A wire of length l and resistance 100 Omega i | vedclass

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(A) The resistance of the original wire is $R = 100 \Omega$.When the wire is divided into $10$ equal parts,the resistance of each part is $r = \frac{R

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$52$

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(A) The resistance of the original wire is $R = 100 \Omega$.When the wire is divided into $10$ equal parts,the resistance of each part is $r = \frac{R}{10} = \frac{100 \Omega}{10} = 10 \Omega$.For the first $5$ parts connected in series,the equivalent resistance is $R_S = 5 	imes r = 5 	imes 10 \Omega = 50 \Omega$.For the next $5$ parts connected in parallel,the equivalent resistance $R_P$ is given by $\frac{1}{R_P} = \frac{1}{r} + \frac{1}{r} + \frac{1}{r} + \frac{1}{r} + \frac{1}{r} = \frac{5}{r}$.Thus,$R_P = \frac{r}{5} = \frac{10 \Omega}{5} = 2 \Omega$.Since these two combinations are connected in series,the final equivalent resistance is $R_{eq} = R_S + R_P = 50 \Omega + 2 \Omega = 52 \Omega$.

$A$ 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: (in $\Omega$)

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