Current through capacitors in parallel

[Krushnaraj Pandya]

Homework Statement

In the given circuit at t=0 S is closed then at time t, i1/i2 a) is constant b) increases c) decreases d) first increases then decreases

Homework Equations

V=IR, Q=CV

The Attempt at a Solution

At t=0, capacitors act like closed switches, so I2=2I1, where I2 and I1 are the currents as shown in figure. gradually current in both starts decreasing but since the upper capacitor gets more current initially, it attains its capacity faster so I2 drops faster than I1 so I1/I2 increases- then I2 stops falling off since capacitor is fully charged while I1 is still dropping so it decreases. But my textbook says it keeps increasing with time- where am I wrong?

 

[BvU]

In the given circuit

I see two whole and two half circuits. Which one ?

Did you ever read the guidelines ? Your shoddy pictures cause a hurdle for those wanting to help !

 

[Krushnaraj Pandya]

I see two whole and two half circuits. Which one ?

Did you ever read the guidelines ? Your shoddy pictures cause a hurdle for those wanting to help !

I'm really sorry. I have read the guidelines- but I seem to forget some time and again. I'll put up a better picture asap.

 

[Krushnaraj Pandya]

I see two whole and two half circuits. Which one ?

Did you ever read the guidelines ? Your shoddy pictures cause a hurdle for those wanting to help !

I hope this one's better- I don't have a very fancy camera, just a 6 MP webcam so pardon the quality; I suppose it's clear enough this time though

 

[BvU]

Very good, but now things like

where I2 and I1 are the currents as shown in figure

run out of context.

Your relevant equations: how do you derive $i_1(t)$ and $i_2(t)$ from them ?

 

[CWatters]

But my textbook says it keeps increasing with time- where am I wrong?

Can you post the exact wording in the book.

 

[Krushnaraj Pandya]

Very good, but now things like
run out of context.

Your relevant equations: how do you derive $i_1(t)$ and $i_2(t)$ from them ?

I was aiming to reach the answer intuitively instead of using mathematical rigor since the options are set that way. But I suppose we can find time constants for charging of a capacitor then use standard equations...but I'm just in high school so I don't have a very deep knowledge or familiarity with that, I'd be happy to learn though

 

[Krushnaraj Pandya]

Can you post the exact wording in the book.

What I wrote under problem statement is the only and exact wording in the book. The option b) is given as the correct answer in the answer key.

 

[BvU]

just in high school

Well, the exercises are tough enough, so you must know something. You already have a starting value 1/2 for the ratio $i_1/i_2$. Which one drops off faster ? As you said, $i_2$. And it goes on doing that (dropping faster), making the book answer the logical answer.

I still recommend that you write out the fulll expression for $i_1(t)$ and $i_2(t)$ and do the division -- very instructive !

 

[Krushnaraj Pandya]

the exercises are tough enough

That's not my school coursework. They taught me upto capacitors in series and parallel in school-I learned the rest myself :D

You already have a starting value 1/2 for the ratio i1/i2i1/i2i_1/i_2. Which one drops off faster ? As you said, i2i2i_2. And it goes on doing that (dropping faster), making the book answer the logical answer.

It is logical that it increases at first. But what's wrong with the second part of my logic in which it then decreases?

I still recommend that you write out the fulll expression for i1(t)i1(t)i_1(t) and i2(t)i2(t)i_2(t) and do the division -- very instructive !

Alright, I will give it a try- I think there are some examples related to this in the book; I'll read them and give it an attempt

 

[Krushnaraj Pandya]

Well, the exercises are tough enough, so you must know something. You already have a starting value 1/2 for the ratio $i_1/i_2$. Which one drops off faster ? As you said, $i_2$. And it goes on doing that (dropping faster), making the book answer the logical answer.

I still recommend that you write out the fulll expression for $i_1(t)$ and $i_2(t)$ and do the division -- very instructive !

Is the time constant 8RC/3? all the examples I read in the book were either very symmetric or had only one capacitor, this came from (the equivalent resistance x equivalent capacitance) but the time constants for both must be different obviously so how do I know the individual time constant- if I short the battery, ignore the other capacitor and take (equivalent resistance x 2C) it gives 4RC/3 but I don't think that's correct either

 

[BvU]

the examples I read in the book were either very symmetric or had only one capacitor

Would it make a difference for $i_1$ if the upper capacitor 2C and resistor R were not present ?
Conversely, would it make a difference for $i_2$ if the lower capacitor 2C and resistor 2R were absent ?

