# Why wouldn't activating the switch make a difference?

1. Aug 12, 2017 at 12:38 PM

### EddiePhys

1. The problem statement, all variables and given/known data

What is the heat generated on closing the switch?
2. Relevant equations

3. The attempt at a solution

On closing the switch, the potential difference across each capacitor should change which would cause it to charge and hence cause heat loss through the capacitors.

But there is no lots of heat. Why?

2. Aug 12, 2017 at 12:43 PM

### Staff: Mentor

Before the switch is closed, what are the voltages at all of the nodes?

3. Aug 12, 2017 at 10:29 PM

### EddiePhys

20 to the left of 4uF, 20-100/9 to its right and 0 to the right of 5uF.
I don't know how to do the calculations once the switch is closed

4. Aug 13, 2017 at 12:50 AM

### haruspex

How do you arrive at that? Is there a current flowing?

5. Aug 13, 2017 at 2:11 AM

### EddiePhys

Initially, yes, while the capacitors were getting charged(before reaching steady state)

6. Aug 13, 2017 at 2:48 AM

### cnh1995

What about the steady state? What are the voltages across the two capacitors?

7. Aug 13, 2017 at 8:42 AM

### SammyS

Staff Emeritus
What is the steady state prior to the switch being closed, after the 4μF capacitor is charged ?

8. Aug 13, 2017 at 9:42 AM

### EddiePhys

I think I got it.
In steady state before the switch is closed, since no current is flowing, the potential difference across the 5uF capacitor is zero and the entire potential difference appears across the 4uF capacitor.

Once the switch is closed, it makes no difference i.e no current would flow because the part of the circuit connected to the negative terminal is already at -20V. i.e the entire potential will still appear across the 4uf capacitor

Is this correct?

9. Aug 13, 2017 at 9:43 AM

### cnh1995

Yes.

10. Aug 13, 2017 at 9:46 AM

### EddiePhys

Okay. Thanks! :D