Why is resistance lower in a parallel configuration and therefore current is more?
Shouldn't resistance be addition of all resistor's resistance?
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the reason is that in parallel the electricity has more path to more (the extra resistance). The same current still goes through the first resistor, but now you opened another path or "gate" and an extra amount will flow through it. If the two resistances are equal then you will have double the current, and this half the resistance: $V=I/R$