AI-generated practice question — model-generated for extra practice, not a previous-year CBSE board question.
$\frac{1}{R_p} = \frac{1}{6}+\frac{1}{12}+\frac{1}{4} = \frac{2+1+3}{12} = \frac{6}{12}$, so $R_p = 2\ \Omega$; total current $I = 12/2 = 6\ \text{A}$. The equivalent resistance (2 Ω) is less than the smallest resistor (4 Ω), as expected in parallel combination.
Source: Chapter 11, Section 11.6.2
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This is actually a multi-part calculation question compressed into 1 mark, so keep each step brief. Examiners look for: correct parallel formula, correct $R_p = 2\ \Omega$, correct $I = 6\ \text{A}$, and the concluding observation that parallel equivalent resistance is always less than the smallest individual resistor. Show the reciprocal calculation clearly — it's the key formula for parallel circuits.