# Current as a function of time in an RC circuit

## Homework Statement

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

## Homework Equations

I = V/R
I_capacitor = I e^(-t/RC)

## The Attempt at a Solution

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gneill
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?

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

gneill
Mentor
The voltage source is 5V.
Actually I just want to know how to get the function, not about the value.
No, show an attempt first. That's the rule here.

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

gneill
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.

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

gneill
Mentor
Before I continue, why is there voltage drop across the capacitor? Isn't it supposed to be voltage rise across the capacitor?
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.

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.
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?

gneill
Mentor
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?
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.

terryds
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.
Oh I see.. Thanks a lot

Is this differential equation correct?

gneill
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.

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gneill
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.

terryds
@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.
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:
gneill
Mentor
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?
It's the initial current. Go back to the circuit and find the initial current value for ##i_2##.

terryds
NascentOxygen
Staff Emeritus
And then, I get

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

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

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

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

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

gneill
Mentor
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
According to your figure (and equations) in post 12, ##i_2## is the capacitor current. I noticed at the time that you had changed your current definitions from your initial post (where i2 went through the 100 Ohm resistor) , but since you wrote correct equations with this new definition I didn't bother to mention the change.

terryds
According to your figure (and equations) in post 12, ##i_2## is the capacitor current. I noticed at the time that you had changed your current definitions from your initial post (where i2 went through the 100 Ohm resistor) , but since you wrote correct equations with this new definition I didn't bother to mention the change.
Oh, I'm sorry.

i2 initial is V/R1, right?
So,
##I_{2i} = \frac{V}{R_1}e^{\frac{t}{C_1 R}}##
Is it correct now?

gneill
Mentor
i2 initial is V/R1, right?
So,
##I_{2i} = \frac{V}{R_1}e^{\frac{t}{C_1 R}}##
Is it correct now?
Yes.

Now that you've found the expression for the capacitor current the hard way (solving the differential equation), let's take a look at what I hinted at in post #14 when I mentioned that the resistor expression, when reduced, would look familiar. You subsequently noted that it was the expression for parallel resistors, which is correct.

If you are looking for the current (or voltage) for a capacitor that is being driven by a source and resistor network, you can treat the capacitor as a load and reduce the source network to its Thevenin equivalent. That will yield the trivial series RC circuit and you should be able to write the solution equation for it by inspection. The time constant ##\tau## and the initial current can both be obtained simply inspection using the Thevenin resistance and short circuit current for the Thevenin model.

terryds