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

1. Mar 20, 2006

### Lisa...

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.

2. Mar 20, 2006

### Lisa...

Okay I've figured out 4 & 5 & 6 just now, but from 7 and on I still don't get anything of the explanation....

Last edited: Mar 20, 2006
3. Mar 20, 2006

### nrqed

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. Mar 20, 2006

### Lisa...

But how does he solve the differential equation? I don't get that part (haven't had it yet in math courses)

5. Mar 20, 2006

### nrqed

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

Ok. An equation of the form ${di \over dt} = A + B \,i(t)$ has a solution of the form $i(t) = a + b e^{ c t}$ 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. Mar 20, 2006

### Lisa...

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:

$$bc e^{ c t}$$ = $$\frac{E}{R_1 R_2 C} - a \frac{R_1 + R_2}{R_1 R_2 C}- b e^{ c t}\frac{R_1 + R_2}{R_1 R_2 C}$$

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

7. Mar 20, 2006

### Lisa...

I've tried this:

$$a \frac{R_1 + R_2}{R_1 R_2 C}$$ = $$\frac{E}{R_1 R_2 C} - b e^{ c t}\frac{R_1 + R_2}{R_1 R_2 C} - bc e^{ c t}$$

Then I've multiplied by $$R_1 R_2 C$$ providing:

$$a (R_1 + R_2)$$ = $$E - b e^{ c t}(R_1 + R_2) - bc e^{ c t} (R_1 R_2 C)$$

8. Mar 20, 2006

### nrqed

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

$$bc e^{ c t} = - b e^{ c t}\frac{R_1 + R_2}{R_1 R_2 C}$$

which tells you what "c" is, and then

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

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. Mar 20, 2006

### Lisa...

Wow Pat!!! Thanks a lot for your great explanation!!! It sure clears up a hell of a lot!! :)

10. Mar 20, 2006

### nrqed

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!!