Question

Two wires of same length and same area of cross-section but made of different materials of resistivities \( \rho_1 \) and \( \rho_2 \) are connected in series. The equivalent resistivity of the combination is :

• A. \(\dfrac{1}{2}(\rho_1+\rho_2)\)
• B. \(\rho_1+\rho_2\)
• C. \(\rho_1\rho_2\)
• D. \(2(\rho_1+\rho_2)\)

Hint:

For wires connected end-to-end (series combination), first find the equivalent resistance and then compare it with \(R=\rho L/A\). Remember that resistivity is a material property and does not simply add like resistances.

Answer 1

The Correct Option is A

Concept: The resistance of a conductor depends upon its resistivity, length, and area of cross-section. The relation is \[ R=\rho\frac{L}{A} \] where \[ \rho = \text{resistivity}, \qquad L = \text{length}, \qquad A = \text{area of cross-section}. \] When two resistors are connected in series, the equivalent resistance is equal to the sum of their individual resistances. To determine the equivalent resistivity of a combination, we first calculate the equivalent resistance and then compare it with the standard relation \[ R_{\text{eq}}=\rho_{\text{eq}}\frac{L_{\text{eq}}}{A}. \]

Step 1: Write the resistance of each wire. Let each wire have length \(L\) and area of cross-section \(A\). For the first wire, \[ R_1=\rho_1\frac{L}{A}. \] Similarly, for the second wire, \[ R_2=\rho_2\frac{L}{A}. \]

Step 2: Determine the equivalent resistance of the series combination. Since the wires are connected in series, \[ R_{\text{eq}}=R_1+R_2. \] Substituting the values, \[ R_{\text{eq}} = \rho_1\frac{L}{A} + \rho_2\frac{L}{A}. \] Taking \(\frac{L}{A}\) common, \[ R_{\text{eq}} = \frac{L}{A}(\rho_1+\rho_2). \]

Step 3: Express the combination as a single wire. The total length of the combination is \[ L_{\text{eq}}=L+L=2L. \] The area of cross-section remains \[ A_{\text{eq}}=A. \] Therefore, \[ R_{\text{eq}} = \rho_{\text{eq}} \frac{2L}{A}. \]

Step 4: Compare both expressions of equivalent resistance. We have \[ \rho_{\text{eq}} \frac{2L}{A} = \frac{L}{A}(\rho_1+\rho_2). \] Cancelling \(\frac{L}{A}\) from both sides, \[ 2\rho_{\text{eq}} = \rho_1+\rho_2. \] Hence, \[ \rho_{\text{eq}} = \frac{\rho_1+\rho_2}{2}. \]

Step 5: Identify the correct option. Thus the equivalent resistivity of the series combination is \[ \boxed{\rho_{\text{eq}}=\frac{\rho_1+\rho_2}{2}}. \] Therefore, the correct answer is \[ \boxed{\text{(A)}} \]