Thanks for the reply. But Im not sure how to answer part B to your question. I get your point across the battery, so A would be 5V correct? I need some more help with the B

Notice that Q1 did NOT ask for the voltage at A, it asked for the voltage at A relative to the voltage at C. This is a very important distinction.

I think part of the problem is that you are making a common mistake about what "voltage" means.

A voltage is ALWAYS a DIFFERENCE between potential energies. So when we say that the voltage AT a point is 12V, that ONLY makes sense if we have an agreed upon reference point AND it is understood that what we are really saying is that the voltage at that point is 12V greater than the voltage at our reference point.

This is something we do all the time in real life. When we talk about a building being 300 ft tall, we are talking about the difference in height between the top of the building and the ground upon which it sits. While if we talk about a mountain being 14,000 ft tall, we are talking about the difference in height between the top of the mountain and mean sea level. If we mean anything else, say the height of a mountain relative to the surrounding plains, then we have to be more explicit.

In circuits, we pretty much always need to be more explicit. So, by convention, we choose one node in the circuit as our reference point and declare that node to be 0V. This is called the "common" node (often called "ground", though this is technically something different) and we place a special symbol (a "common" or a "ground" symbol) there to indicate this choice.

Now we can talk about the voltages at the other nodes because we have an agreed upon reference.

So the answer to Q1 is 5V, but the voltage at Node A is only 5V if we happen to choose Node C as the reference node (or, if it should turn out that Node C just ends up happening to be equal to 0V relative to whatever reference point we choose).

We could choose Node A as our reference point, in which case the voltage at node A would be (by definition) 0V and the voltage at node C would be -5V.

When all else fails, fall back on the definitions.

Putting a battery of voltage Vbatt between nodes X and Y (positive terminal on X) means that

Vx - Vy = Vbatt

Similarly, connecting a resistor between nodes X and Y means that

Vx - Vy = Ixy * R

where Ixy is the current flowing from X to Y through R.

See if that helps you at all.

 

For Part B.
You have just closed the switch. The capacitor is charging, which makes it a short circuit. So you have 5 V from battery and a voltage drop across resistor. To me, Voltage from A to B=5 volts + IR. If I stick a voltmeter from A to B, I will see voltage from the battery plus voltage across resistor.

Just because a capacitor is charging does NOT make it a short circuit.

 

Just because a capacitor is charging does NOT make it a short circuit.

Except at t = 0, which I was thinking was mentioned in the problem.

 

Just because a capacitor is charging does NOT make it a short circuit.

That is true. But we are talking about text book problem, and it looks to me from 200 level college course. In this cases, in my classes, we assumed two things, the capacitor is fully discharged and in the instant t=0 when switch is closed it is short circuit. The other side of these assumptions is that full charged capacitor is an open circuit.

 

Except at t = 0, which I was thinking was mentioned in the problem.

Even at t=0 that is not the reason that the voltage across the capacitor is zero.

 

That is true. But we are talking about text book problem, and it looks to me from 200 level college course. In this cases, in my classes, we assumed two things, the capacitor is fully discharged and in the instant t=0 when switch is closed it is short circuit. The other side of these assumptions is that full charged capacitor is an open circuit.

Text book problem or not, a capacitor does not look like a short just because it is charging. It looks like a short because it is fully discharged, regardless of whether it is charging or not. And if it is charging, it only looks like a short if it also happens to be fully discharged. Charging and being fully discharged are NOT synonomous -- they are two very different and separate things. So saying that "The capacitor is charging, which makes it a short circuit" is simply wrong and is very likely to lead the person into making a misconception about how to go about solving these problems.

 

Text book problem or not, a capacitor does not look like a short just because it is charging. It looks like a short because it is fully discharged, regardless of whether it is charging or not. And if it is charging, it only looks like a short if it also happens to be fully discharged. Charging and being fully discharged are NOT synonomous -- they are two very different and separate things. So saying that "The capacitor is charging, which makes it a short circuit" is simply wrong and is very likely to lead the person into making a misconception about how to go about solving these problems.

Ok. Thank you.

 

Just for the sake of the problem, the homework hints is that "at T=0 the capacitor can be modeled as a short"

 

And for C: "Hint: at T=∞ the capacitor can be modeled as a open"

 

A voltage is ALWAYS a DIFFERENCE between potential energies.

I think we should be a little careful here.

I am not convinced that this is what the OP is being taught.

a) What is the voltage difference between node A and node B when T<0?

There are two quantities in physics measured in volts.

EMF and Potential Difference

They are not the same. Understanding that there is this difference and what that difference is causes a great deal of confusion, especially when both are colloquially referred to as 'voltage'.

 

Even at t=0 that is not the reason that the voltage across the capacitor is zero.

Yes, that was my mistake. That was under the assumption that the capacitor was fully discharged when the switch was thrown.

 

I think we should be a little careful here.

I am not convinced that this is what the OP is being taught.



There are two quantities in physics measured in volts.

EMF and Potential Difference

They are not the same. Understanding that there is this difference and what that difference is causes a great deal of confusion, especially when both are colloquially referred to as 'voltage'.

If he's not being taught that, then what IS he being taught.

