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    Attachment 594236594238Name:  IMG_0405.jpg
Views: 58
Size:  349.1 KBHi everyone

    I have recently started doing Higher Physics without ever doing it previously.
    I passed the Dynamic Universe unit and am now on the Electricity unit.

    However, I am struggling with the recent homework I was given (see attached). I only need help on question 2, as I have already completed question 1.

    Any help is much appreciated

    Thanks in advance
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    That picture is upside down :rolleyes:

    To get the resistance between A and C you need to simplify the resistor network

    1. calculate the resistance of R1 and R2 in series
    2 calculate the resistance of R3 and R4 in series
    3. calculate the resistance of Rs on it's own (ok no calculation required, it's just Rs)

    you have simplified the resistor network to an equivalent network...
    Name:  net111116.jpg
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    now you need to combine the 3 remaining resistances in parallel into a single value of resistance...

    the rule for this is 1/Reff=1/R1 + 1/R2 +1/R3 +... (or that's how it's shown in the A level formula sheets)
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    Okay, thanks for your help. I just realised it the picture was upside down as soon as I posted, probably should have fixed that!

    Would you be able to help me on the rest of question 2? Physics is difficult because I've never done it before.

    Thanks for your help
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    (Original post by merryroversrule)
    Okay, thanks for your help. I just realised it the picture was upside down as soon as I posted, probably should have fixed that!

    Would you be able to help me on the rest of question 2? Physics is difficult because I've never done it before.

    Thanks for your help
    You've had no lessons?

    The relevant definitions for electrical power are

    P= V I = I2 R = V2/R

    so probably the simplest thing to do would be work out the current in each branch - then the power dissipation for the each resistor in that branch will be I2R
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    Ok, thanks again for your help.

    It's not that I don't know what to do, it's just that I can never remember what strategy / formula to use in each situation.

    Would you be able to help me with 2c and 2d?

    Any help is much appreciated
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    Would you be able to go through a step by step example of 2b to enable me to check my answer?

    Once again, thanks very much for your help, it is much appreciated!
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    (Original post by merryroversrule)
    Would you be able to go through a step by step example of 2b to enable me to check my answer?

    Once again, thanks very much for your help, it is much appreciated!
    Several approaches work.

    based on the earlier result we can immediately see that the power dissipated in R5 is the greatest - the central branch has the lowest resistance 10 ohms and the other two branches are both 20 ohms total resistance.- all branches have the same pd across them so current is greatest in R5.

    The current passing through R5 is V/R
    =12/10
    =1.2 A

    The power dissipated by R5 is I2R
    =(1.2)2 R
    =1.44 * 10
    =14.4 Watt

    (or you could have got the same answer by saying
    P=VI
    =12*1.2
    =14.4)



    --
    in the more complicated situation where the resistors all had different values you would have to work out the current in each branch using the total resistance for each branch and the pd across each branch... in this case the upper and lower branches are identical values.

    upper branch
    I=V/R
    =12/20
    =0.6 (half the current going through the R5 branch btw)

    power dissipated in R1 = I2R
    =(0.6)2 R
    =0.36 *10
    =3.6 W

    (or alternatively by treating R1 and R2 as a potential divider...
    P=VI
    =6 * 0.6
    =3.6 W)

    and so on for the remaining resistors until you know the power dissipated in all of them.

    ---
    It's just about practising and following the rules.
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    Ok, thanks for your help. I did do it correctly!

    Would you be able to help me with 2c and 2d please?

    As always, any help is much appreciated.
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    (Original post by merryroversrule)
    Ok, thanks for your help. I did do it correctly!

    Would you be able to help me with 2c and 2d please?

    As always, any help is much appreciated.
    Current supplied by the battery is the sum of the currents in the 3 parallel branches

    can't see which resistor it's asking about... but the important relationship probably is Ohm's law
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    Can you do a step by step guide for questions 2c and 2d please?

    Thanks in advance
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    (Original post by merryroversrule)
    Can you do a step by step guide for questions 2c and 2d please?

    Thanks in advance
    1. work out the current in the top branch using the values for resistance we've calculated earlier and the PD across it that we've been given in the question.

    2. then do the same for the middle branch

    3. and then the lower branch

    4. add together all three

    5. write down total

    ----
    1. plug the original numbers into the formula for ohms law and see what the original current is

    2. plug the new PD into the formula for ohms law and see what the new current is

    3. compare the new to the original current

    4. write down result
 
 
 
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