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The total resistance of three resistances connected in parallel is

  • A.\(R = R_{1} + R_{2} + R_{3}\)
  • B.\(\frac{1}{R} = \frac{2}{R_{1}} + \frac{2}{R_{2}} + \frac{2}{R_{3}}\)
  • C.\(R = R_{1} \times R_{2} \times R_{3}\)
  • D.\(\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}\)

Answer: D

For resistances in parallel, \(\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}\).

In a parallel connection every resistor has the same voltage across it, and the total current is the sum of the branch currents:

\[ \begin{aligned} \frac{V}{R} &= \frac{V}{R_{1}} + \frac{V}{R_{2}} + \frac{V}{R_{3}} \\ \frac{1}{R} &= \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} \end{aligned} \]

\(R = R_{1} + R_{2} + R_{3}\) is the rule for resistances in series. The version with 2 in every numerator has no basis, and multiplying the resistances gives the wrong unit.

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