Question

Show how you would connect three bulbs, each of resistance 6 $\Omega$, so that the combination has a resistance of

(i) 9 $\Omega$

(ii) 4 $\Omega$

Answer 1

(i)
In order to get a resistance of 9 $\Omega$ from three bulbs, each of resistance 6 $\Omega$, we connect two bulbs in parallel.
$\frac{1}{R}$ = $\frac{1}{6}+\frac{1}{6}=\frac{1+1}{6}=\frac{2}{6}=\frac{1}{3}$
$R=3\Omega$
this parallel combination (or resistance 3 $\Omega$) in series with the third bulb of 6 $\Omega$ as shown in figure. which in result gives 9 $\Omega$ of resistance.

(ii) In order to get a resistance of 4 $\Omega$ from three bulbs, each of 6 $\Omega$ resistance. We need to connect two bulbs in series which makes 12 $\Omega$. And now this combination of two bulbs is connected in parallel to the remaining bulb of 6 $\Omega$ resistance. So the final equivalent resistance is:
$\frac{1}{R}$ = $\frac{1}{12}+\frac{1}{6}=\frac{1+2}{12}=\frac{3}{12}=\frac{1}{4}$

$R=4\Omega$