### 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.

• #### 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.

• #### 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

• #### 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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