Science (086) — AI-generated practice question

In a parallel combination, the same potential difference V acts across each resistor. Each branch provides an additional path for current, so total current increases. Since $R_p = V/I$, a larger total current means a smaller equivalent resistance.

The formula confirms this:
$$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots$$

Since $\frac{1}{R_p}$ is greater than any individual term (e.g., $\frac{1}{R_1}$), it follows that $R_p < R_1$ and $R_p < R_2$.

Numerical Example: Take $R_1 = 6\ \Omega$, $R_2 = 3\ \Omega$ (smallest = 3 Ω).

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

$$R_p = 2\ \Omega$$

$R_p = 2\ \Omega < 3\ \Omega$ (smallest resistor). ✓

Source: Chapter 11, Section 11.6.2 — Resistors in Parallel

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• Key idea: Parallel connection adds extra current paths without changing voltage, so total current rises → equivalent resistance falls. This is the physical reason, and examiners expect it.
• Formula logic: $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}$ means $\frac{1}{R_p} > \frac{1}{R_{\min}}$, so $R_p < R_{\min}$. State this explicitly.
• Always verify with numbers when the question asks for it — full marks require both reasoning and the example.
• Keep the calculation neat and boxed/highlighted so the examiner spots it quickly.