1 R_2 C[/tex] providing:

[tex]a (R_1 + R_2)[/tex] = [tex]E - b e^{ c t}(R_1 + R_2) - bc e^{ c t} (R_1 R_2 C)[/tex]
 
  • #8
Lisa... said:
Yeah I understand how he got to the equation, thanks for helping btw...

I guess the suggestion is just a rule from maths, but I plugged it in the equation and I can't seem to solve it:

I have:

[tex]bc e^{ c t}[/tex] = [tex]\frac{E}{R_1 R_2 C} - a \frac{R_1 + R_2}{R_1 R_2 C}- b e^{ c t}\frac{R_1 + R_2}{R_1 R_2 C} [/tex]

and don't know what to do next...


Ok.
Well, there is the initial condition I(0) = 0 which implies right awat that b= -a. It's easier to use that before plugging in the differential equation to simplify things as much as possible, but we can also use it afterward.

All you have to do is to impose that the terms with no exponential on both sides of your equation must be equal and that the terms with an exponential must be equal, which gives


[tex]bc e^{ c t} = - b e^{ c t}\frac{R_1 + R_2}{R_1 R_2 C} [/tex]

which tells you what "c" is, and then

[tex] 0 = [tex]\frac{E}{R_1 R_2 C} - a \frac{R_1 + R_2}{R_1 R_2 C} [/tex]

which tells you what "a" is.

All this doesn't tell you what b is and for that you need to use the initial condition I(0) = 0.

Hope this helps.

Pat
 
  • #9
Wow Pat! Thanks a lot for your great explanation! It sure clears up a hell of a lot! :)
 
  • #10
Lisa... said:
Wow Pat! Thanks a lot for your great explanation! It sure clears up a hell of a lot! :)

You are welcome Lisa. The key trick was to consider separately the terms with exponentials and the terms without exponentials, as you saw. (and to use the initial condition).

Best luck for your test!
 

1. What is a capacitor circuit?

A capacitor circuit is an electronic circuit that uses a capacitor to store and release electrical charge. It is used to temporarily store energy and smooth out voltage in a circuit.

2. How does a capacitor circuit work?

A capacitor circuit works by storing electrical charge on two conductive plates separated by an insulating material. When a voltage is applied, electrons are transferred from one plate to the other, creating a potential difference across the plates. This stored energy can then be released when needed.

3. What is current in a capacitor circuit?

Current in a capacitor circuit refers to the flow of electrical charge through the circuit. In a capacitor circuit, current is initially high when the capacitor is charging, but decreases as the capacitor becomes fully charged.

4. How is current affected by a capacitor in a circuit?

In a capacitor circuit, current is affected by the capacitance (ability to store charge) of the capacitor and the rate at which the voltage changes across the capacitor. As the capacitor charges, the current decreases until it reaches a steady state where the capacitor is fully charged and there is no more flow of current.

5. How do I calculate the current in a capacitor circuit?

To calculate the current in a capacitor circuit, you can use the formula I = C*(dV/dt), where I is the current, C is the capacitance of the capacitor, and dV/dt is the rate of change of voltage across the capacitor. Alternatively, you can use Ohm's law (I = V/R) if there is a resistor in the circuit.

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