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Question

Two wires of same length and thickness having specific resistances 6Ω-cm and 3Ω-cm respectively are connected in parallel. The effective resistivity is ρΩ-cm. The value of ρ, to the nearest integer, is ________.


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Solution

Step 1: Given

Specific resistance of a first wire: ρ1=6Ω-cm

Specific resistance of a second wire: ρ2=3Ω-cm

Effective resistivity: ρΩ-cm

Resistance of first wire is R1.

Resistance of second wire is R2.

Step 2: Formula Used

The resistance of a wire is given by R=ρlA, where ρ is resistivity, l is length of wire and A is area of cross-section.

Equivalent resistance when connected in parallel: R=R1R2R1+R2

Step 3: Find the effective resistivity

Find an expression for the effective resistivity using the formula R=R1R2R1+R2. The area of the system will be twice the area of wires as 2 wires are connected together, so the area for effective resistance will be 2A but the length will remain the same as they are connected in parallel.

Substitute ρl2A for R, ρ1lA for R1 and ρ2lA for R2.

R=R1R2R1+R2ρl2A=ρ1lAρ2lAρ1lA+ρ2lAρ2lA=ρ1ρ2ρ1+ρ2lAρ2=ρ1ρ2ρ1+ρ2

Calculate the effective resistivity by substituting the values in the obtained expression

ρ2=ρ1ρ2ρ1+ρ2ρ2=6×36+3ρ=189×2ρ=4

Hence, the value of ρ is 4.


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