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Q.

Two resistor R1=(24±0.5)Ω and R(8±0.3)Ω are joined in series. The equivalent resistance is

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a

32±0.33Ω

b

32±0.8Ω

c

32±0.2Ω

d

32±0.5Ω

answer is B.

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Detailed Solution

R1=(24±0.5)ΩR2=(8±0.3)ΩR3=R1+R2=(32±0.8)Ω
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Two resistor R1=(24±0.5)Ω and R(8±0.3)Ω are joined in series. The equivalent resistance is