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So my teacher set this question to do for prep for next week but idk how to do it at all, any help wld be appreciated![Name: Screenshot 2021-01-18 at 17.47.16.png
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Last edited by firewithasoul; 1 month ago
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#2
For the first part: resistivity = R*A / L
The question has given you R/L, so all you need to do is multiply it by the area (A).
It says the diameter is 0.5mm, firstly convert that to metres. Then calculate the area: (pi x d^2) divided by 4.
Then just multiply your answer by R/L (6.30 ohms).
I’d guess the appropriate amount of significant figures is 3sf as all other values are quoted to that many
(I’ll have a look at the other bits in a min)
The question has given you R/L, so all you need to do is multiply it by the area (A).
It says the diameter is 0.5mm, firstly convert that to metres. Then calculate the area: (pi x d^2) divided by 4.
Then just multiply your answer by R/L (6.30 ohms).
I’d guess the appropriate amount of significant figures is 3sf as all other values are quoted to that many
(I’ll have a look at the other bits in a min)
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(Original post by 11lightsinthesky)
For the first part: resistivity = R*A / L
The question has given you R/L, so all you need to do is multiply it by the area (A).
It says the diameter is 0.5mm, firstly convert that to metres. Then calculate the area: (pi x d^2) divided by 4.
Then just multiply your answer by R/L (6.30 ohms).
I’d guess the appropriate amount of significant figures is 3sf as all other values are quoted to that many (I’ll have a look at the other bits in a min)
For the first part: resistivity = R*A / L
The question has given you R/L, so all you need to do is multiply it by the area (A).
It says the diameter is 0.5mm, firstly convert that to metres. Then calculate the area: (pi x d^2) divided by 4.
Then just multiply your answer by R/L (6.30 ohms).
I’d guess the appropriate amount of significant figures is 3sf as all other values are quoted to that many
(I’ll have a look at the other bits in a min)
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#4
(Original post by firewithasoul)
thank you so much!
thank you so much!
•Power dissipated is energy lost per second
•This energy is lost (dissipated) as thermal energy in the wire.
•The fixed resistor reduces the potential difference across the wire, therefore reducing the current in the wire.
•This means that the resistance is reduced in the wire, thus reducing the thermal energy emitted.
•Overall reducing power dissipated in the wire.
Now that’s probably over kill but I write all my answers in bullet points (even in the exam) so I can logically sort through my response
For the last bit: you aren’t given any length. It says a larger diameter but doesn’t give a value, so all we know is the cross-sectional area is greater than before. There’s more room in the wire meaning the electron collisions are less frequent, which reduces the resistance.
So with that in mind I’d draw an identical line (same gradient) just underneath the line on the graph.
I hope that helps!
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(Original post by 11lightsinthesky)
No problem! For 2b, as its three marks I’m not too sure where each point is coming from but this is my best guess:
•Power dissipated is energy lost per second
•This energy is lost (dissipated) as thermal energy in the wire.
•The fixed resistor reduces the potential difference across the wire, therefore reducing the current in the wire.
•This means that the resistance is reduced in the wire, thus reducing the thermal energy emitted.
•Overall reducing power dissipated in the wire.
Now that’s probably over kill but I write all my answers in bullet points (even in the exam) so I can logically sort through my response
For the last bit: you aren’t given any length. It says a larger diameter but doesn’t give a value, so all we know is the cross-sectional area is greater than before. There’s more room in the wire meaning the electron collisions are less frequent, which reduces the resistance.
So with that in mind I’d draw an identical line (same gradient) just underneath the line on the graph.
I hope that helps!
No problem! For 2b, as its three marks I’m not too sure where each point is coming from but this is my best guess:
•Power dissipated is energy lost per second
•This energy is lost (dissipated) as thermal energy in the wire.
•The fixed resistor reduces the potential difference across the wire, therefore reducing the current in the wire.
•This means that the resistance is reduced in the wire, thus reducing the thermal energy emitted.
•Overall reducing power dissipated in the wire.
Now that’s probably over kill but I write all my answers in bullet points (even in the exam) so I can logically sort through my response
For the last bit: you aren’t given any length. It says a larger diameter but doesn’t give a value, so all we know is the cross-sectional area is greater than before. There’s more room in the wire meaning the electron collisions are less frequent, which reduces the resistance.
So with that in mind I’d draw an identical line (same gradient) just underneath the line on the graph.
I hope that helps!
))))
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