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Question

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

(i) 9 Ω

(ii) 4 Ω

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Solution

(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

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