how is Q negative and P positive? (1 Viewer)

[anonymoushehe]

Am i tripping, for part B we would care about the gradient for 10 -12 s and get the voltage which I get. but since flux is decreasing, the induced current should be clockwise, so current goes from Q to P; so assuming conventional current, Q would be positive and P would be negative? but the marking guidelines saying the opposite?

 

[cheesynooby]

current indeed goes clockwise.
buuut the V on the diagram is not a power source - so if P is positive and Q is negative current will go +ve to -ve and take the shortest route (through V), matching the clockwise current.
for a battery, current will go the long way around from +ve to -ve, and then the battery can be thought of as using energy to 'push' the electrons from -ve to +ve for them to begin their journey again.

 

[anonymoushehe]

current indeed goes clockwise.
buuut the V on the diagram is not a power source - so if P is positive and Q is negative current will go +ve to -ve and take the shortest route (through V), matching the clockwise current.
for a battery, current will go the long way around from +ve to -ve, and then the battery can be thought of as using energy to 'push' the electrons from -ve to +ve for them to begin their journey again.

wait my brain is not braining can you like dumbify it a little

 

[cheesynooby]

current goes from +ve to -ve.
since current goes clockwise, it goes from P to Q and so P is +ve, Q is -ve.

note this reasoning doesnt work for power sources such as batteries, because inside the battery the current goes the opposite direction.

 

[wizzkids]

Q would be positive and P would be negative? but the marking guidelines saying the opposite?

I agree with everything you have said. Where did you find this question?

so if P is positive and Q is negative current will go +ve to -ve and take the shortest route (through V), matching the clockwise current.

Wrong. Voltmeters have a high resistance. There is no such thing as a shortest route in electricity. Electric current follows the path of least resistance, which would be clockwise around the loop.

 

[nonya2000]

Current should travel from Q to P

 

[anonymoushehe]

I agree with everything you have said. Where did you find this question?

Wrong. Voltmeters have a high resistance. There is no such thing as a shortest route in electricity. Electric current follows the path of least resistance, which would be clockwise around the loop.

2017 HSC q27

 

[cheesynooby]

I agree with everything you have said. Where did you find this question?

Wrong. Voltmeters have a high resistance. There is no such thing as a shortest route in electricity. Electric current follows the path of least resistance, which would be clockwise around the loop.

there would be an E field going clockwise by Faraday's Law, correct?
regardless of current or resistance of the voltmeter, would not the positive and negative terminals for a component always match the direction of the E field across it (which would be clockwise hence up from P to Q)?
just like for, say, a light globe in a circuit, the side the current enters through will be positive because that aligns with the E field across the component?

also, this is an hsc physics question from the 2017 paper.

 

[nonya2000]

2017 HSC q27

thats the gay old syllabus, try finding stuff from vce physics papers

 

[wizzkids]

Oh hang on, wait a minute, I just realised the question is ambiguous. Remember Kirchoff's Circuit Rules which says, "the sum of the emfs around a circuit must come to zero."
The question asked, "indicate the polarity of the terminals P and Q when this occurs." This is ambiguous. There are two ways to interpret that part because there are two parts to the circuit.

(1) which branch of the conductor will appear to be more positive, and which branch will appear to be more negative.
OR
(2) which way will the voltmeter need to be connected to correctly indicate the induced emf.
According to interpretation (1) branch Q is more positive than branch P and conventional current flows from Q to P.
According to interpretation (2) the voltmeter will see P as more positive than Q, because that's the way the current is flowing between the terminals of the voltmeter.

 

[anonymoushehe]

Oh hang on, wait a minute, I just realised the question is ambiguous. Remember Kirchoff's Circuit Rules which says, "the sum of the emfs around a circuit must come to zero."
The question asked, "indicate the polarity of the terminals P and Q when this occurs." This is ambiguous. There are two ways to interpret that part because there are two parts to the circuit.

(1) which branch of the conductor will appear to be more positive, and which branch will appear to be more negative.
OR
(2) which way will the voltmeter need to be connected to correctly indicate the induced emf.
According to interpretation (1) branch Q is more positive than branch P and conventional current flows from Q to P.
According to interpretation (2) the voltmeter will see P as more positive than Q, because that's the way the current is flowing between the terminals of the voltmeter.

is it possible if you could like dumbify it like a little bit

 

[nonya2000]

is it possible if you could like dumbify it like a little bit

i believe this shitty question's answer is wrong

 

[Socialism]

i believe this shitty question's answer is wrong

what the fuck nonya has 400 messages now???

 

[Sorghum Sneedley]

View attachment 47139View attachment 47140
Am i tripping, for part B we would care about the gradient for 10 -12 s and get the voltage which I get. but since flux is decreasing, the induced current should be clockwise, so current goes from Q to P; so assuming conventional current, Q would be positive and P would be negative? but the marking guidelines saying the opposite?

I think electrons will be pushed against the flow of current towards q, with the large resistance from the voltmeter stopping it there, so the electron will bunch up there making it negative and so p is positive