Question

A uniform wire of resistance \(R\) is cut into four equal parts. These parts are then connected in parallel. The equivalent resistance of the combination is:

• A. \(R\)
• B. \(\dfrac{R}{4}\)
• C. \(\dfrac{R}{16}\)
• D. \(4R\)

Hint:

If a wire of resistance \(R\) is cut into \(n\) equal parts: \[ R_{\text{each}}=\frac{R}{n} \] If all \(n\) pieces are connected in parallel: \[ R_{\text{eq}} = \frac{R}{n^2} \] For \(n=4\): \[ R_{\text{eq}} = \frac{R}{16} \]

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The resistance ($R$) of a uniform conductor depends directly on its physical dimensions according to the formula $R = \rho \frac{l}{A}$, where $l$ is the length, $A$ is the cross-sectional area, and $\rho$ is the material's resistivity. Cutting a wire into equal parts reduces the length of each individual segment, which proportionally scales down its resistance. Once these segments are wired together in a parallel configuration, the total equivalent resistance drops further because the combination creates multiple paths for the electric current to flow through.
Step 2: Key Formula or Approach:
1. Resistance of a cut piece: Since the wire is cut into 4 identical parts, the length of each piece becomes $\frac{l}{4}$. Because resistance is directly proportional to length ($R \propto l$), the resistance of each piece ($R'$) is: $$ R' = \frac{R}{4} $$ 2. Parallel combination formula: When $n$ identical resistors each of resistance $R'$ are connected in parallel, their net equivalent resistance ($R_{\text{eq}}$) is given by: $$ R_{\text{eq}} = \frac{R'}{n} $$
Step 3: Detailed Explanation:
Let's calculate the final equivalent resistance by combining our values: - Number of identical parts ($n$) = $4$ - Resistance of each single part ($R'$) = $\frac{R}{4}$ Substitute these values directly into the parallel combination formula: $$ R_{\text{eq}} = \frac{\left(\frac{R}{4}\right)}{4} $$ To simplify this fraction, multiply the denominators together: $$ R_{\text{eq}} = \frac{R}{4 \times 4} = \frac{R}{16} $$ Thus, the total equivalent resistance of the parallel circuit network becomes $\frac{R}{16}$. This corresponds to option (C).
Step 4: Final Answer:
The equivalent resistance of the combination is R/16.