Theory of Electric Circuit: Potential Difference and Internal Resistance

[songoku]

Homework Statement

1. The diagram shows a simple circuit.

Which statement is correct?
a. When switch S is closed, the potential difference across the battery falls because work is done against the internal resistance of the battery
b. When switch S is closed, the potential difference across the battery falls because work is done against the resistance R.

2. A cell of e.m.f E and internal resistance r is connected in series with a switch S and an external resistor R.

The p.d. between P and Q is V. When S is closed,
a. V decreases because there is a p.d. across R
b. V decreases because there is a p.d. across r

Homework Equations

The Attempt at a Solution

1. I answered a and wrong. The answer should be b
2. I answered b and wrong. The answer should be a

I think the potential difference across the battery will fall because there is internal resistance. Lest say if we measure the e.m.f to be 10 V, then the potential difference of the battery when the switch is closed will still 10 V if there is no internal resistance. If there is internal resistance, the potential difference measured will not be 10 V, maybe 9 V; the other 1 V goes to internal resistance. Am I wrong?

Thanks

 

[SammyS]

1. I answered a .
2. I answered b .

Your answers look right to me !

 

[crashdirty86]

1. I answered a and wrong. The answer should be b
2. I answered b and wrong. The answer should be a

I think the potential difference across the battery will fall because there is internal resistance. Lest say if we measure the e.m.f to be 10 V, then the potential difference of the battery when the switch is closed will still 10 V if there is no internal resistance. If there is internal resistance, the potential difference measured will not be 10 V, maybe 9 V; the other 1 V goes to internal resistance. Am I wrong?

Thanks

So, I see no internal resistance in the first diagram. This would lead me straight to answering b.

For the second the p.d. across R would cause a drop in V coming out of the battery. Even though the internal resistance would drop the immediate, its the p.d. that drops the V applied to the while circuit.

 

[songoku]

So, I see no internal resistance in the first diagram. This would lead me straight to answering b.

So if there is internal resistance, would the answer be a?

For the second the p.d. across R would cause a drop in V coming out of the battery. Even though the internal resistance would drop the immediate, its the p.d. that drops the V applied to the while circuit.

I don't quite get it. Why would the p.d. across R cause a drop in V coming out of the battery?

Assume e.m.f of battery = 10 V, p.d. across internal resistance is 1 V. So the value of V measured will be 9 V. This is the V coming out of the battery. The p.d. across R will also be 9 V.

You mean the value of V (which is 9 Volt) will decrease to value lower than 9 V because there is p.d. across R? The whole circuit here consists only of cell, internal resistor, and R

Thanks

 

[crashdirty86]

When you talk about a pd it is always some final minus initial, thus the potential difference between two points. So when you look at a pd across a battery with internal resistance, the pd will be that between the two points measured across the battery. Thus, when an EMF from a battery has a voltage drop, it would be due to some other source like the resistance found in your diagrams.

 

[crashdirty86]

I would also keep in mind that W=-qV=- change in electric potential energy.

 

[songoku]

When you talk about a pd it is always some final minus initial, thus the potential difference between two points. So when you look at a pd across a battery with internal resistance, the pd will be that between the two points measured across the battery.

I get this part

Thus, when an EMF from a battery has a voltage drop, it would be due to some other source like the resistance found in your diagrams.

Sorry I still don't get this part. I still see r as the cause of voltage drop.

When the switch is open, the value of V = 10 V.
a. When the switch is closed and there is no internal resistance, the value of V will still 10 V
b. When the switch is closed and there is internal resistance, the value of V will be lower than 10 V

Based on that, the cause of the drop is r not R. Am I wrong? Or maybe I misinterpret the question somehow?

Thanks

 

[CWatters]

It's a badly worded almost a trick question in my opinion.

Both answers a and b are correct and I would probably have answered b.

Answer a is also correct because "V" and the "pd across R" are the same (they are between the same two nodes).

Consider what happens if R is replaced by a voltage source. Clearly V would be independent of r in that case. I believe that's the point they are trying to make.

Consider what happens if R is replaced by a current source. V is then independent on r but it's still equal to the pd across the current source.

 

[CWatters]

What happens if R is replaced by a diode?

 

[songoku]

It's a badly worded almost a trick question in my opinion.

Both answers a and b are correct and I would probably have answered b.

Answer a is also correct because "V" and the "pd across R" are the same (they are between the same two nodes).

Consider what happens if R is replaced by a voltage source. Clearly V would be independent of r in that case. I believe that's the point they are trying to make.

Consider what happens if R is replaced by a current source. V is then independent on r but it's still equal to the pd across the current source.

What happens if R is replaced by a diode?

Still don't get it clearly. I'll think about it and borrow my friend's book for more reference.

Thanks a lot for all the help here