Show how you would connect three resistors each of resistance 6 Ω, so that the combination has a resistance of 9 Ω. Also justify your answer.

Given,

• Number of resistors = 3
• Resistance of each resistor ($\text R$) = 6 Ω
• Combined resistance ($\text R_\text {net}$) = 9 Ω

First combine two resistor in parallel combination and their effective resistance is given by,

$\dfrac{1}{\text R$\text P} = \dfrac{1}{\text R} + \dfrac{1}{\text R} \\[1em] = \dfrac{1}{6} + \dfrac{1}{6} \\[1em] = \dfrac{2}{6} \\[1em] \Rightarrow \text R\text P = \dfrac{6}{2} \\[1em] \Rightarrow \text R_\text P = 3\ \text Ω

Now, combine one more resistor in series with $\text R$\text P to get $\text R\text {net}$ which is given by,

$\text R$\text {net} = \text R\text P + \text R \\[1em] = 3 + 6 \\[1em] \Rightarrow \text R_\text {net} = 9\ \text Ω

Justification : When two resistors are connected in parallel, their effective resistance decreases and adding another resistor in series increases the total resistance. By this arrangement, the overall resistance becomes exactly 9 Ω, as required.

Hence, combine two 6 Ω resistors in parallel and the third in series to give 9 Ω total resistance.