Parallel Combination Of Resistors

For resistors joined between the same pair of nodes, each has the full voltage across it and the equivalent resistance follows $\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$, as detailed in NCERT Class 12, Chapter 3 (Current Electricity). Substituting the values, $\frac{1}{R_{eq}}=\frac{1}{6}+\frac{1}{3}=\frac{1}{6}+\frac{2}{6}=\frac{3}{6}=\frac{1}{2}$, so $R_{eq}=2$ ohm. Alternatively, the product-over-sum shortcut gives $\frac{6\times 3}{6+3}=\frac{18}{9}=2$ ohm. The option 9 ohm is wrong because that would be the series sum of the two resistors, not the parallel result, and series and parallel behave oppositely. The option 18 ohm forgets to divide by the sum in the product-over-sum shortcut and reports only the numerator product. The option 4.5 ohm is merely the arithmetic average of the two values, which has no basis for a parallel network. A plausibility check confirms correctness: a parallel combination must always be smaller than the smallest branch resistor, and 2 ohm is indeed less than 3 ohm, since adding a parallel path always provides extra room for current to flow and thereby lowers the overall opposition to the circuit.