Calculate current in one resistor in RC circuit

[hopkinmn]

Homework Statement

The capacitor is initially uncharged when the switch is closed for a long time. Calculate the current through R2.
Then find the potential difference across R1, R2, and R3

Homework Equations

V=IR
Iin=Iout

The Attempt at a Solution

I first set up current going into top node (I1) equal to the sume of the currents leaving that node (I2 and I3). So I1=I2+I3. I set voltage of the battery equal to V.
Then, I used Kirchoff's rules and the following equations 0=-R1*I1-R3*I3+V and R3*I3=R2*I2 to get I2=10/13, I3=5/13, and I1=15/13.

For the potential differences I found V2=V3=40/13V and V1=90/13V
So the sum of the potential differences across the resistors is equal to the voltage on the battery.

But if this is true, that means that the capacitor does not have a potential difference. But is this true? Or would there be no voltage on R2 and instead have voltage on the capacitor?

 

[gneill]

Homework Statement

The capacitor is initially uncharged when the switch is closed for a long time. Calculate the current through R2.
Then find the potential difference across R1, R2, and R3

Homework Equations

V=IR
Iin=Iout

The Attempt at a Solution

I first set up current going into top node (I1) equal to the sume of the currents leaving that node (I2 and I3). So I1=I2+I3. I set voltage of the battery equal to V.
Then, I used Kirchoff's rules and the following equations 0=-R1*I1-R3*I3+V and R3*I3=R2*I2 to get I2=10/13, I3=5/13, and I1=15/13.

For the potential differences I found V2=V3=40/13V and V1=90/13V
So the sum of the potential differences across the resistors is equal to the voltage on the battery.

But if this is true, that means that the capacitor does not have a potential difference. But is this true? Or would there be no voltage on R2 and instead have voltage on the capacitor?

Hi hopkinmn, welcome to Physics Forums.

Can you clarify the problem statement a bit? It appears that the capacitor should have a charge when the switch is closed for a long time. So how can it be uncharged when the switch is closed for a long time?

At what time is the current through R2 supposed to be calculated: when the switch is open, when it's first closed, or a long time after the switch is closed?

 

[hopkinmn]

Yes, it does have the charge shown in the diagram. I meant to say that before the switch is closed, the capacitor is uncharged.

R2 is supposed to be calculated long after the switch is closed.

Thanks!

 

[gneill]

Okay, so after a long time has passed the capacitor will have reached its final voltage as determined by the surrounding circuit elements. What will be the current in the capacitor branch?

 

[NascentOxygen]

R2 is supposed to be calculated long after the switch is closed.

You are wanting to determine the DC current in R2 long after the switch has been closed?

 

[hopkinmn]

Okay, so after a long time has passed the capacitor will have reached its final voltage as determined by the surrounding circuit elements. What will be the current in the capacitor branch?

Wouldn't it be the same as the current in R2?

 

[hopkinmn]

You are wanting to determine the DC current in R2 long after the switch has been closed?

Yes, I believe so

 

[NascentOxygen]

Wouldn't it be the same as the current in R2?

As the R and C are in series, their currents will be the same.

 

[hopkinmn]

As the R and C are in series, their currents will be the same.

Yes, but the way I found R2 was by saying R2*I2=R3*I3, since parallel currents are equal. But is this true? Would there be voltage from the capacitor that I need to include, making the equation R2*I2 + V(capacitor)=R3*I3?

If so, I'm not sure how to solve the problem (using Vcapacitor=Q/C), since charge is not given

 

[gneill]

Wouldn't it be the same as the current in R2?

Yes, but that observation is not particularly helpful What value will it have?

What have you learned about the behavior of capacitors and inductors when they reach steady state?

 

[hopkinmn]

Yes, but that observation is not particularly helpful What value will it have?

What have you learned about the behavior of capacitors and inductors when they reach steady state?

Doesn't the capacitor have a constant voltage after it's reached a steady state?

 

[gneill]

Yes, but the way I found R2 was by saying R2*I2=R3*I3, since parallel currents are equal. But is this true? Would there be voltage from the capacitor that I need to include, making the equation R2*I2 + V(capacitor)=R3*I3?

If so, I'm not sure how to solve the problem (using Vcapacitor=Q/C), since charge is not given

Yes, but you're talking about different circuit conditions. In your first post you had assumed that the capacitor was uncharged (and so behaved as a short circuit --- as though it were just a piece of wire), so that put R2 and R3 in parallel. In the present case the circuit has reached steady state so the capacitor is charged to some constant voltage. NO CURRENT will flow into or out of the capacitor at steady state.

 

[hopkinmn]

Yes, but you're talking about different circuit conditions. In your first post you had assumed that the capacitor was uncharged (and so behaved as a short circuit --- as though it were just a piece of wire), so that put R2 and R3 in parallel. In the present case the circuit has reached steady state so the capacitor is charged to some constant voltage. NO CURRENT will flow into or out of the capacitor at steady state.

So if no current flows into or out of the capacitor at a steady state, does that mean the current in R2 is zero as well?

 

[gneill]

So if no current flows into or out of the capacitor at a steady state, does that mean the current in R2 is zero as well?

Can the current through two components in series be different?

 

[NascentOxygen]

Yes, but the way I found R2 was by saying R2*I2=R3*I3, since parallel [strike]currents[/strike] voltages are equal. But is this true?

It's true now. This applies only to the start-up situation, when the capacitor is uncharged.

 

[hopkinmn]

Can the current through two components in series be different?

No, I guess not. Thanks for helping me work this out!

 

[hopkinmn]

It's true now. This applies only to the start-up situation, when the capacitor is uncharged.

Oops, yes, you are right! Thanks!

 

[gneill]

No, I guess not. Thanks for helping me work this out!

No problem, glad to help.

The basic things to know for these sorts of problem include:

1. Current is the same in series components
2. Voltage is the same across parallel components
3. Capacitors don't change their charges (or potential differences) instantaneously, so capacitors initially behave like short circuits to sudden changes.
4. Inductors don't change their currents instantaneously, so inductors initially behave like open circuits to sudden changes.
5. At steady state capacitors behave like open circuits
6. At steady state inductors behave like short circuits