# Current as a function of time in an RC circuit

1. Feb 26, 2017

### terryds

1. The problem statement, all variables and given/known data

How to determine the current as a function of time that flows in 10 ohm resistor, 100 ohm resistor, and 15 microFarad capacitor?

2. Relevant equations
I = V/R
I_capacitor = I e^(-t/RC)

3. The attempt at a solution

2. Feb 26, 2017

### Staff: Mentor

If you are studying RC circuits you must have already covered other circuit analysis methods, laws, and formulas.

You should be able to make some attempt either by writing the differential equation for the circuit and solving, or by determining initial and final conditions and applying the known response functions that govern RC circuits in general.

I note that your problem doesn't specify a value for the supply voltage. Is this intentional?

3. Feb 26, 2017

### terryds

The voltage source is 5V.
Actually I just want to know how to get the function, not about the value.

4. Feb 26, 2017

### Staff: Mentor

No, show an attempt first. That's the rule here.

5. Feb 26, 2017

### terryds

Alright,

I'll use KVL on loop 1, let the 10 ohm resistor be R1, and 100 ohm resistor be R2
the current goes through r1 is i1
the current through r2 is i2
the current through capacitor is i3

For loop 1 (assume current is clockwise)
V - i1 R1 - i2 R2 = 0

For loop 2 (assume current is clockwise)
-i2 R2 + (integral i3 dt)/C = 0 (is it correct? im not sure about the positive sign, but I think it is since there will be voltage rise across the capacitor)

Is it correct?

6. Feb 26, 2017

### Staff: Mentor

I see that you've chosen to write the differential equations. Okay, what you've done so far is fine. You''ve got three unknowns though (i1, i2, and i3) so you'll need three equations. You've got two so far.

7. Feb 26, 2017

### terryds

8. Feb 26, 2017

### terryds

Please look at my reply above. I forgot to quote you before

9. Feb 26, 2017

### Staff: Mentor

It is true that the potential across the capacitor will increase over time, but this is not what "rise" or "drop" terms refer to in this context. They refer to the change in potential seen when you sum the potential changes around the circuit according to KVL.

If you do a "KVL walk" around the loop in a clockwise direction (the direction of the current flow) there's a potential rise across the battery, then potential drops across the resistor and capacitor.

10. Feb 26, 2017

### terryds

So, the voltage drop refers to the voltage of something - voltage source?
Is it correct?
I mean, even if potential across capacitor is increasing, but since the value of it is still less than the voltage source, so we can say that it is voltage drop, right?

11. Feb 26, 2017

### Staff: Mentor

The value of the voltage source doesn't matter. It's the polarity of the change in potential when you "walk over" the component that matters. If you did your KVL walk in the counterclockwise direction (against the current flow in this case) then there would be potential rises on the the capacitor and resistor and a potential drop across the voltage source.

12. Feb 26, 2017

### terryds

Oh I see.. Thanks a lot

Is this differential equation correct?

13. Feb 26, 2017

### Staff: Mentor

Somehow the $V_1$ (source voltage) disappeared from the last two lines of your work. Better check those steps.

Edit: Never mind! I see that it disappeared with the differentiation. It'll come back again when you supply the initial condition for $i_2$ when you solve the differential equation.

Last edited: Feb 26, 2017
14. Feb 26, 2017

### Staff: Mentor

@terryds, See my edit to my last post.

Your differential equation looks fine. You can simplify that first term a bit more; You should see a familiar expression involving resistors when you do.

15. Feb 27, 2017

### terryds

It becomes

(R1*R2/(R1+R2)) di2/dt - i2/C1 = 0

Let (R1*R2)/(R1+R2) be R_parallel

And then, I get

$i_2 = k e^{\frac{t}{C_1 R}}$

Is it correct?
But, how to know the value of k?

Last edited: Feb 27, 2017
16. Feb 27, 2017

### Staff: Mentor

It's the initial current. Go back to the circuit and find the initial current value for $i_2$.

17. Feb 27, 2017

### Staff: Mentor

When t=0, this simplifies to $i_2 = k$
So k will be the initial value of $i_2$ before current has had a chance to add charge to the capacitor plates.

18. Feb 27, 2017

### terryds

Using current divider, I get

$I_{2i} = \frac{VR_1}{(R_1+R_2)^2}$

So,
$i_2 = \frac{VR_1}{(R_1+R_2)^2} e^{\frac{t}{C_1 R}}$
Is it right?

19. Feb 27, 2017

### Staff: Mentor

No. At time = 0+, what does an uncharged capacitor "look like" in a circuit? What is it equivalent to?

20. Feb 27, 2017

### terryds

It is short circuit.

So i2 initial is zero, since all the current will rather go through the capacitor than r2 right?
But, it's very funny to say that k is 0