Discharging Capacitor in an RC Circuit w/ Battery

[deftonix]

Homework Statement

The bit I'm struggling with is to determine the time constant $$\tau$$ for the referenced circuit after the switch is closed. Constants are given in the image, which can be found at https://www.physicsforums.com/attachment.php?attachmentid=12737&d=1203627574.

Homework Equations

$$\tau=RC$$

The Attempt at a Solution

I know that the answer is $$100k\Omega \cdot 10.0\mu F = 1.00s$$ and that it is because current is going to flow in the loop that goes from the cap, through the switch, through the $$100k\Omega$$ resistor, and back to the cap; since the only resistor that affects it is $$100k\Omega$$ one, it's just that times C. I also understand that there should be a similar loop on the battery side of the circuit. What I don't understand is why the loop that doesn't go through the switch(i.e., the one containing everything) doesn't affect the time constant. Wouldn't the battery be pushing current into that loop as the capacitor's voltage dropped?

 

[berkeman]

Homework Statement

The bit I'm struggling with is to determine the time constant $$\tau$$ for the referenced circuit after the switch is closed. Constants are given in the image, which can be found at https://www.physicsforums.com/attachment.php?attachmentid=12737&d=1203627574.

Homework Equations

$$\tau=RC$$

The Attempt at a Solution

I know that the answer is $$100k\Omega \cdot 10.0\mu F = 1.00s$$ and that it is because current is going to flow in the loop that goes from the cap, through the switch, through the $$100k\Omega$$ resistor, and back to the cap; since the only resistor that affects it is $$100k\Omega$$ one, it's just that times C. I also understand that there should be a similar loop on the battery side of the circuit. What I don't understand is why the loop that doesn't go through the switch(i.e., the one containing everything) doesn't affect the time constant. Wouldn't the battery be pushing current into that loop as the capacitor's voltage dropped?

The switch is a short circuit, so there is no voltage drop across it as the left and right loop currents go through it. So the current in the left loop cannot affect the circuit loop on the right -- there is no voltage to drive any extra current.

 

[deftonix]

The switch is a short circuit, so there is no voltage drop across it as the left and right loop currents go through it. So the current in the left loop cannot affect the circuit loop on the right -- there is no voltage to drive any extra current.

OK, so if I'm getting this correctly, the current from the capacitor reaches the junction and it all turns into that segment because there's no voltage to drive some charge away. Is that right?

Thanks for the quick reply, by the way.

 

[berkeman]

OK, so if I'm getting this correctly, the current from the capacitor reaches the junction and it all turns into that segment because there's no voltage to drive some charge away. Is that right?

Thanks for the quick reply, by the way.

Welcome.

A better way to picture it, is to have the left battery+resistor loop moved more over to the left, and the right RC loop over on the right, and just connect the two of them with a single wire that goes between the two loops (this is after the switch is closed, obviously).

With that kind of picture, it's easier to see that there is no reason for any current to flow through that single wire between the two loops.

 

[deftonix]

Welcome.

A better way to picture it, is to have the left battery+resistor loop moved more over to the left, and the right RC loop over on the right, and just connect the two of them with a single wire that goes between the two loops (this is after the switch is closed, obviously).

With that kind of picture, it's easier to see that there is no reason for any current to flow through that single wire between the two loops.

So what happens if there is a resistor on the branch with the switch?

 

[berkeman]

So what happens if there is a resistor on the branch with the switch?

You tell me. What equation(s) would you write to solve for the new time constant?

 

[deftonix]

You tell me. What equation(s) would you write to solve for the new time constant?

Would I have to write a Kirchhoff equation?

 

[berkeman]

Would I have to write a Kirchhoff equation?

Yep, a KCL equation at the top node, and you would use the differential equation for the I-V relationship for the cap. You would solve the overall differential equation assuming an exponential form for the solution, and use all of that to figure out your new time constant.

Give it a try, and post your work here if you have trouble with it.

 

[deftonix]

Yep, a KCL equation at the top node, and you would use the differential equation for the I-V relationship for the cap. You would solve the overall differential equation assuming an exponential form for the solution, and use all of that to figure out your new time constant.

Give it a try, and post your work here if you have trouble with it.

I'm having a bit of trouble getting started... So I've labeled $$I_1$$ as the current in the batt-resistor loop, $$I_2$$ as the current in the RC loop, and $$I_3$$ as the current in the resistor/switch branch. My small amount of work so far:

$$I_1+I_2=I_3 \\$$

$$\frac{V_1}{R_1}+\frac{V_2}{R_2} e^\frac{-t}{\tau}=I_3$$

I'm not sure where to go with it because I don't know $$\tau$$ or $$I_3$$.

 

[berkeman]

I'm having a bit of trouble getting started... So I've labeled $$I_1$$ as the current in the batt-resistor loop, $$I_2$$ as the current in the RC loop, and $$I_3$$ as the current in the resistor/switch branch. My small amount of work so far:

$$I_1+I_2=I_3 \\$$

$$\frac{V_1}{R_1}+\frac{V_2}{R_2} e^\frac{-t}{\tau}=I_3$$

I'm not sure where to go with it because I don't know $$\tau$$ or $$I_3$$.

I'd still recommend using KCL, and not the KVL equations that you are trying to write.

Also, what is the relationship between current I through a capacitor and the voltage V across it? It involves a derivative. Don't try to put in the exponential stuff until you have the differential equation written out in the KCL.

 

[deftonix]

I'd still recommend using KCL, and not the KVL equations that you are trying to write.

Also, what is the relationship between current I through a capacitor and the voltage V across it? It involves a derivative. Don't try to put in the exponential stuff until you have the differential equation written out in the KCL.

Isn't KCL Kirchhoff's Current Law which states that currents that join at a junction/node must be combine to equal the third current? And KVL is Kirchhoff's Voltage Law which is the summing of all the voltage changes in each object in the circuit? Because the equation I wrote is certainly a current law equation.

I'm completely blanking on the relationship for current through a capacitor and the voltage, and I can't find it in my notes...

 

[berkeman]

Isn't KCL Kirchhoff's Current Law which states that currents that join at a junction/node must be combine to equal the third current? And KVL is Kirchhoff's Voltage Law which is the summing of all the voltage changes in each object in the circuit? Because the equation I wrote is certainly a current law equation.

I'm completely blanking on the relationship for current through a capacitor and the voltage, and I can't find it in my notes...

Ah, sorry, I misread your equations. The form of KCL that I usually use is the sum of all currents leaving a node must be zero. That's the easiest form for me to use.

And for the capacitor:

$$I(t) = C \frac{dV(t)}{dt}$$

So write the KCL for the top middle node with a resistor in series with the switch, and you'll end up with a differential equation. Do the rest of what I suggested about how to solve it, and see where you end up...

 

[gomunkul51]

RI + Q/C = 0 ?

(the right square closed loop)

can make an ODE from it? :)

 

[berkeman]

RI + Q/C = 0 ?

(the right square closed loop)

can make an ODE from it? :)

Which R is that? Did you see where the OP is asking a different question now? He is asking what happens to the time constant if there is a resistor in series with the switch itself.