Problem — three resistors in series and parallel

Problem

Three resistors of $2{,}0\ \Omega$, $4{,}0\ \Omega$ and $6{,}0\ \Omega$ are first connected in series, then in parallel. Calculate the equivalent resistance in both cases.

Solution

In series the resistances add: $R_\text{tot} = R_1 + R_2 + R_3 = 2{,}0 + 4{,}0 + 6{,}0 = 12\ \Omega$

In parallel the inverses add: $\frac{1}{R_\text{tot}} = \frac{1}{2{,}0} + \frac{1}{4{,}0} + \frac{1}{6{,}0} = \frac{6}{12} + \frac{3}{12} + \frac{2}{12} = \frac{11}{12}\ \Omega^{-1}$ $R_\text{tot} = \frac{12}{11} \approx 1{,}09\ \Omega$

Note: the parallel resistance ($\approx 1{,}09\ \Omega$) is smaller than the smallest of the three ($2{,}0\ \Omega$), as always happens in parallel combinations.

$\ev{R_\text{tot} = 12\ \Omega \text{ (series)} \qquad R_\text{tot} \approx 1{,}09\ \Omega \text{ (parallel)}}$

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Topics: Circuiti elettrici Concepts: Resistenze in serie e parallelo Skills: Semplificazione di reti