 

[Krushnaraj Pandya]

Would it make a difference for $i_1$ if the upper capacitor 2C and resistor R were not present ?
Conversely, would it make a difference for $i_2$ if the lower capacitor 2C and resistor 2R were absent ?

That'd short-circuit the resistance and no current would flow through the capacitor under consideration

 

[jbriggs444]

That'd short-circuit the resistance and no current would flow through the capacitor under consideration

The idea is that not only would one capacitor and one resistor be removed, but that the wires connecting to them would be either left as disconnected stubs or removed as well. No short would result.

[Note that if you short an ideal power source with an ideal wire, the situation is indeterminate. It is not proper to make any assertion at all about how much current would flow between two terminals that are forced both to a fixed non-zero potential difference and also to a fixed zero potential difference.]

 

[Krushnaraj Pandya]

The idea is that not only would one capacitor and one resistor be removed, but that the wires connecting to them would be either left as disconnected stubs or removed as well. No short would result.

[Note that if you short an ideal power source with an ideal wire, the situation is indeterminate. It is not proper to make any assertion at all about how much current would flow between two terminals that are forced both to a fixed non-zero potential difference and also to a fixed zero potential difference.]

Oh, alright, in that case if we are ignoring the other branch entirely-the time constants for lower one is is 2C x 2R =4CR and for the upper branch 2CR; is this correct?

 

[jbriggs444]

Oh, alright, in that case if we are ignoring the other branch entirely-the time constants for lower one is is 2C x 2R =4CR and for the upper branch 2CR; is this correct?

Right. Now when both RC paths are in there, as in the actual setup, what happens?

 

[Krushnaraj Pandya]

Right. Now when both RC paths are in there, as in the actual setup, what happens?

current starts flowing in according to the equation I=(initial-i2)e^(-t/2RC) for the upper branch. and I=(initial-i1)e^(-t/4RC)

 

[jbriggs444]

current starts flowing in according to the equation I=(initial-i2)e^(-t/2RC) for the upper branch. and I=(initial-i1)e^(-t/4RC)

Let us rewrite that using $I_1(t)$ to represent the upper current and $I_2(t)$ and some superscripts to make it easier on the eyes.
$$I_1(t) = I_1(0)e^{-t/2RC}$$
$$I_2(t) = I_2(0)e^{-t/4RC}$$

Keep going. What is I_1(0)? What is I_2(0)?

 

[Krushnaraj Pandya]

What is I_1(0)? What is I_2(0)?

V/2R and V/R respectively. let's just say V/R=k (constant)
so,
$$I_1(t) = ke^{-t/2RC}/2$$
$$I_2(t) = {k}e^{-t/4RC}$$

 

[Krushnaraj Pandya]

at t=infinite the ratio i1/i2 would be indeterminate but an intuition of limits since we know one falls faster implies definitely that it will keep increasing

 

[jbriggs444]

at t=infinite the ratio i1/i2 would be indeterminate but an intuition of limits since we know one falls faster implies definitely that it will keep increasing

I agree. The ratio has an indeterminate form. Though that does not prevent it having a limit. [Edit: you write that the ratio will increase? I am not sure I agree with that]

We should be able to do better than that and actually write a formula for the ratio. Recall that $e^{2c} = (e^c)^2$

 

[gneill]

Recall also that:
$$\frac{e^a}{e^b} = e^{(a-b)}$$

 

[Krushnaraj Pandya]

the ratio is $\frac{1}{2} e^{\frac{-t}{4RC}}$

 

[Krushnaraj Pandya]

I agree. The ratio has an indeterminate form. Though that does not prevent it having a limit. [Edit: you write that the ratio will increase? I am not sure I agree with that]

We should be able to do better than that and actually write a formula for the ratio. Recall that $e^{2c} = (e^c)^2$

Is that expression correct or am I making a mistake?

 

[jbriggs444]

Is that expression correct or am I making a mistake?

It looks close. Some of the right pieces, but I think there are two errors. Can you show your work?