# AS Physics Electrical Circuit Help Watch

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#2

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Could somebody please explain part (ii) please?

**Jpw1097**)Could somebody please explain part (ii) please?

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(Original post by

What don't you understand? It's usually helpful if you say what you've already done for the question.

**lerjj**)What don't you understand? It's usually helpful if you say what you've already done for the question.

But the mark scheme says that when V is large, R is large which I understand, but it also says that V=e.m.f when R is infinite. I don't understand how that has come about

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#4

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Well, I've already stated that as current increases, voltage decreases due to resistance decreases.

But the mark scheme says that when V is large, R is large which I understand, but it also says that V=e.m.f when R is infinite. I don't understand how that has come about

**Jpw1097**)Well, I've already stated that as current increases, voltage decreases due to resistance decreases.

But the mark scheme says that when V is large, R is large which I understand, but it also says that V=e.m.f when R is infinite. I don't understand how that has come about

What happens to how the voltage splits when you change the resistance of the variable resistor?

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(Original post by

Ok, so the situation is basically a potential divider and the voltage is being split between the variable resistor and the internal resistance of the cell.

What happens to how the voltage splits when you change the resistance of the variable resistor?

**lerjj**)Ok, so the situation is basically a potential divider and the voltage is being split between the variable resistor and the internal resistance of the cell.

What happens to how the voltage splits when you change the resistance of the variable resistor?

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#6

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I understand that, but it specifically states that V is also the terminal p.d.

**Jpw1097**)I understand that, but it specifically states that V is also the terminal p.d.

That's not the same as the emf, which is a little larger because of the voltage drop due to the internal resistance. As you make the variable resistance go to infinity, the current goes to zero and the lost volts from the internal resistance also goes to zero- making the terminal p.d. equal the emf.

Does that make sense?

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(Original post by

It is, the only resistance you've got connected is the variable resistor, hence the resistance across it must be the terminal p.d.

That's not the same as the emf, which is a little larger because of the voltage drop due to the internal resistance. As you make the variable resistance go to infinity, the current goes to zero and the lost volts from the internal resistance also goes to zero- making the terminal p.d. equal the emf.

Does that make sense?

**lerjj**)It is, the only resistance you've got connected is the variable resistor, hence the resistance across it must be the terminal p.d.

That's not the same as the emf, which is a little larger because of the voltage drop due to the internal resistance. As you make the variable resistance go to infinity, the current goes to zero and the lost volts from the internal resistance also goes to zero- making the terminal p.d. equal the emf.

Does that make sense?

Also, how is there a p.d. across the terminals if there is no current flowing?

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#8

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Doesn't the terminal p.d. + p.d. across the variable resistor = e.m.f though?

Also, how is there a p.d. across the terminals if there is no current flowing?

**Jpw1097**)Doesn't the terminal p.d. + p.d. across the variable resistor = e.m.f though?

Also, how is there a p.d. across the terminals if there is no current flowing?

So if the resistance placed across the supply terminals approaches infinity, the current flowing must also reduce to approach zero. Which means that as the load resistance tends to infinity, the potential lost across the (fixed) internal resistance will also drop to approach zero (no loss).

Taken to it's conclusion, with an infinite resistance across the supply terminals (open circuit) no volts will be lost across the internal resisitance and hence the voltage appearing at the terminals must be that of the supply e.m.f. N.B. this is a logical and mathematical deduction.

In reality, the measuring device (voltmeter) has a very high but finite resistance and hence a very small current must still flow. However, the current is so small (<nano amps) that the voltage dropped (lost potential) across the internal resistance is equally negligible and can be ignored for most useful purposes. The voltage measured is therefore so close to the e.m.f. of the supply that losses in the internal resistance can be safely ignored.

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[QUOTE=uberteknik;54414021]You are right, the voltages (potential energy) dropped around a circuit must sum to the supply voltage in each and every case.

So if the resistance placed across the supply terminals approaches infinity, the current flowing must also reduce to approach zero. Which means that as the load resistance tends to infinity, the potential lost across the (fixed) internal resistance will also drop to approach zero (no loss).

Taken to it's conclusion, with an infinite resistance across the supply terminals (open circuit) no volts will be lost across the internal resisitance and hence the voltage appearing at the terminals must be that of the supply e.m.f. N.B. this is a logical and mathematical deduction.

In reality, the measuring device (voltmeter) has a very high but finite resistance and hence a very small current must still flow. However, the current is so small (

You say that as the internal resistance approaches infinity there will be no lost volts across it....but how, since if there is an infinite resistance and therefore no current?

Also, the question says that terminal p.d. = the p.d. across the variable resistor.....so wouldn't that mean that they are always the same??

So if the resistance placed across the supply terminals approaches infinity, the current flowing must also reduce to approach zero. Which means that as the load resistance tends to infinity, the potential lost across the (fixed) internal resistance will also drop to approach zero (no loss).

Taken to it's conclusion, with an infinite resistance across the supply terminals (open circuit) no volts will be lost across the internal resisitance and hence the voltage appearing at the terminals must be that of the supply e.m.f. N.B. this is a logical and mathematical deduction.

