# Circuits Problem

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#1
I have been trying to solve this problem:

https://isaacphysics.org/questions/m...5-ebf80dc47798

However I am not sure how to use the information they have given me... Can I combine the batteries? I am not sure how to start...

Can anyone give me a hint?
0
3 years ago
#2
I can't do this question or the other level 4 one either XD.
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#3
I can't do this question or the other level 4 one either XD.
Yeah same. I always have difficulties with potential divider/two batteries problems...What are your ideas so far?
0
3 years ago
#4
(Original post by FXLander)
Yeah same. I always have difficulties with potential divider/two batteries problems...What are your ideas so far?
No idea, I normally watch the video and solve the equations but that isn't an option here. Issac physics has made me realize how bad at circuits I am XD.
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#5
No idea, I normally watch the video and solve the equations but that isn't an option here. Issac physics has made me realize how bad at circuits I am XD.
Yeah I have a really bad understanding of circuits 0
3 years ago
#6
(Original post by FXLander)
Yeah I have a really bad understanding of circuits Yeah I can do the normal problems in the AS exams but not these harder questions.
0
3 years ago
#7
(Original post by FXLander)
I have been trying to solve this problem:

https://isaacphysics.org/questions/m...5-ebf80dc47798

However I am not sure how to use the information they have given me... Can I combine the batteries? I am not sure how to start...

Can anyone give me a hint?
hi. use the formula of resistivity to get the resistance at which R could be in both cases. clue: resistivity is not resistance.
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#8
(Original post by ibhelper)
hi. use the formula of resistivity to get the resistance at which R could be in both cases. clue: resistivity is not resistance.
If I work out for example which resistance does that represent? Is that the unknown R?
0
3 years ago
#9
(Original post by FXLander)
I have been trying to solve this problem:

https://isaacphysics.org/questions/m...5-ebf80dc47798

However I am not sure how to use the information they have given me... Can I combine the batteries? I am not sure how to start...

Can anyone give me a hint?
I can't do this question or the other level 4 one either XD.
At position B, the ammeter reads zero which means no current flows through it. i.e. the potential difference across the ammeter must also be zero.

The only way this can happen is when the resistance ratios of the two potential dividers (100 and unknown resistor is one divider, the other is the linear potentiometer) are in exactly the same ratios.

This means the 400mm length of potentiometer track resistance equates to the 100 ohms reference resistor.

588/400 = (100 + x)/100

x = ?

Spoiler:
Show

47.0 ohms
0
#10
(Original post by uberteknik)
At position B, the ammeter reads zero which means no current flows through it. i.e. the potential difference across the ammeter must also be zero.

The only way this can happen is when the resistance ratios of the two potential dividers (100 and unknown resistor is one divider, the other is the linear potentiometer) are in exactly the same ratios.

This means the 400mm length of potentiometer track resistance equates to the 100 ohms reference resistor.

588/400 = (100 + x)/100

x = ?

Spoiler:
Show

47.0 ohms

I would like some clarification on a couple of things:

1. Ratio of lengths = Ratio of Resistances. Is this always true?
2. "The only way this can happen is when the resistance ratios of the two potential dividers (100 and unknown resistor is one divider, the other is the linear potentiometer) are in exactly the same ratios." How does one know this?(I think if they weren't equal then there would be a p.d. between the any the the potential dividers and the resistors so there would be a current flowing through the ammeter, which can't be true, however I am not sure whether this explanation is valid.)
0
3 years ago
#11
(Original post by FXLander)
I would like some clarification on a couple of things:

1. Ratio of lengths = Ratio of Resistances. Is this always true?
As long as the resistivity and cross sectional area of the resistive material is constant throughout the length of the potentiometer, then yes, the ratio of the lengths will be equal to the ratio of resistances.

(Original post by FXLander)
2. "The only way this can happen is when the resistance ratios of the two potential dividers (100 and unknown resistor is one divider, the other is the linear potentiometer) are in exactly the same ratios." How does one know this?(I think if they weren't equal then there would be a p.d. between the any the the potential dividers and the resistors so there would be a current flowing through the ammeter, which can't be true, however I am not sure whether this explanation is valid.)
You are correct.

For the experiment to work, the potentiometer power supply, must produce a voltage equal to or greater than that of the supply voltage across the series combination of the reference resistor and the unknown resistor.

The actual resistance or resistivity of the potentiometer does not matter (provided it complies with my answer above). It's the ratio of the two lengths (and thus the ratio of resistances set up with the potentiometer) where the ammeter reading is zero that matters.

The potentiometer wiper arm can now be varied such that the p.d. picked off the potentiometer, will match exactly the p.d.'s at the test positions B and C. i.e. no p.d. is developed across the ammeter because the p.d's picked off on either side are exactly the same.

Under these conditions, the ratio of resistances either side of the ammeter must be the same. And if the resistivity of the potentiometer is constant, then the lengths measured must also be in the same ratio as the resistances under test.
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