A wire of length l and resistance 100 Ω is divided into ten equal parts. First five parts are in series and remaining five in parallel. Two combinations are again connected in series. Total resistance is:
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Question

A wire of length l and resistance 100 Ω is divided into ten equal parts. First five parts are in series and remaining five in parallel. Two combinations are again connected in series. Total resistance is:

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G1 Physics · Accepted Answer

Step 1: Each equal part has resistance 100/10=10 Ω. Step 2: Five parts in series: Rₛ = 5×10=50 Ω. Step 3: Five parts in parallel: Rₚ =10/5=2 Ω. Step 4: Total R=50+2=52 Ω.