Derive an expression for equivalent resistance when three resistors of resistance $R_{1}, R_{2}$ and $R_{3}$ are connected in parallel.

(N/A) Consider three resistors $R_{1}, R_{2}$ and $R_{3}$ connected in parallel as shown in the figure.
When the total current $I$ reaches point $a$,it splits into three parts: $I_{1}$ flowing through $R_{1}$,$I_{2}$ flowing through $R_{2}$,and $I_{3}$ flowing through $R_{3}$.
Since charge must be conserved,the total current $I$ entering point $a$ must be equal to the sum of the currents leaving that point. Therefore,we have:
$I = I_{1} + I_{2} + I_{3}$ $...(1)$
Since the resistors are connected in parallel,the potential difference $V$ across each resistor is the same. According to Ohm's law,the current through each resistor is:
$I_{1} = \frac{V}{R_{1}}, I_{2} = \frac{V}{R_{2}}, I_{3} = \frac{V}{R_{3}}$
Substituting these values into equation $(1)$,we get:
$I = \frac{V}{R_{1}} + \frac{V}{R_{2}} + \frac{V}{R_{3}}$ $...(2)$
If $R_{P}$ is the equivalent resistance of the parallel combination,then the total current $I$ can be written as:
$I = \frac{V}{R_{P}}$ $...(3)$
Comparing equations $(2)$ and $(3)$,we have:
$\frac{V}{R_{P}} = \frac{V}{R_{1}} + \frac{V}{R_{2}} + \frac{V}{R_{3}}$
Dividing both sides by $V$,we get the expression for equivalent resistance:
$\frac{1}{R_{P}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}$