# How do I find value of U on R2 long time after switch is closed?

Discussion in 'Homework Help' started by thexy, Dec 24, 2015.

1. ### thexy Thread Starter Member

Dec 13, 2015
130
0 How to find value of U on R2 long time after switch is being closed?

Capacitors are charged so there is no current going through? Or?
I used kirchof rule so i get
Uo+Uc1-U=0
U=Uc1+Uo
U=R2/(R3+R2+(R1R/R1+R))*Uo

Is this correct or?

If not can someone do this example and explain me 2. ### hsazerty2 New Member

Sep 25, 2015
25
1
The answer is in the meaning of the sentence: "long time after switch is closed".

3. ### thexy Thread Starter Member

Dec 13, 2015
130
0
Explain me more?

4. ### thexy Thread Starter Member

Dec 13, 2015
130
0
I know that the voltage drop across the resistor after long time is 0 – no current is flowing, but i don't think that's good answer.

5. ### shteii01 AAC Fanatic!

Feb 19, 2010
4,490
710
You are correct in your assumption. After the switch is closed for a long time, all the capacitors are fully charged. In dc circuit a fully charged capacitor act as an open circuit, you have approached it from the point of current being zero, which is correct. What does it mean:
- C1 is gone
- C3 is gone
- C4 is gone
- R is gone
- R3 is gone

What do you have left:
- U0
- R1
- R2
All the components are in series. If you pay attention, you will notice that R1 and R2 form Voltage Divider. Apply voltage divider formula to find U.
BAM!
You are done.

6. ### thexy Thread Starter Member

Dec 13, 2015
130
0
But still how do I know that R1 is left? In my case i think it's gone because there C1 belove

7. ### shteii01 AAC Fanatic!

Feb 19, 2010
4,490
710
Draw the circuit with pencil. All the capacitors are fully charged, this means they act as open circuit. Do you know what Open Circuit is? It means there is no wire, there is Nothing there. So erase all the capacitors. Now try to follow the current. Any branch that has open circuit in it will have no current though it.

R3 has an open circuit in its branch, so there is no current though R3.
R has open circuit in its branch, so there is no current though R.

R1 is connected to voltage source on one end, and it is connected to the voltage source on the other end through R2. As you can see R1 is part of a Complete circuit, which means it will have current through it.

8. ### thexy Thread Starter Member

Dec 13, 2015
130
0
So U is r2/R1+R2 * Uo

9. ### shteii01 AAC Fanatic!

Feb 19, 2010
4,490
710
You can simulate the circuit to check your answer.

10. ### thexy Thread Starter Member

Dec 13, 2015
130
0
How to stimulate circuit?

11. ### shteii01 AAC Fanatic!

Feb 19, 2010
4,490
710
You can download Pspice 9.1 Student Edition or some such, it is old, it has been around for the past 15 or 18 years, it is about 65 MB, I used to carry it on 100 MB zip disk. Or you can download LTspice, it is free and is current. I also have MultiSim 2001 that I got with my textbook.

You build the circuit in whatever simulator you have, place voltmeters and current meters into the circuit, let the simulation run. If the problem does not give values for the parts, choose your own, like use 1 kOhm resistors for R, R1, R2, R3, 10 nF for C1, C3, C4. Use something for U0, 5 volts or 10 volts or whatever you like.

Then apply the equation that you derived. Does the simulation result agrees with the result calculated from the formula that you derived?

12. ### shteii01 AAC Fanatic!

Feb 19, 2010
4,490
710
Here is simulation of your circuit.

As you can see R and R3 have 12 nanoVolts across them, for all practical reasons it is 0 volts. Applying Ohm's Law tells us that the only reason they can have 0 volts is when the current though them is 0 A too. The only way for current though them to be zero is when they are disconnected from the circuit. Which means that capacitors act as open circuits and are disconnecting the R and R3 from the rest of the circuit. Which is correct behavior for fully charged capacitors in a DC circuit.

The other two resistors show normal voltages. What we would expect for resistors connected to a DC voltage source. 13. ### WBahn Moderator

Mar 31, 2012
23,704
7,284
A long time after the switch is closed, things have settled down so that nothing is changing. This does not mean that everything is zero, just that whatever it is, it isn't changing. This is called steady state. For a capacitor if there is any current flowing in it then the voltage across it would have to be changing. Thus, in steady state, the current through all capacitors is zero. If the current through a component is zero, that means that it is behaving like an open circuit, which means that it is behaving just is if it weren't even there. So erase all of the capacitors from the circuit and see what you are left with. Do you still have a complete path from the source out into the circuit and back to the other side of the source? If so, then current can flow in those components. But it can't flow in any components that are not part of a complete circuit, so you can erase those components, too. What is left?