The increase of loads in a parallel circuit * decreases*the total resistance

*.*

If the resistance of the original circuit is `R_0` *, *and an additional load of the resistance *R* is added in parallel, then the effective resistance of the resultant circuit will be determined by the formula:

...

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The increase of loads in a parallel circuit * decreases *the total resistance

*.*

If the resistance of the original circuit is `R_0` *, *and an additional load of the resistance *R* is added in parallel, then the effective resistance of the resultant circuit will be determined by the formula:

`1/R_(eff) = 1/R_0 + 1/R`

(This formula can be derived from the Ohm's Law and the fact that when the branches of a circuit connected in parallel have the same voltage - please see the reference link.)

Notice that if we consider this formula without the extra load, it would look like

`1/R_(eff) = 1/R_0`

But, because the extra load is added, the 1/R term is added to the right side of the equation, so this means that the left side has to increase. The quantity `1/R_(eff)` becomes **larger** because of the addition of the extra load *R.*

This quantity `1/R_(eff)` , however, is the * inverse*of the effective resistance `R_(eff)`

so this means that the effective resistance `R_(eff)` became* smaller*.

So, whenever an extra load is added in parallel, **the total resistance decreases**.

This fact can be also understood conceptually. If there are more parallel branches in a circuit, then there are more pathways for the current to go through, so the total resistance of the circuit decreases. This is similar to a crowd of people passing through a hallway: if the hallway widens, it will be easier for people to go through.

**Further Reading**