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

Question

A wire of length ' $l$ ' 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

Professor PC Thomas Classes · Answer

(b) : As, wire having resistance $R=100 \Omega$, is divided into 10 equal parts, so, each part will have resistance $\frac{R}{10}$. Equivalent resistance of series combination of first 5 parts, $$ \begin{aligned} & R_{e q}^{\prime}=R_1+R_2+R_3+R_4+R_5 \ & =\frac{R}{10}+\frac{R}{10}+\frac{R}{10}+\frac{R}{10}+\frac{R}{10} ; R_{e q}^{\prime}=\frac{R}{2} \end{aligned} $$ Equivalent resistance of parallel combination of next 5 parts, $$ \begin{aligned} & \frac{1}{R_{e q}}=\frac{1}{R_6}+\frac{1}{R_7}+\frac{1}{R_8}+\frac{1}{R_9}+\frac{1}{R_{10}}=\frac{10}{R}+\frac{10}{R}+\frac{10}{R}+\frac{10}{R}+\frac{10}{R}=\frac{50}{R} \ & R_{e q}=\frac{R}{50} \end{aligned} $$ Resistance of final series combination is, $R_{	ext {final }}=R_{e q}^{\prime}+R_{e q}$ $$ \begin{aligned} & =\frac{R}{2}+\frac{R}{50}=\frac{26 R}{50}=\frac{26 	imes 100}{50} \ & 	herefore \quad R_{	ext {final }}=52 \Omega \end{aligned} $$