A place of wire of resistance $R$ is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of these of this combination is $R^{'}$ , then the ratio $R / R^{'}$ is-

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Solution

Since resistance is directly proportional to length, resistance of each part is $\frac{R}{5}$

Using $\frac{1}{R_{e q}}$ $= \frac{1}{R_{1}}$ $+ \frac{1}{R_{2}}$ $+ \frac{1}{R_{3}} + . . .$ for parallel combination, equivalent resistance of the combination $(R^{'}) =$

$\frac{R}{5} \times (\frac{1}{5}) = \frac{R}{25}$

Therefore, $R / R^{'} = 25$

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Question 1
A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R', then the ratio $\frac{R}{R^{'}}$ is:

(a) $\frac{1}{25}$
(b) $\frac{1}{5}$
(c) 5
(d) 25

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A place of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of these of this combination is R′, then the ratio R/R′ is-

Since resistance is directly proportional to length, resistance of each part is R5Using 1Req =1R1 +1R2 +1R3+... for parallel combination, equivalent resistance of the combination (R′)=R5×(15)=R25Therefore, R/R′=25

A place of wire of resistance $R$ is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of these of this combination is $R^{'}$ , then the ratio $R / R^{'}$ is-