A wire of resistance R is cut into four equal parts. These parts are t

A wire of resistance R is cut into four equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio $\frac{\text{R’}}{\text{R}}$ is :

$rac{1}{16}$
$rac{1}{4}$
4
16
This question was previously asked in
UPSC CAPF – 2023
The ratio $\frac{\text{R’}}{\text{R}}$ is $\frac{1}{16}$.
Let the original resistance of the wire be R. When the wire is cut into four equal parts, the resistance of each part becomes $\frac{R}{4}$ (assuming uniform material and cross-section). Let these four parts be $R_1, R_2, R_3, R_4$, where $R_1 = R_2 = R_3 = R_4 = \frac{R}{4}$. These parts are then connected in parallel. The equivalent resistance R’ of resistances connected in parallel is given by the formula: $\frac{1}{R’} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4}$. Substituting the values, we get: $\frac{1}{R’} = \frac{1}{R/4} + \frac{1}{R/4} + \frac{1}{R/4} + \frac{1}{R/4} = \frac{4}{R} + \frac{4}{R} + \frac{4}{R} + \frac{4}{R} = \frac{4+4+4+4}{R} = \frac{16}{R}$. Therefore, $R’ = \frac{R}{16}$. The required ratio $\frac{\text{R’}}{\text{R}}$ is $\frac{R/16}{R} = \frac{1}{16}$.
The resistance of a wire is directly proportional to its length. Cutting a wire into four equal parts reduces the length of each part to one-fourth of the original length, thus reducing the resistance of each part to R/4. Connecting resistors in parallel decreases the total equivalent resistance compared to the individual resistances. This principle is used in electrical circuits to control current flow.
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