(i)
In order to get a resistance of 9
Ω from three bulbs, each of resistance 6
Ω, we connect two bulbs in parallel.
1R =
16+16=1+16=26=13
R=3Ω
this parallel combination (or resistance 3
Ω) in series with the third bulb of 6
Ω as shown in figure. which in result gives 9
Ω of resistance.
(ii) In order to get a resistance of 4
Ω from three bulbs, each of 6
Ω resistance. We need to connect two bulbs in series which makes 12
Ω. And now this combination of two bulbs is connected in parallel to the remaining bulb of 6
Ω resistance. So the final equivalent resistance is:
1R =
112+16=1+212=312=14
R=4Ω