Question

The effective resistance of a parallel connection that consists of four wires of equal length, equal area of cross-section and same material is $0.25\, \Omega$. What will be the effective resistance if they are connected in series?

• A. $0.25\, \Omega$
• B. $0.5\, \Omega$
• C. $1\, \Omega$
• D. $4\, \Omega$

Answer 1

The Correct Option is D

To determine the effective resistance when the wires are connected in series, we need to first understand the difference in resistance calculation for parallel and series circuits.

Given: The effective resistance of a parallel connection of four wires is \(0.25\, \Omega\).

Let's denote the resistance of each wire as \(R\). In a parallel circuit, the total resistance \(R_{\text{parallel}}\) is given by:

\(\frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4}\)

Since all wires have the same resistance \(R\), the equation becomes:

\(\frac{1}{R_{\text{parallel}}} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{4}{R}\)

Given that \(R_{\text{parallel}} = 0.25\, \Omega\), we can substitute: 

\(\frac{1}{0.25} = \frac{4}{R}\)

Solving for \(R\):

\(4 = \frac{4}{R} \cdot 0.25\)

\(R = 1\, \Omega\)

This means each wire has a resistance of \(1\, \Omega\).

When these wires are connected in series, the total resistance \(R_{\text{series}}\) is simply the sum of individual resistances:

\(R_{\text{series}} = R_1 + R_2 + R_3 + R_4 = 4 \cdot R = 4 \cdot 1 \, \Omega = 4\, \Omega\)

Thus, the effective resistance when the wires are connected in series is \(4\, \Omega\).

Therefore, the correct answer is: \(4\, \Omega\).