Problem — resistors in parallel

Problem

Three resistors $R_1 = 60\ \Omega$, $R_2 = 30\ \Omega$, $R_3 = 20\ \Omega$ are connected in parallel. Calculate the equivalent resistance.

Three resistors in parallel between the two common nodes.

Solution

Parallel — the reciprocals of the resistances add up: $\frac{1}{R_\text{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{60} + \frac{1}{30} + \frac{1}{20}$

Reducing to the common denominator $60$: $\frac{1}{R_\text{eq}} = \frac{1}{60} + \frac{2}{60} + \frac{3}{60} = \frac{6}{60} = \frac{1}{10}\ \Omega^{-1}$

Hence $R_\text{eq} = 10\ \Omega$

Physical note: in parallel the equivalent resistance is smaller than the smallest of the three ($20\ \Omega$), because more paths are offered to the current.

$\ev{R_\text{eq} = 10\ \Omega}$

Links

Topics: Circuiti elettrici Concepts: Resistenze in serie e parallelo