27 Wtf (1 Viewer)

frostysnow

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Pwnage101 said:
lol yeh im pretty sure i got that wrong, i changed my asnwer from that to 'the ammeter goes up to 2.50 only' LMAO i cant stop laughing at how stoopid my asnwer was, oh well, it fit pretty well at the time ----> 1 mark

and did we have to subtract 2 ohms from our answers (ie inverso fo the gradient of graph) caus ethere was a 2 ohm resistor in teh circuit??
hey i wrote the same thing. i know it sounds stupid, but this can happen... hopefully
 

GilesG

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timmiitippii said:
About the minusing 2 ohm resistor thingy...
They gave us the current and voltage already so im assuming its the overall current which already included the resistance due to the 2 ohm resistor...

uhh dw i lost my train of thought.
Does that mean we lose marks twice for calculating wrong resistance of both X and Y...?
ARgh should of known the last question they were gonna chuck in bullshit like a random resistor that ppl would ignore = =
The voltmeter only measured the voltage drop across X/Y as adnan said, and so you don't need to subtract 2 ohms. The resistor will result in decreased current through the circuit compared to if it weren't there, but the voltage drop across X/Y will decrease in proportion. I doubt that they would have asked a question involving adding or subtracting resistances as this is part of year 11 that isn't on the formula sheet at all.
 

Kearnzo

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Ah well. Another mark gone. Whether they wanted it to or not, year 11 work would have effected people.

Why put the resistor there? Meh
 

Aaron.Judd

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Tha Wishkah said:
I was unsure and decided that the 2 ohms was a tarp.

i also said the wire was incapable of transmitting a higher voltage due to it being too thin/damaged.

lawl
I said about it being too thin etc as well.
 

GilesG

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Kearnzo said:
Ah well. Another mark gone. Whether they wanted it to or not, year 11 work would have effected people.

Why put the resistor there? Meh
I think it's because before testing X and Y you don't know whether they have appreciable resistance. If not, then without the resistor it will short circuit.
 

dp624

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GilesG said:
I think it's because before testing X and Y you don't know whether they have appreciable resistance. If not, then without the resistor it will short circuit.
IIRC the Voltmeter has a very high degree of resistance.
Maybe it's to trick those who think they're looking out for silly stuff, but are overanalysing.
 

GilesG

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dp624 said:
IIRC the Voltmeter has a very high degree of resistance.
Maybe it's to trick those who think they're looking out for silly stuff, but are overanalysing.
Yes, but all the current will go through the wire X/Y instead of the voltmeter, which is connected in parallel.

I know the Board is crap for almost all of the syllabus, and this question was certainly verging on being year 11 (and identifying which wire was which probably shouldn't have been asked at all), but they aren't supposed to try to trick students and I think in this case it actually was just a rare glimmer of decent physics principles. Unfortunately, that required having stuff in the circuit that made the question seem more complicated than it was.
 

iEdd

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I don't know how many answers they will accept, as most are plausible, however, what I put, which is correct is that the higher current in the wire will heat it, due to it have finite resistance. The heat will increase resistance, in turn causing a drop in current.

Malfunctioning resistor is unlikely if the same resistor was used for all other values and all of them formed a straight line.

I'd imagine they may accept it if you said that one wire sample was bent until it work hardened, where resistance increases.

Voltmeters have very high internal resistance, so only a small current (microamps) flows through the voltmeter, leaving the overall current of the circuit largely unchanged, as most will still travel through the component that the meter is in parallel with.
 

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