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  • A wire of resistance $R$ is cut into 8 equal pieces. From these pieces two equivalent resistances are made by adding four of these together in parallel. Then these two sets are added in series.

Q.12) A wire of resistance $R$ is cut into 8 equal pieces. From these pieces two equivalent resistances are made by adding four of these together in parallel. Then these two sets are added in series. The net effective resistance of the combination is

A) $\frac{R}{8}$

B) $\frac{R}{64}$

C) $\frac{R}{32}$

D) $\frac{R}{16}$

 

Answers (1)

best_answer


Solution:
1. The wire is cut into 8 equal pieces, so each piece has resistance:

$$
R_{\text {piece }}=\frac{R}{8}
$$

2. From these, form two groups of 4 resistors, each group has 4 pieces in parallel:

For one group:

$$
\begin{aligned}
& \frac{1}{R_{\text {goup }}}=\frac{1}{R / 8}+\frac{1}{R / 8}+\frac{1}{R / 8}+\frac{1}{R / 8}=\frac{4}{R / 8}=\frac{32}{R} \\
& \Rightarrow R_{\text {group }}=\frac{R}{32}
\end{aligned}
$$

3. Now, there are two such groups of resistance $R / 32$, and they are connected in series:

$$
R_{\text {net }}=\frac{R}{32}+\frac{R}{32}=\frac{2 R}{32}=\frac{R}{16}
$$
 

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Dimpy

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