In reality, the measuring device (voltmeter) has a very high but finite resistance and hence a very small current must still flow. However, the current is so small (

You say that as the internal resistance approaches infinity there will be no lost volts across it....but how, since if there is an infinite resistance and therefore no current?

Also, the question says that terminal p.d. = the p.d. across the variable resistor.....so wouldn't that mean that they are always the same??

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#10

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You say that as the internal resistance approaches infinity there will be no lost volts across it....but how, since if there is an infinite resistance and therefore no current?

Also, the question says that terminal p.d. = the p.d. across the variable resistor.....so wouldn't that mean that they are always the same??

**Jpw1097**)You say that as the internal resistance approaches infinity there will be no lost volts across it....but how, since if there is an infinite resistance and therefore no current?

Also, the question says that terminal p.d. = the p.d. across the variable resistor.....so wouldn't that mean that they are always the same??

The current will fall but, more and more p.d. will develop across the increasing load resistance and less will appear across the internal resistance.

The battery can be modelled by a voltage source (E) in series with a resistor (r) representing its internal resistance.

i.e. the battery terminals will then be as shown by the dotted boundary lines in the diagram.

This is identical to a potential divider:

Where R1 represents the battery internal resistance and R2 represents the load resistance.

If R1 remains constant and R2 increases towards infinity, the equation for the p.d. developed across the load shows that

This represents the asymptotic limit as the current falls to zero when the load resistance approaches infinity.

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#11

(Original post by

If R1 remains constant and R2 increases towards infinity, the equation for the p.d. developed across the load shows that

This represents the asymptotic limit as the current falls to zero when the load resistance approaches infinity.

**uberteknik**)If R1 remains constant and R2 increases towards infinity, the equation for the p.d. developed across the load shows that

This represents the asymptotic limit as the current falls to zero when the load resistance approaches infinity.

?

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(Original post by

Not the internal resistance (that can be assumed to stay constant). It's the case of when the load resistance approaches infinity.

The current will fall but, more and more p.d. will develop across the increasing load resistance and less will appear across the internal resistance.

The battery can be modelled by a voltage source (E) in series with a resistor (r) representing its internal resistance.

i.e. the battery terminals will then be as shown by the dotted boundary lines in the diagram.

This is identical to a potential divider:

Where R1 represents the battery internal resistance and R2 represents the load resistance.

If R1 remains constant and R2 increases towards infinity, the equation for the p.d. developed across the load shows that

This represents the asymptotic limit as the current falls to zero when the load resistance approaches infinity.

**uberteknik**)Not the internal resistance (that can be assumed to stay constant). It's the case of when the load resistance approaches infinity.

The current will fall but, more and more p.d. will develop across the increasing load resistance and less will appear across the internal resistance.

The battery can be modelled by a voltage source (E) in series with a resistor (r) representing its internal resistance.

i.e. the battery terminals will then be as shown by the dotted boundary lines in the diagram.

This is identical to a potential divider:

Where R1 represents the battery internal resistance and R2 represents the load resistance.

If R1 remains constant and R2 increases towards infinity, the equation for the p.d. developed across the load shows that

This represents the asymptotic limit as the current falls to zero when the load resistance approaches infinity.

I understand that as load resistance approaches infinity, the voltage across the resistor will approach the e.m.f and so terminal p.d.=0. But I just don't understand why it says terminal p.d. = V, that is all.

And thanks!

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#14

(Original post by

I thought that initially, it's just the wording of the question confused me since it said the terminal p.d. = the voltage across the variable resistor. Could somebody please explain what the question means by that.

**Jpw1097**)I thought that initially, it's just the wording of the question confused me since it said the terminal p.d. = the voltage across the variable resistor. Could somebody please explain what the question means by that.

The terminal voltage simply means the voltage appearing at the electrodes (+ve and -ve terminals) of the battery. The question wording means R2 (which is a variable load resistor) is placed across the cell electrodes.

(Original post by

I understand that as load resistance approaches infinity, the voltage across the resistor will approach the e.m.f and so

**Jpw1097**)I understand that as load resistance approaches infinity, the voltage across the resistor will approach the e.m.f and so

**terminal p.d.=0**. But I just don't understand why it says terminal p.d. = V, that is all.The p.d. developed across the internal resistance approaches 0.

The terminal p.d. approaches the e.m.f. of the cell.

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(Original post by

Look in the smaller diagram I posted above. Notice the dotted lines showing the outer casing of the cell (battery).

The terminal voltage simply means the voltage appearing at the electrodes (+ve and -ve terminals) of the battery. The question wording means R2 (which is a variable load resistor) is placed across the cell electrodes.

No, the terminal p.d. does not = 0.

The p.d. developed across the internal resistance approaches 0.

The terminal p.d. approaches the e.m.f. of the cell.

**uberteknik**)Look in the smaller diagram I posted above. Notice the dotted lines showing the outer casing of the cell (battery).

The terminal voltage simply means the voltage appearing at the electrodes (+ve and -ve terminals) of the battery. The question wording means R2 (which is a variable load resistor) is placed across the cell electrodes.

No, the terminal p.d. does not = 0.

The p.d. developed across the internal resistance approaches 0.

The terminal p.d. approaches the e.m.f. of the cell.

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