Which combination of three resistors R1, R2 and R3 has minimum total resistance? (Consider R1 = 1 Ω, R2 = 2 Ω and R3 = 3 Ω)
When all are connected in parallel
For resistors in series, the equivalent resistance is the sum \(R_1 + R_2 + R_3 = 1 + 2 + 3 = 6\ \Omega\), which is the largest possible combination.
For all three in parallel: \(\frac{1}{R_{eq}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} = \frac{6+3+2}{6} = \frac{11}{6}\), giving \(R_{eq} = \frac{6}{11} \approx 0.545\ \Omega\).
For R1 and R2 in parallel plus R3 in series: parallel part \(=\frac{1\times2}{1+2}=\frac{2}{3}\ \Omega\), total \(=\frac{2}{3}+3 \approx 3.67\ \Omega\).
For R1 and R3 in parallel plus R2 in series: parallel part \(=\frac{1\times3}{1+3}=0.75\ \Omega\), total \(=0.75+2=2.75\ \Omega\).
Comparing all four values (6, 0.545, 3.67, 2.75 Ω), the all-parallel combination gives the minimum total resistance, since connecting resistors in parallel always yields an equivalent resistance smaller than the smallest individual resistor.
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