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.
Important Questions from Current Electricity
The property of electric current which is applicable in the fuse wires is
An electric current is expressed by a unit called .
What do you call a component of identical size that offers a higher resistance to electricity?
Identify the correct statement.
_______ is a simple device that is used to either break the electric circuit, or to complete it.
