# Physics question

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#1
abcd
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#2
need help for this one
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1 week ago
#3
What have you tried/what do you know already?
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#4
(Original post by Callicious)
What have you tried/what do you know already?
Well TBH, the answer is B and i got it right but not sure about the method. Actually i have some trouble understanding how a galvanometer works. I know that at some point along the length Galvanometer reads zero. If i remember correctly it's due to the fact PD across the branches is same. If you could maybe explain why this happens that would be great. : )
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1 week ago
#5
(Original post by papie)
Well TBH, the answer is B and i got it right but not sure about the method. Actually i have some trouble understanding how a galvanometer works. I know that at some point along the length Galvanometer reads zero. If i remember correctly it's due to the fact PD across the branches is same. If you could maybe explain why this happens that would be great. : )
Here's a good video on Galvanometer's (at least the rotating coil ones that I expect you get at A-Level) https://www.khanacademy.org/science/...ometer-working

That's the only one I know the operating principle of so if you have any Q's after watching the video I can answer.
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1 week ago
#6
so we have the power supply here positive and positive meeting at a point negative and negative meeting at the other the potentiometer where pq is 100 cm long so his pqe is 100 cm long and has a resistance of 5 ohm so that means 100 centimeters centimeter corresponds to resistance of 5 ohm the power supply has an emf 2 volt 2 volt and the solar cell has an emf of 5.0 millivolt. (idk if that makes sense to you lmao)

100cm- 5 ohms
vps=es=0.005 v
40cm=5/100 x 40= 2 ohms
rps = 2 ohms
rsq= 3 ohms
vps=rps/rps+rsq+r
0.005=2/2+3+r x 2
5+r=4/0/0.005 or r= 4/0.005 - 5 which is 795 ohms

this is probably right but just check just in case as im a gcse student (year 11) hope it helps lmao
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#7
(Original post by gcse2021student)
so we have the power supply here positive and positive meeting at a point negative and negative meeting at the other the potentiometer where pq is 100 cm long so his pqe is 100 cm long and has a resistance of 5 ohm so that means 100 centimeters centimeter corresponds to resistance of 5 ohm the power supply has an emf 2 volt 2 volt and the solar cell has an emf of 5.0 millivolt. (idk if that makes sense to you lmao)

100cm- 5 ohms
vps=es=0.005 v
40cm=5/100 x 40= 2 ohms
rps = 2 ohms
rsq= 3 ohms
vps=rps/rps+rsq+r
0.005=2/2+3+r x 2
5+r=4/0/0.005 or r= 4/0.005 - 5 which is 795 ohms

this is probably right but just check just in case as im a gcse student (year 11) hope it helps lmao
okay, that eqn you used: vps=rps/rps+rsq+r, Is it an application of Kirchhoff's second law ?
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