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    Name:  IMG_0407.jpg
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Size:  275.0 KBHi everyone

    I am currently crashing Higher Physics and have already passed the Dynamic Universe unit.

    I am currently doing the electricity unit and am struggling on some homework.

    Can you help me with questions 2c and 2d? (Sorry about poor picture quality).

    Any help is much appreciated.

    Thanks in advance
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    (Original post by merryroversrule)
    Name:  IMG_0407.jpg
Views: 52
Size:  275.0 KBHi everyone

    I am currently crashing Higher Physics and have already passed the Dynamic Universe unit.

    I am currently doing the electricity unit and am struggling on some homework.

    Can you help me with questions 2c and 2d? (Sorry about poor picture quality).

    Any help is much appreciated.

    Thanks in advance
    You may appreciate the help.

    Everyone who wishes to answer will appreciate your courteous effort to post legible questions the right way up!

    part c) look at the circuit carefully. It's in a configuration called a 'Wheatstone Bridge'.

    This simply means two potential dividers in parallel with a resistance between them forming the bridge part. If all resistance values are equal (assuming they are without the full question), both potential dividers will form the same potential differences across their respective divider pairs. with respect to the common battery terminals.

    That means there will be no p.d. developed across the bridge resistor and hence no current can flow through it. So for all purposes, the bridge resistor (in this instance) can be ignored.

    http://www.bbc.co.uk/bitesize/higher...rs/revision/3/

    part d) Changing the battery voltage will change the total current in the circuit but will have no effect on the ratio of the resistances and hence the current through the bridge resistor remains unchanged.
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    Thanks for your help. Sorry about the picture being upside down!
 
 
 
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