We have four resistors A, B, C and D of resistance 3 Ω, 6 Ω, 9 Ω and 12 Ω respectively. Find out the lowest resistance which can be obtained by combining these four resistors.

Given,

• Resistance of resistor A ($\text R_\text A$) = 3 Ω
• Resistance of resistor B ($\text R_\text B$) = 6 Ω
• Resistance of resistor C ($\text R_\text C$) = 9 Ω
• Resistance of resistor D ($\text R_\text D$) = 12 Ω

To minimize equivalent resistance, connect all the resistors in parallel (parallel combinations always give a resistance less than the smallest individual resistor).

When resistors A, B, C and D are connected in parallel then net resistance of the circuit is given by,

$\dfrac {1}{\text R$\text P} = \dfrac {1}{\text R\text A} + \dfrac {1}{\text R\text B} + \dfrac {1}{\text R\text C} + \dfrac {1}{\text R\text D} \\[1em] = \dfrac {1}{3} + \dfrac {1}{6} + \dfrac {1}{9} + \dfrac {1}{12} \\[1em] = \dfrac {12 + 6 + 4 + 3}{36} \\[1em] \Rightarrow \dfrac {1}{\text R\text P} = \dfrac {25}{36} \\[1em] \Rightarrow \text R\text P = \dfrac {36}{25} \\[1em] \Rightarrow \text R\text P = 1.44\ \text Ω

Hence, the lowest resistance of the circuit is 1.44 Ω.