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Step-by-Step Solution
Step 1: Understand the Problem
We have two wires of equal length and thickness but different specific resistances, 6 Ω⋅cm and 3 Ω⋅cm. They are connected in parallel, and we want to find the effective resistivity of this arrangement.
Step 2: Recall the Resistance–Resistivity Relationship
The resistance R of a wire is related to its resistivity ρ by the formula:
R = \rho \frac{l}{A}
where l is the length of the wire and A is its cross-sectional area.
Step 3: Express Each Wire’s Resistance
Let both wires have the same length l and cross-sectional area A. Then their resistances are:
R_1 = \rho_1 \frac{l}{A} and R_2 = \rho_2 \frac{l}{A}
Given:
\rho_1 = 6\,\Omega\text{⋅cm}
\rho_2 = 3\,\Omega\text{⋅cm}
Step 4: Calculate the Equivalent Resistance in Parallel
When two resistors are connected in parallel, their equivalent resistance R_{\text{eq}} is given by:
\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} \quad \Longrightarrow \quad R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2}
Substituting R_1 and R_2 in terms of ρ, l, and A:
R_1 = 6 \frac{l}{A} \quad,\quad R_2 = 3 \frac{l}{A}
R_{\text{eq}} = \frac{\left(6 \frac{l}{A}\right)\left(3 \frac{l}{A}\right)}{6 \frac{l}{A} + 3 \frac{l}{A}} = \frac{18 \frac{l^2}{A^2}}{9 \frac{l}{A}} = 2 \frac{l}{A}
Step 5: Relate Equivalent Resistance to the Effective Resistivity
The new arrangement (two wires in parallel) can be viewed as a single wire of the same length l but with double the cross-sectional area, 2A. Let the effective resistivity be ρ. Then the equivalent resistance is:
R_{\text{eq}} = \rho \frac{l}{2A}
We know from our parallel calculation that R_{\text{eq}} = 2\,\frac{l}{A} . Hence,
\rho \frac{l}{2A} = 2 \frac{l}{A}
Canceling l and A from both sides (assuming l ≠ 0, A ≠ 0):
\frac{\rho}{2} = 2 \quad \Longrightarrow \quad \rho = 4
Step 6: State the Final Answer
The effective resistivity of the two wires connected in parallel is 4 Ω⋅cm.
