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Two wires of same metal have same length, but their cross sections are in the ratio 3 : 1. They are joined in series. The resistance of thicker wire is 10Ω.The total resistance of the combination will be

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a

52Ω

b

403Ω

c

40Ω

d

100Ω

answer is C.

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Detailed Solution

We know that R= p(l/A)For same material and same lengthR1R2=A2A1=31 or R1=3R˙2Given, R2=10Ω, hence R1=30ΩIn series, R = R1 +R2=30Ω+10Ω=40Ω

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