Hey there! Sign in to join this conversationNew here? Join for free

Edexcel Physics Unit 2 "Physics at work" June 2013 Watch

  • View Poll Results: The last question - Does resistance increase or decrease?
    It increases ( using V=IR or some other method)
    70.73%
    It decreases using the 'lattice vibrations' theory
    29.27%

    Offline

    0
    ReputationRep:
    (Original post by StUdEnTIGCSE)
    Now its difference to producing standing waves in a string. The harmonics are produced for odd multiples of half wavelength. Like this
    Ive never understood why its lambda/4, why not lambda/2 for the fundamental?
    Offline

    11
    ReputationRep:
    (Original post by Branny101)
    Ive never understood why its lambda/4, why not lambda/2 for the fundamental?
    But you are BRANNY101

    Only kidding
    Offline

    0
    ReputationRep:
    Name:  Captura.PNG
Views: 105
Size:  18.2 KBName:  Captura2.PNG
Views: 87
Size:  7.8 KBOn part (c)(iii) How do i do it? why is resistance of R greater than of parallel ?
    Offline

    0
    ReputationRep:
    (Original post by krisshP)
    But you are BRANNY101

    Only kidding
    maaaaaate :ahee:

    Posted from TSR Mobile
    Offline

    3
    ReputationRep:
    (Original post by Branny101)
    Ive never understood why its lambda/4, why not lambda/2 for the fundamental?
    I don't know. My teacher said that its about resonance and that will learn about it in A2. I guess this is the theory behind it

    Spoiler:
    Show

    At the end of the tube the sound wave undergoes reflection and another wave is produced. Now at this point of incidence there should be node, a point with zero amplitude that undergoes no displacement at all. This is impossible if the wave has undergone half wavelength where it produces an antinode in the end which is impossible! Notice also that there should always be an antinode at the opening.
    A hard reflection takes place. The sound wave tries to displace the particles of the tube at the end in one direction but its way to small a magnitude. According to Newton's Third law the tube then exerts a equal and opposite force, a force at the opposite direction, so the resultant displacement is zero because of complete destruction of anti phase points.

    Its good to remember that at the opening an anti node is necessary and at the end a node is necessary for a standing wave. And if both are openings then both should be antinodes.:yes:



    Posted from TSR Mobile
    Offline

    0
    ReputationRep:
    (Original post by StUdEnTIGCSE)
    I don't know. My teacher said that its about resonance and that will learn about it in A2. I guess this is the theory behind it

    Spoiler:
    Show

    At the end of the tube the sound wave undergoes reflection and another wave is produced. Now at this point of incidence there should be node, a point with zero amplitude that undergoes no displacement at all. This is impossible if the wave has undergone half wavelength where it produces an antinode in the end which is impossible! Notice also that there should always be an antinode at the opening.
    A hard reflection takes place. The sound wave tries to displace the particles of the tube at the end in one direction but its way to small a magnitude. According to Newton's Third law the tube then exerts a equal and opposite force, a force at the opposite direction, so the resultant displacement is zero because of complete destruction of anti phase points.

    Its good to remember that at the opening an anti node is necessary and at the end a node is necessary for a standing wave. And if both are openings then both should be antinodes.:yes:



    Posted from TSR Mobile
    Right, but in an open-ended tube , both sides are at anti-nodes, right? ( e.g. a tube )

    Posted from TSR Mobile
    Offline

    3
    ReputationRep:
    (Original post by Daniel Atieh)
    Thank you sooo much really!
    one more question (from CIE) but who cares ...it's still electricity!
    http://vvcap.net/db/rhNqORlDMtAxemgMkbWe.htp http://vvcap.net/db/DlCJFeu-fbrM6OA7lLmZ.htp
    the second last one when it says : closed open open .... i thought it's 3 ?
    Where's the circuit diagram?:confused:

    Posted from TSR Mobile
    Offline

    3
    ReputationRep:
    (Original post by Branny101)
    Right, but in an open-ended tube , both sides are at anti-nodes, right? ( e.g. a tube )

    Posted from TSR Mobile
    Yes that's what I said in the last sentence

    Posted from TSR Mobile
    Offline

    0
    ReputationRep:
    (Original post by StUdEnTIGCSE)
    Yes that's what I said in the last sentence

    Posted from TSR Mobile
    Ahh sorry - reading this off of my phone and the last sentence was cut-off.

