Three wires each of length $L$, cross-sectional area $A$ and resistivity $\rho$ are connected as shown in the figure.These are to be replaced by another wire of same resistivity such that the resistance between points $X$ and $Z$ does not change. If $L_1$ is the length and $A_1$ is the cross-sectional area of the new wire, then which one among the following is correct?

Question

Prepp Team · Accepted Answer

To find the equivalent wire that has the same resistance between points $X$ and $Z$, we need to analyze the given setup and calculate the total resistance.

The three wires in the setup are arranged in such a way that two wires, each with resistance $R$, are in parallel, and this combination is in series with another wire of resistance $R$.

Step-by-step Calculation:

Calculate the resistance of each wire:
- The resistance of a wire is given by the formula: $R = \frac{\rho L}{A}$.
- For each wire, $R = \frac{\rho L}{A}$.
Calculate the equivalent resistance of the two parallel resistors:
- The formula for two resistors, $R$ in parallel is: $\frac{1}{R_{\text{parallel}}} = \frac{1}{R} + \frac{1}{R}$.
- Thus, $R_{\text{parallel}} = \frac{R}{2} = \frac{\rho L}{2A}$.
Calculate the overall series resistance:
- Add the series resistance: $R_{\text{total}} = R_{\text{parallel}} + R$.
- Substituting values: $R_{\text{total}} = \frac{\rho L}{2A} + \frac{\rho L}{A} = \frac{3\rho L}{2A}$.
For the equivalent single wire with resistance $R_{\text{new}}$, we have:
- $R_{\text{new}} = \frac{\rho L_1}{A_1}$ should equal $R_{\text{total}} = \frac{3\rho L}{2A}$.
- Equating the resistances: $\frac{\rho L_1}{A_1} = \frac{3\rho L}{2A}$.
- This simplifies to: $\frac{L_1}{A_1} = \frac{3L}{2A}$.
- One possibility is $L_1 = 3L$ and $A_1 = 2A$.