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

If we connect the resistors in series, then the equivalent resistance will be the sum of the resistors,

\(\Rightarrow \) 6 \(\Omega \) + 6 \(\Omega \) + 6 \(\Omega \) = 18 \(\Omega \),

which is not desired. If we connect the resistors in parallel, then the equivalent resistance will be \(\frac{6}{2} \) = 3 \(\Omega \) is also not desired.

Hence, we should either connect the two resistors in series or parallel.

(a) Two resistor in parallel


Two 6 \(\Omega \) resistors are connected in parallel. Their equivalent resistance will be
$$ \frac{1}{(\frac{1}{6} + \frac{1}{6} )}= \frac{6}{2} = 3 \Omega . $$

The third 6 \(\Omega \) resistor is in series with 3 \(\Omega \). Hence, the equivalent resistance of the circuit is 6 \(\Omega \)+ 3 \(\Omega \) = 9 \(\Omega \).

(b) Two resistor in parallel



Two 6 \(\Omega \) resistors are in series. Their equivalent resistance will be the sum 6 + 6 = 12 \(\Omega \).
The third 6 \(\Omega \) resistor is in parallel with 12 \(\Omega \). Hence, equivalent resistance will be

$$ 12 x \frac{6}{12} + 6 = 4 \Omega . $$

Therefore, the total resistance is 4\(\Omega \)