    Thanks

    Posted from TSR Mobile
    Offline

    3
    ReputationRep:
    (Original post by KBenzema)
    On part (c)(iii) How do i do it?

    When lamp is removed the total resistance of the circuit increases. So the current decreases. So the voltage of R decreases.
    So the voltage of P should now increase , giving a higher voltmeter reading, as P now receives a greater share of the total voltage.

    This might be a better explanation

    (Original post by muonz)
    P is a fixed resistor, when the lamp is in parallel with the fixed resistor, the resistance of the parallel combination is smaller than the resistance of P. Therefore now that the lamp is removed you have a potential divider, the ratio of the resistance between P and R changes so that P has a higher proportion of the voltage (as the resistance of P > the resistance of P and the lamp in parallel), hence the voltmeter reading increases.


    Posted from TSR Mobile
    Offline

    0
    ReputationRep:
    (Original post by StUdEnTIGCSE)
    Now its difference to producing standing waves in a string. The harmonics are produced for odd multiples of half wavelength. Like this
    I see! Thanks
    Offline

    0
    ReputationRep:
    could someone please explain to me the difference between a potential divider and rheostat

    I know a rheostat varies the current of the whole circuit, but I'm not sure how a potential divider works??

    is it just if you increase it's resistance it requires a larger potential difference therefore less for the other components? (If so, I'm not sure how that works because aren't they connected in parallel therefore same p.d?)
    Offline

    0
    ReputationRep:
    Please help with this question.

    Name:  ImageUploadedByStudent Room1369994742.615203.jpg
Views: 137
Size:  45.9 KB

    I don't know how to interpret the question to a diagram and then answer the question.please help.The answer is D according to mark scheme.


    Posted from TSR Mobile
    Offline

    0
    ReputationRep:
    (Original post by StUdEnTIGCSE)
    Well. When the bulb is removed the total resistance of the circuit decreases. This is because now there's no parallel connection to the component P. As the total V is constant more R means less I (V=IR). So as current decreases and the resistance of P is assumed to be constant then V decreases.




    Posted from TSR Mobile
    Hm? P is a fixed resistor, when the lamp is in parallel with the fixed resistor, the resistance of the parallel combination is smaller than the resistance of P. Therefore now that the lamp is removed you have a potential divider, the ratio of the resistance between P and R changes so that P has a higher proportion of the voltage (as the resistance of P > the resistance of P and the lamp in parallel), hence the voltmeter reading increases.
    Offline

    11
    ReputationRep:
    (Original post by blacknightking)
    Please help with this question.

    Name:  ImageUploadedByStudent Room1369994742.615203.jpg
Views: 137
Size:  45.9 KB

    I don't know how to interpret the question to a diagram and then answer the question.please help.The answer is D according to mark scheme.


    Posted from TSR Mobile
    Name:  1369997239085.jpg
Views: 105
Size:  24.3 KB
    Sorry for it being messy. Basically there are 2 resistors in series, R2 and R3, with resistor R2 in parallel with both of them. Then there's a voltmeter across the whole set of 3 resistors. In parallel, components have the same pd, so pd across R1=PD across R2 and R3. All .resistors are identical, R2=R3=R1. Now let consider just the circuit's series part. PD across R2 and R3=PD across R2+PD across R3. So PD across say R2 or even just across R3 is HALF of PD across series set of R2 and R3, so therefore PD across R2 or even just across R3 is HALF of PD across whole parallel combination, which is HALF of PD across R1. By P=V^2/R, power dissipated by R2 alone or even R3 alone is equal to a quarter of power dissipated by R1, making D the answer
    Offline

    0
    ReputationRep:
    (Original post by StUdEnTIGCSE)
    Well phase difference is the fraction of difference of oscillation of one oscillator to another. It measured as an angle in radians or in degrees. It actually sees how the sine curve of one wave has been translated to get another along the x axis. A translation or phase difference of 0,2π,4π
    ...2nπ (multiples of 2π) gives the same curve back again so its thought to be in phase, while a phase angle difference of π,3π....2n+1π gives a sine curve that is totally opposite so its out of phase.
    You can work with degrees also.

    Path difference is the difference in the distance travelled between two waves from their source. It can be measured as distance or in wavelength. If the difference shows a multiple of one wavelength they are in phase while a odd multiple of half wavelengths(0.5lamda,1.5lambda) shows an exactly out of phase points in the two wave. The significance of this can be seen in the principle of superposition of waves, diffraction and interference patterns etc

    The relationship between these two variables are simple. A phase angle difference of 2π or 360° corresponds to 1 lambda path difference so use this ratio method

    Spoiler:
    Show

     \dfrac{phase difference}{2\pi} =\dfrac{path difference}{\lambda}


    You can substitute for path difference or wavelength to find for the other. Use the lambda and the 2π as numbers which can be cancelled.
    THANK YOU SO MUCH!!!!!!!! Get it now haha. Thanks again


    Posted from TSR Mobile
    Offline

    0
    ReputationRep:
    (Original post by blacknightking)
    Please help with this question.

    Name:  ImageUploadedByStudent Room1369994742.615203.jpg
Views: 137
Size:  45.9 KB

    I don't know how to interpret the question to a diagram and then answer the question.please help.The answer is D according to mark scheme.


    Posted from TSR Mobile
    I was doing this yesterday... it was exploding my mind. Sorry if I'm wrong but I'll try my best to help you. This is the diagram I did:

    Name:  3856f15187.png
Views: 103
Size:  2.2 KB

    Basically, let's call the top path, path 1, and the bottom path, path 2.
    Path 1 has double the resistance of path 2, and since both paths are in parallel, they have the same potential difference across them. So:

     V=IR

    path 1 has half the current flowing through it than path 2 (current takes the path of least resistance).

    Power dissipated is:

     P=I^2R

    So the power dissipated for ONE of the resistors in path 1 is:

     Current = \dfrac{1}{3}I; Resistance = R



Power = (\dfrac{1}{3}I)^2R

         = \dfrac{1}{9}I^2R

    Power dissipated for the resistor in path 2 is:

     Current = \dfrac{2}{3}I; Resistance = R



Power = (\dfrac{2}{3}I)^2R

= \dfrac{4}{9}I^2R

    Hence, the power dissipated by the resistor in path 2 is four times the power dissipated by a resistor in path 1.

    I hope this helped.
    Offline

    0
    ReputationRep:
    (Original post by krisshP)
    Name:  1369997239085.jpg
Views: 105
Size:  24.3 KB
    Sorry for it being messy. Basically there are 2 resistors in series, R2 and R3, with resistor R2 in parallel with both of them. Then there's a voltmeter across the whole set of 3 resistors. In parallel, components have the same pd, so pd across R1=PD across R2 and R3. All .resistors are identical, R2=R3=R1. Now let consider just the circuit's series part. PD across R2 and R3=PD across R2+PD across R3. So PD across say R2 or even just across R3 is HALF of PD across series set of R2 and R3, so therefore PD across R2 or even just across R3 is HALF of PD across whole parallel combination, which is HALF of PD across R1. By P=V^2/R, power dissipated by R2 alone or even R3 alone is equal to a quarter of power dissipated by R1, making D the answer
    Thanx a lot.Now I understood.


    Posted from TSR Mobile
    Offline

    0
    ReputationRep:
    Edexcels gonna **** us over again. Gosh, how i hate edexcel physics.
    Offline

    0
    ReputationRep:
    (Original post by muonz)
    I was doing this yesterday... it was exploding my mind. Sorry if I'm wrong but I'll try my best to help you. This is the diagram I did:

    Name:  3856f15187.png
Views: 103
Size:  2.2 KB

    Basically, let's call the top path, path 1, and the bottom path, path 2.
    Path 1 has double the resistance of path 2, and since both paths are in parallel, they have the same potential difference across them. So:

     V=IR

    path 1 has half the current flowing through it than path 2 (current takes the path of least resistance).

    Power dissipated is:

     P=I^2R

    So the power dissipated for ONE of the resistors in path 1 is:

     Current = \dfrac{1}{3}I; Resistance = R



Power = (\dfrac{1}{3}I)^2R

         = \dfrac{1}{9}I^2R

    Power dissipated for the resistor in path 2 is:

     Current = \dfrac{2}{3}I; Resistance = R



Power = (\dfrac{2}{3}I)^2R

= \dfrac{4}{9}I^2R

    Hence, the power dissipated by the resistor in path 2 is four times the power dissipated by a resistor in path 1.

    I hope this helped.
    Your so amazing man!<3 why are you so clever!!
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Has a teacher ever helped you cheat?
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Write a reply...
    Reply
    Hide
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.