A wire of length ' 1 and resistance $100 \Omega$ is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:

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Aicharya · Accepted Answer

Divided into 10 parts $\begin{aligned} R & =\frac{\rho l}{A} \\ R^{\prime} & =\frac{\rho l}{10 A}=\frac{R}{10} \\ R_S & =5 \times \frac{R}{10} \quad \text { [series] } \\ R_S & =50 \\ R_P & =\frac{R}{50} \quad \text { [parallel] } \\ R_{\text {eq }} & =R_S+R_P \\ & =52 \Omega \end{aligned}$ /

A wire of length ' 1 and resistance $100 \Omega$ is divided into 10 equal parts. The first 5 parts are…

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