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 :
[amp_mcq option1=”$\frac{1}{16}$” option2=”$\frac{1}{4}$” option3=”4″ option4=”16″ correct=”option1″]
This question was previously asked in
UPSC CAPF – 2023
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 $\frac{R/16}{R} = \frac{1}{16}$.