# How does this circuit work?Watch

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#1
The question (Jan 2011, 11.c) asks to explain the advantage of the circuit in Fig. 11.3 over the circuit in Fig. 11.2 (shown below). According to the mark scheme, 11.3 gives a greater range of p.d. from 0 to the maximum (12V). I understand how the circuit in 11.2 works, but I don't understand how the circuit in 11.3 works. Could someone explain this to me? Thanks!
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5 years ago
#2
(Original post by Chlorophile)
The question (Jan 2011, 11.c) asks to explain the advantage of the circuit in Fig. 11.3 over the circuit in Fig. 11.2 (shown below). According to the mark scheme, 11.3 gives a greater range of p.d. from 0 to the maximum (12V). I understand how the circuit in 11.2 works, but I don't understand how the circuit in 11.3 works. Could someone explain this to me? Thanks!
The potentiometer is just a potential divider and you can adjust the ratio. The output is the middle point between the two imaginary resistors.
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#3
(Original post by NightStrider)
The potentiometer is just a potential divider and you can adjust the ratio. The output is the middle point between the two imaginary resistors.
I don't understand how that works though. If the arrow is at the top of the potentiometer, then it's basically a parallel circuit and there should be an equal voltage across both the potentiometer and the lamp. If it's at the bottom of the potentiometer, then it's basically a series circuit and the potential difference is shared between the lamp and the potentiometer. So I still don't understand how the p.d. across the lamp can ever be 0.
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5 years ago
#4
I don't understand either :/
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5 years ago
#5
(Original post by Chlorophile)
I don't understand how that works though. If the arrow is at the top of the potentiometer, then it's basically a parallel circuit and there should be an equal voltage across both the potentiometer and the lamp. If it's at the bottom of the potentiometer, then it's basically a series circuit and the potential difference is shared between the lamp and the potentiometer. So I still don't understand how the p.d. across the lamp can ever be 0. This is essentially what the circuit will look like when the arrow is at the bottom. There is a path of no resistance, all the current will flow through there. If the current through the lamp is 0, V=IR and so pd is 0.
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#6
(Original post by NightStrider) This is essentially what the circuit will look like when the arrow is at the bottom. There is a path of no resistance, all the current will flow through there. If the current through the lamp is 0, V=IR and so pd is 0.
Thanks, that does make sense, but why does that happen? I thought the potential difference is the same across all branches.
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5 years ago
#7
(Original post by Chlorophile)
Thanks, that does make sense, but why does that happen? I thought the potential difference is the same across all branches.
PD is the same across the separate branch on the bottom. You can consider those two resistors, and because there is only one component all the voltage drop will occur across that.
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5 years ago
#8
(Original post by Chlorophile)
Thanks, that does make sense, but why does that happen? I thought the potential difference is the same across all branches.
I think the main advantage of using a potential divider rather than a variable resistor is due to the fact that you can get 0v across the component. There isn't a limitation as to how low the potential difference you can achieve across the component. With a variable resistor, the resistor is always going to have resistance (that is if it isn't a super conductor with zero resistivity) hence there is always going to be a potential difference across it and hence always a pd across the compnent.
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#9
(Original post by Protoxylic)
I think the main advantage of using a potential divider rather than a variable resistor is due to the fact that you can get 0v across the component. There isn't a limitation as to how low the potential difference you can achieve across the component. With a variable resistor, the resistor is always going to have resistance (that is if it isn't a super conductor with zero resistivity) hence there is always going to be a potential difference across it and hence always a pd across the compnent.
I know that that's the advantage, the issue is that I don't understand how it's possible to get 0v across the component in the potential divider circuit.
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5 years ago
#10
(Original post by Chlorophile)
I know that that's the advantage, the issue is that I don't understand how it's possible to get 0v across the component in the potential divider circuit.
The lamp is connected to the wiper of the potentiometer.

When the wiper is at the lowest point then the lamp is effectively connected at both ends to the same point in the circuit (neglecting the very small resistances of the connecting leads).

If both ends of the lamp are connected to the same point, they must also both be at the same voltage potential.

i.e. the difference in potential across the lamp (potential difference) must be zero.

It matters not if the voltage potential wrt some other point in the circuit is 10V or 10,000,000V. The difference in potential between the ends of the lamps will still be 10V -10V = 0; or 10,000,000V - 10,000,000V = 0.
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#11
(Original post by uberteknik)
The lamp is connected to the wiper of the potentiometer.

When the wiper is at the lowest point then the lamp is effectively connected at both ends to the same point in the circuit (neglecting the very small resistances of the connecting leads).

If both ends of the lamp are connected to the same point, they must also both be at the same voltage potential.

i.e. the difference in potential across the lamp (potential difference) must be zero.

It matters not if the voltage potential wrt some other point in the circuit is 10V or 10,000,000V. The difference in potential between the ends of the lamps will still be 10V -10V = 0; or 10,000,000V - 10,000,000V = 0.
Oh great, that makes sense. Thanks!
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