|Sep21-04, 02:51 PM||#1|
can anyone please explain to me, in terms of maths, physics and all other elements involved, why the resistance decreases/current increases in a parallel circuit as the amount of loads increase. i just need to have it in words that i can put on paper, i know the rough idea. thanks
|Sep21-04, 06:07 PM||#2|
Everytime you add an additional load (in parallel), you create a new path for current to travel. Now the circuit is not limited by the value of resistance of the first pathway. It's been bypassed in a way limited by the resistance value of the second pathway. So, the total value of resistance for the entire circuit has been reduced. If you add another pathway, the overall resistance of the circuit will be reduced even further as a result of that new pathway.
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...
|Sep21-04, 07:42 PM||#3|
To add to Metallicbeing's response:
Recall (or realize) that the resistance is inversely proportional to the cross-sectional area [of a cylindrical ohmic resistor]:
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.
|Sep21-04, 09:11 PM||#4|
It's like bypass surgery.
|Sep21-04, 09:23 PM||#5|
If you want to simplify it, think about this:
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:
|Sep22-04, 04:50 AM||#6|
thanks guys, that's helped me alot
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