How Does a Switch Affect Voltage and Current in a Circuit?

[Eitan Levy]

Homework Statement

This circuit is given:
Also: V_C=18, V_D=0[/B]

When the switch is open, what is the voltage AB?
When the switch is closed, what is V_B?

Homework Equations

The Attempt at a Solution

I am completely stuck, I don't really know how the switch affects the circuit. I think that maybe between two points there is no current, and therefore no voltage between them and A/B. Maybe the current from C goes only to the capacitor and not to the resistor? I don't really know, any help would be appreciated. [/B]

 

[gneill]

Also: VC=18, VD=0

Are these meant to be defining the starting potential at node C and a reference node as node D? Since the diagram shows no voltage source that would maintain node C at a given value, I suppose one is to assume that node C just happens to have that given starting potential with respect to node D.

I am completely stuck, I don't really know how the switch affects the circuit. I think that maybe between two points there is no current, and therefore no voltage between them and A/B. Maybe the current from C goes only to the capacitor and not to the resistor? I don't really know, any help would be appreciated.

With the switch open it can be completely ignored; Redraw your circuit without the switch branch. Then ask yourself how the top node (C) can be at 18 V with respect to node (D). What must the potentials across the capacitors be for this to be the case? If V_C is constant (steady state) in this situation, what can you say about currents in the circuit?

When the switch is closed, a short circuit is formed from A to B. Take a look at the loops that are formed and choose one for a closer look. What will happen to the capacitor charge?

The question does not make it explicit whether they are looking for the steady state solution for V_B (a "long time" after the switch is closed), or the transient solution that occurs immediately after the switch closes.

 

[jbriggs444]

As presented here, the question is not well posed. I see two difficulties.

1. The circuit is floating -- it has no ground. A better posed question would either include a grounding wire before specifying that V_D = 0 or would specify only the voltage difference between A and D and then ask for the result as a potential difference from D.

2. The question does not make it clear whether the asked-for result is for the instant the switch is closed, for the steady state that is eventually achieved or whether one is expected to produce the voltage as a function of time after the switch is closed.

Let's fix up the easy one first. Assume that point D is grounded. The reference potential at D will always be zero.

Now then. Let's get you started. Leave the switch open for the moment. The circuit is as given, all charged up. [Assume that someone put an 18 volt power supply on it for a long while, let the circuit stabilize and then disconnected the leads]

Question: Is any current flowing at the moment?

[Edit: @gneill beat me to most of the same points]

 

[Eitan Levy]

Are these meant to be defining the starting potential at node C and a reference node as node D? Since the diagram shows no voltage source that would maintain node C at a given value, I suppose one is to assume that node C just happens to have that given starting potential with respect to node D.With the switch open it can be completely ignored; Redraw your circuit without the switch branch. Then ask yourself how the top node (C) can be at 18 V with respect to node (D). What must the potentials across the capacitors be for this to be the case? If V_C is constant (steady state) in this situation, what can you say about currents in the circuit?

When the switch is closed, a short circuit is formed from A to B. Take a look at the loops that are formed and choose one for a closer look. What will happen to the capacitor charge?

The question does not make it explicit whether they are looking for the steady state solution for V_B (a "long time" after the switch is closed), or the transient solution that occurs immediately after the switch closes.

Alright, thank you.
I think that at the beginning, the current will go to the capacitors until they are "full", and then there won't be any current no more. So V_a=18V and V_b=0.
After that, the current will be able to go in CABD, so the current will be 2A, and then we can easily see that V_A=V_B=6V. It gives the correct answer, however, what I don't understand is how the current will be able to go from A to B if there is no voltage between them?

 

[Eitan Levy]

As presented here, the question is not well posed. I see two difficulties.

1. The circuit is floating -- it has no ground. A better posed question would either include a grounding wire before specifying that V_D = 0 or would specify only the voltage difference between A and D and then ask for the result as a potential difference from D.

2. The question does not make it clear whether the asked-for result is for the instant the switch is closed, for the steady state that is eventually achieved or whether one is expected to produce the voltage as a function of time after the switch is closed.

Let's fix up the easy one first. Assume that point D is grounded. The reference potential at D will always be zero.

Now then. Let's get you started. Leave the switch open for the moment. The circuit is as given, all charged up. [Assume that someone put an 18 volt power supply on it for a long while, let the circuit stabilize and then disconnected the leads]

Question: Is any current flowing at the moment?

[Edit: @gneill beat me to most of the same points]

I think there isn't, because it doesn't have anywhere to go?

 

[jbriggs444]

I think there isn't, because it doesn't have anywhere to go?

Right. Which, as you correctly concluded above means that V_a = 18V and V_b = 0V before the switch is closed.

And you have then correctly concluded that the new flow pattern after the switch closes is C to A to B to D.

Based on the "correct" answer, I would conclude that the intended interpretation is that a power supply is continuously connected so that the V_C = 18V and V_D condition is maintained. At the moment the switch is thrown, momentary and infinite current flows as the capacitors discharge to the new equilibrium condition and the circuit stabilizes in its new flow pattern.

 

[Eitan Levy]

Right. Which, as you correctly concluded above means that V_a = 18V and V_b = 0V before the switch is closed.

And you have then correctly concluded that the new flow pattern after the switch closes is C to A to B to D.

Based on the "correct" answer, I would conclude that the intended interpretation is that a power supply is continuously connected so that the V_C = 18V and V_D condition is maintained. At the moment the switch is thrown, momentary and infinite current flows as the capacitors discharge to the new equilibrium condition and the circuit stabilizes in its new flow pattern.

Thank you, however I still don't understand how the charge is able to flow from A to B if the voltage is 0.

 

[jbriggs444]

Thank you, however I still don't understand how the charge is able to flow from A to B if the voltage is 0.

It is an idealization.

With zero voltage and zero resistance, the amount of current that flows is 0/0 = undefined. Any current flow is consistent with that condition. If you find undefined quantities offensive (a reasonable position), then consider that any real switch will have some small finite resistance. So there will be some small finite voltage required to push the current across.

If the resistance is very small and the voltage drop is very small, we can ignore them both and pretend that it is just an ideal wire with zero resistance carrying non-zero current with a zero voltage drop. Hence the idealization.

 

[Eitan Levy]

It is an idealization.

With zero voltage and zero resistance, the amount of current that flows is 0/0 = undefined. Any current flow is consistent with that condition. If you find undefined quantities offensive (a reasonable position), then consider that any real switch will have some small finite resistance. So there will be some small finite voltage required to push the current across.

If the resistance is very small and the voltage drop is very small, we can ignore them both and pretend that it is just an ideal wire with zero resistance carrying non-zero current with a zero voltage drop. Hence the idealization.

Oh, my teacher never mentioned this. Thanks!