Let's consider two resistors only for simplicity. When the resistors are connected in parallel, the potential difference on each resistor is the same, and it is equal to the voltage of the battery, V:
[tex]V_1 = V_2 = V[/tex] (1)
Instead, the current splits in the two resistors:
[tex]I=I_1 + I_2[/tex] (2)
where I is the current in the circuit, I1 and I2 are the currents in the two resistors.
By using Ohm's law, we can rewrite (2) as
[tex] \frac{V}{R_{eq}} = \frac{V_1}{R_1}+ \frac{V_2}{R_2} [/tex] (3)
where [tex]R_{eq}[/tex] is the equivalent resistance of the circuit. But we also said that
[tex]V_1 = V_2 = V[/tex]
So we can rewrite (3) as
[tex] \frac{V}{R_{eq}}= \frac{V}{R_1}+ \frac{V}{R_2} [/tex]
which becomes
[tex] \frac{1}{R_{eq}}= \frac{1}{R_1}+ \frac{1}{R_2} [/tex]
So, the equivalent resistance of the parallel of n resistors is given by
[tex] \frac{1}{R_{eq}}= \frac{1}{R_1}+ \frac{1}{R_2}+...+ \frac{1}{R_n} [/tex]
