Q49 of 330 Page 43

What is (a) highest, and (b) lowest, resistance which can be obtained by combining four resistors having the following resistances?

4Ω, 8Ω, 12Ω, 24Ω

(a) To obtain the highest resistance we must connect the given resistances in series.

Highest resistance R = 4 + 8 + 12 + 24 = 48ohms


(b) To obtain the lowest resistance we must connect the given resistances in parallel.


1/R=1/4+1/8+1/12+1/24


On solving we get, R=2ohms


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