Science (086) — AI-generated practice question

In a parallel combination, the reciprocal of equivalent resistance is:

$$\frac{1}{R_p} = \frac{1}{4} + \frac{1}{6} + \frac{1}{12} = \frac{3+2+1}{12} = \frac{6}{12} = \frac{1}{2}$$

So, $R_p = 2\ \Omega$.

This is less than the smallest individual resistance (4 Ω) because adding more parallel paths provides additional routes for current, increasing total current flow. As per the formula $\frac{1}{R_p} = \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$, each extra resistor increases $\frac{1}{R_p}$, thereby decreasing $R_p$ below any individual resistance.

Source: Chapter 11, Section 11.6.2 – Resistors in Parallel

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• Examiners expect the formula, the calculation showing $R_p = 2\ \Omega$, and a one-line conceptual reason (additional current paths → lower equivalent resistance).
• The key phrase to remember: "reciprocal of equivalent resistance equals sum of reciprocals" — each added branch always makes $\frac{1}{R_p}$ larger, so $R_p$ is always smaller than the smallest individual resistor.
• Don't just state the result — show the arithmetic to earn full marks.