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AQA P2 Current, Voltage & Resistance

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Messages: 1 - 4 of 4
  • Message 1. 

    Posted by U15589753 (U15589753) on Thursday, 24th January 2013

    What are the rules for current, voltage and resistance in series and parallel circuits? (Actually, I don't need to know the one for resistance in parallel.) I'm really struggling with physics :S

    Any help is appreciated, thank you.

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  • Message 2

    , in reply to message 1.

    Posted by Tp (U15610827) on Wednesday, 20th February 2013

    What are the rules for current, voltage and resistance in series and parallel circuits? (Actually, I don't need to know the one for resistance in parallel.) I'm really struggling with physics :S

    Any help is appreciated, thank you. 
    A voltmeter, which measures voltage must always be connected in parellel to a component in a circuit eg in parellel to a variable resistor.
    Ammeter which measures current must be connected in series in a circuit.

    Current = the flow of electrons
    Potential difference: The difference of electrical difference between two points in a circuit.
    smiley - smiley

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  • Message 3

    , in reply to message 1.

    Posted by William (U15005823) on Wednesday, 20th February 2013

    for simple series and parallel circuits you can use Ohm's Law:

    V = IR

    where
    V is voltage or potential difference, measured in Volts (V)
    I is current, measured in Amperes (A)
    R is Resistance, measured in Ohms (Ω)


    The rule for calculating total resistance is simple
    In a SERIES circuit, you just add them up!

    So R(total) = R(1) + R(2) + R(...)

    For example the total resistance of a series circuit, R(total) with three resistors each of value 10Ω will be

    R (total) = 10Ω + 10Ω + 10Ω = 30Ω


    A PARALLEL circuit is more tricky:
    1/R(total) = 1/R(1) + 1/R(2) + 1/R(...)

    For example, the total resistance of a series circuit (Rtotal) with three resistors in parallel, each with a value of 10Ω will be
    1/R(total) = 1/10Ω + 1/10Ω + 1/10Ω

    so

    1/R(total) = 0.3Ω

    so

    R(total) = 3.33Ω


    You can see that in a parallel circuit the resistance is lower!


    Using Ohms Law, you can now work out the current in each circuit. Say the circuit was powered by a 10V battery; so V = 10 V

    SERIES
    I = V/R = 10V/30Ω = 0.33A

    PARALLEL

    I = V/R = 10V/3.3Ω = 3A

    So because the parallel circuit has a LOWER resistance, the current is HIGHER.

    Try changing the values of the battery, V, and the values of the resistors, R

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  • Message 4

    , in reply to message 3.

    Posted by U15589753 (U15589753) on Saturday, 6th April 2013

    I just realised I forgot to thank you all~
    A very belated thank you - the information was very helpful smiley - smiley

    Report message4

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