As far as I know, even in a non-conservative field EMF and Potential Difference are still intrinsically differences. The only difference is that in a non-conservative electric field the difference becomes path dependent and so you can have the voltage difference between Point A and Point A be non-zero (i.e., KVL doesn't hold in non-conservative fields).

 

Okay so let me take another shot:

a) What is the voltage difference, stored charge, and current at T < 0?

Voltage diff = Vbatt = 5V

Stored charge = 0, assuming this capacitor has never been charged before (?) or if it has, the charge has bled out over the "long time"

Current = 0, since it is an open circuit

b) What is the voltage difference, stored charge, and current at T = 0?

Voltage diff = Vbatt = 5V

Stored charge = 0, since no time as elapsed for charge to be stored (?) Or is it some really small positive number since it has started to accumulate charge?

Current = V/R = 5V/50 ohms = 0.1 Amps

c) What is the voltage difference, stored charge, and current at T --> ∞ 0?

Voltage diff = Vbatt = 5V

Stored charge = approaches 50 coulombs (Q = VC)

Current = V/R = 5V/50 ohms = 0.1 Amps (not sure about this one)

 

c) What is the voltage difference, stored charge, and current at T --> ∞ 0?

Voltage diff = Vbatt = 5V

Stored charge = approaches 50 coulombs (Q = VC)

Current = V/R = 5V/50 ohms = 0.1 Amps (not sure about this one)

For Part C I think current will be zero. A charged capacitor act as an open circuit, so the loop is broken, there is no current.

 

Just for the sake of the problem, the homework hints is that "at T=0 the capacitor can be modeled as a short"

So for this reason the stored charge at the capacitor = 0, correct?

 

So for this reason the stored charge at the capacitor = 0, correct?

That is how I see it. We are looking at capacitor at an instant in time. Also my textbook in several places points out that capacitors do not change instantly! So at T<0, capacitor is discharged, charge=0. At T=0, at that instant, the capacitor has not had time to change, charge still is zero.

At least that is how I would answer those questions.

 

So for this reason the stored charge at the capacitor = 0, correct?

Yes. The hint says that the cap, at t=0, can be viewed as a short. A short can have any current in it but it has zero volts across it. Thus, this hint tells you nothing about the current in it, but it DOES tell you that the charge on the capacitor is zero since V=CQ.

 

Okay so let me take another shot:

a) What is the voltage difference, stored charge, and current at T < 0?

Voltage diff = Vbatt = 5V

Indirectly, yes. The better way to show this would be something like the following:

Vab = Va - Vb = Vac + Vca
Vca = Vbatt
Vac = Iac*R

Vab = Vbatt + Iac*R

This is a general result. It is always true for this circuit. Now all you have to do is find Iac.

Stored charge = 0, assuming this capacitor has never been charged before (?) or if it has, the charge has bled out over the "long time"

This is a very reasonable way to justify the necessary assumption.

Current = 0, since it is an open circuit

b) What is the voltage difference, stored charge, and current at T = 0?

Voltage diff = Vbatt = 5V

Stored charge = 0, since no time as elapsed for charge to be stored (?) Or is it some really small positive number since it has started to accumulate charge?

Current = V/R = 5V/50 ohms = 0.1 Amps

Now take a look back at the general equation developed before. If there is any current, is Vab still equal to Vbatt?

Also, be sure to note that the equation uses Iac, which is very specifically the current flowing from Node A to Node C through the resistor and NOT the current flowing from Node C to Node A (i.e., polarity matters).

c) What is the voltage difference, stored charge, and current at T --> ∞ 0?

Voltage diff = Vbatt = 5V

Stored charge = approaches 50 coulombs (Q = VC)

Current = V/R = 5V/50 ohms = 0.1 Amps (not sure about this one)

The notion of the behavior as time goes to infinity implies steady state operation. This does NOT mean no voltage or no current, only that whatever voltages or currents do exist are not changing. But now you need to consider that if there is any current in a capacitor that it's voltage must be changing as a consequence.

 

Now take a look back at the general equation developed before. If there is any current, is Vab still equal to Vbatt?

Also, be sure to note that the equation uses Iac, which is very specifically the current flowing from Node A to Node C through the resistor and NOT the current flowing from Node C to Node A (i.e., polarity matters).

No, Vab is not equal to Vbatt. So Vab = 5V + Iac*R, but I'm not sure how to calculate that voltage (more specifically, what the current is). My guys would be 0.1 amps, but I'm assuming Ohm's Law

The notion of the behavior as time goes to infinity implies steady state operation. This does NOT mean no voltage or no current, only that whatever voltages or currents do exist are not changing. But now you need to consider that if there is any current in a capacitor that it's voltage must be changing as a consequence.

I'm kind of confused, was I wrong in my answers then? Or the logic

 

Stored charge = approaches 50 coulombs (Q = VC)

Take care with your units.

Does 5 x 10 x .000001 = 50?

I have avoided commenting on the problem before as several are already helping, but here's a tip.

It is a good idea to redraw the circuit in each condition (T<0; T=0; T=∞), replacing the capacitor with either a short or open as hinted in your original material.