Why wouldn't activating the switch make a difference?

[EddiePhys]

Homework Statement

What is the heat generated on closing the switch?

Homework Equations

The Attempt at a Solution

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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?

 

[berkeman]

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

 

[EddiePhys]

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

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

 

[haruspex]

20-100/9 to its right

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

 

[EddiePhys]

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

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

 

[cnh1995]

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

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

 

[SammyS]

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

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

 

[EddiePhys]

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

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?

 

[cnh1995]

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?

Yes.

 

[EddiePhys]

Yes.

Okay. Thanks! :D