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Edexcel Physics Unit 2 "Physics at work" June 2013

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The experiment says:

set the voltage so that the motor can lift a mass of 100g a height of 1m in the time t.
Calculate the power in using P=IV
and the time taken for the mass to be raised at the corresponding voltage.
Repeat at different voltages
calculate efficiency using Power out/Power in
Power out is calculated using mgh/t
Reply 942
Avatar for hamzeh h
hamzeh h
Original post by ALevel96
Power out is calculated using mgh/t

Yup :P
Original post by hamzeh h
Yup :P


Can't believe I was so stupid on the drift velocity question haha.
The answer was indeed V/2!

How did you do in unit 1?
Original post by hamzeh h
i did it in january :P


Oh I see what did you get?
Original post by GCSE-help
No because it was being operated at below it's normal rating! I think people are missing a massive key point. The question asked for the effect of low current ON the resistance. So it's not asking us to consider WHY the current was low, just accept that it was! And then we had to find the effect of the low current on the resistance!


But I based my entire argument on the "fact" that they are in parallel to the mains power supply so the sum of the voltage should be still equal to 230!
Your answer makes more sense but the question didn't mention if they were taken out of the mains or not?!

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Original post by sarah.102
Negative. At least I hope so.


I made a thread here on the physics forums so hopefully some genius can help us its driving me insane. http://www.physicsforums.com/showthread.php?p=4408037&posted=1#post4408037
Original post by Mo_maths
I made a thread here on the physics forums so hopefully some genius can help us its driving me insane. http://www.physicsforums.com/showthread.php?p=4408037&posted=1#post4408037


This came a LOT in previous papers.
Its obvious its negative!

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Original post by StUdEnTIGCSE
This came a LOT in previous papers.
Its obvious its negative!

Posted from TSR Mobile


Some guy before said it was positive because if you look were the voltmeter is its not over the load resistance, not sure if he's right or wrong its just making me worry
Original post by Mo_maths
Some guy before said it was positive because if you look were the voltmeter is its not over the load resistance, not sure if he's right or wrong its just making me worry


The voltmeter was across the internal resistance.
V=E-rI

So why worry?

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Original post by StUdEnTIGCSE
The voltmeter was across the internal resistance.
V=E-rI

So why worry?

Posted from TSR Mobile


Its usually above the load resistance so the resistor if you look at the pics online such as
http://www.s-cool.co.uk/a-level/assets/learn_its/alevel/physics/Resistance/internal-resistance-emf-and-potential-difference/Finding%20the%20internal%20resistance.gif
I read online this

"For an external load resistor, you connect a voltmeter's (say) negative terminal to the terminal of the battery, and measure the voltage across the resistor. You get Ohm's Law: V=IR
For an internal source resistance, you connect a voltmeter's negative terminal to the negative terminal of the battery, and the positive terminal of the voltmeter to the positive terminal of the battery. Increasing the current drawn from the battery causes the voltage of the battery to decrease. In this case, the voltage seems change like -IR, the opposite of Ohm's Law above." so its negative right????
Original post by Mo_maths



Well both the circuits are one and the same.
Voltmeter has very high resistance therefore no current flows through it.
In this and the circuit in the exam the voltmeter is measuring the potential difference across the cell AND the load resistance (including that of the ammeter) because in parallel the voltage is the same.

So no worry!

Posted from TSR Mobile
Original post by StUdEnTIGCSE
Well both the circuits are one and the same.
Voltmeter has very high resistance therefore no current flows through it.
In this and the circuit in the exam the voltmeter is measuring the potential difference across the cell AND the load resistance (including that of the ammeter) because in parallel the voltage is the same.

So no worry!

Posted from TSR Mobile


AHHH YEAHHH!!!!!! Thanks man, feel much better now :smile:
On the last question. Wasnt it a series circuit. Cause of it was in series the. Of 1 bulb Is switched off the other one dont light up as it is an incomplete Circuit. :O

Am i wrong?

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Original post by Freddy-Francis
On the last question. Wasnt it a series circuit. Cause of it was in series the. Of 1 bulb Is switched off the other one dont light up as it is an incomplete Circuit. :O

Am i wrong?

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yeah it was series
Reply 956
Avatar for AsimZZZ
AsimZZZ
whats the explanation for the bulbs in series and not the other way?
Reply 957
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Phteven
15
On the E.M.F question did people have a straight or curved gradient? Mine was curved and I think that's wrong :frown: how many marks would I get for doing it negative and stating how to work them out correctly?
Reply 958
Avatar for x0x
x0x
OP
7
Original post by StephenNaulls
On the E.M.F question did people have a straight or curved gradient? Mine was curved and I think that's wrong :frown: how many marks would I get for doing it negative and stating how to work them out correctly?


Mine was a straight line with a negative gradient
the y intercept is the e.m.f
the negative of the gradient is the internal resistance
i guess there were 3 marks for the graph and 2 for explaning how to find e.m.f and r so i think you'll get some credit
Reply 959
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x0x
OP
7
Guys i added a poll for you to vote
i just want to see what the majority went with ...

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