Why does resistance decrease in a parallel combination of resistors ?
Also please describe how to calculate effective resistance. Thank you . .! !
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Resistance (R) depends on the resistivity of the material (p), length of the material (L) and cross sectional area (A) of the material following the equation:
R = p * L / A
When hooking resistors in parallel, you are increasing the cross sectional area of the resistance path. If you like to use the water hose analogy, you are draining the tub (battery - power source) with two hoses (resistors) instead of one. It makes the tub move water (current) faster because you have more paths (bigger cross sectional area) for the water to flow.
If you look at the original equation above you will notice that replacing A with 2A (two of the same resistors in parallel) you will cut the resistance in half.
That's the theory. In practice, to find the resistance you would normally just add the reciprocal of the individual resistors to get the reciprocal of the total resistance
1/R-total = 1/R1 + 1/R2 + 1/R3 + ....
for however many resistors you have in parallel. For example,
If you have only 2 resistors in parallel, a 50 ohm in parallel with a 100 ohm resistor then...
1/R-total = 1/50 ohm + 1/100 ohm = 2/100 ohm + 1/100 ohm
= 3/100 ohm
R-total = 100/3 = 33.3 ohm (most people forget this step)
A good rule to remember is that you total resistance is less than the smallest resistor when they are connected in parallel.
Resistance is directly proportional to the potential difference across the terminals of a circuit. In series, the potential difference is greater, thus offering more resistance. Whereas in the parallel circuit, the potential difference is distributed, thus, in each series less potential difference is observed. That is why lesser resistance is offered by the conductor.
Resistance also depends (inversely) on the length and thickness of the conductor.
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