Question:
What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?
Answer:
(a) To obtain the highest resistance, connect the coils in series,
RS = R1 + R2 + R3 + R4
= 4 + 8 + 12 + 24
= 48 Ω
(b) To obtain the lowest resistance, connect the coils in parallel,
1/RP = 1/R1 + 1/R2 + 1/R3 + 1/R4
1/RP = 1/4 + 1/8 + 1/12 + 1/24
1/RP = 1/4 ( 1 + 1/2 + 1/3 + 1/6)
1/RP = 1/4 ((6 + 3 + 2 + 1) /6)
1/RP = 1/4 (12/6)
1/RP = 1/4 (2)
1/RP = 1/2
RP = 2 Ω
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