Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.

If we connect resistors in series, R = 6 Ω + 6 Ω + 6 Ω = 18 Ω.

If we connect all resistors in parallel,

$\dfrac{1}{\text{R}} = \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{3}{6} = \dfrac{1}{2}$

Hence, R = 2Ω

We can obtain the desired value by connecting two of the resistors in either series or parallel.

Case (i)

If two resistors are connected in parallel, then their equivalent resistance is

$\dfrac{1}{\text{R}_\text{p}} = \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{2}{6} = \dfrac{1}{3}$

Hence, R_p = 3Ω

The third resistor in series, then we get equivalent resistance

R_p + 6 = 3 + 6 = 9Ω

∴ To get a resistance of 9 Ω, two 6 Ω resistors should be connected in parallel and the third 6 Ω resistor should be connected in series with the combination as shown below:

Case (ii)

When two resistors are connected in series, their equivalent resistance is

R_s = 6 Ω + 6 Ω = 12 Ω

The third resistor is connected in parallel with 12 Ω. Hence the equivalent resistance is:

$\dfrac{1}{\text{R}} = \dfrac{1}{12} + \dfrac{1}{6} = \dfrac{1+2}{12} = \dfrac{3}{12} = \dfrac{1}{4}$

⇒ R = 4 Ω

∴ To get a resistance of 4 Ω, two 6 Ω resistors should be connected in series and the third 6 Ω resistor should be connected in parallel with the combination as shown below: