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AQA P2 Current, Voltage & Resistance

• 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.
  

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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

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