- #1

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- Thread starter bob4000
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- #1

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- #2

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RT=1/((1/R1)+(1/R2)+(1/R3)+...)

If you reduce the amount of resistance in the circuit, you increase the amount of amps flowing through the entire circuit.

I = E/RT

Where I= Amps, E= Volts, and RT= Total Ohms

Hope this helps...

- #3

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Recall (or realize) that the resistance is inversely proportional to the cross-sectional area [of a cylindrical ohmic resistor]:

[tex]R=\frac{\rho \ell}{A}[/tex]

Adding a second resistor in parallel to the first effectively increases the cross-sectional area of the combination of resistors. Assuming an ideal constant voltage source across the first resistor [and other resistors in parallel with it], the current through the first resistor is unchanged. However, now there is an additional current through the second resistor.

- #4

Gonzolo

It's like bypass surgery.

- #5

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You have two kinds of filters:

A - light screen mesh, allows a high flow (10L/s)

B - thicker carbon filter, allows a low flow (1L/s)

You put A into a circuit, its allowing 10L/s to move through it. Now if you connect filter B parallel to A you're allowing an additional 1L/s through. A total of 11L/s.

Water analogies always help with simple electronics.

Difference in potential:

voltage

pressure

Flow rate:

current

current

Resistance:

resistance

restrictive orfice

- #6

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thanks guys, that's helped me alot

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