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What is (i) the highest, (ii) the lowest total resistance that can be secured by combinations of four coils of resistance 4\(\Omega \) , 8\(\Omega \), 12\(\Omega \), 24\(\Omega \) ?


Answer :

(i) we can get highest resistance by connecting the four coils in series.
Then, R = 4\(\Omega \) + 8\(\Omega \) + 12\(\Omega \) + 24\(\Omega \) = 48\(\Omega \)

(ii)We can get lowest resistance by connecting the four coils in parallel.

\(\frac{1}R=\frac{1}4 +\frac{1}8+\frac{1}{12} =\frac{12}{24}=\frac{1}2 \)

we get, R=2\(\Omega \)

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