Question

(b) Two metallic rods, each of length L, area of cross section A1 and A2, having resistivity p1 and p2 are connected in parallel across a d,c battery. Obtain the expression for the effective resistivity of this combination​

Answer 1

R(eq) = R1R2/(R1+R2) = (ρ1+ρ2)L / (A1+A2)

Explanation:

Given:

Length of first rod = Length of second rod = L.

Area of cross section of first rod = A1.

Area of cross section of second rod = A2.

Resistivity of first rod = p1.

Resistivity of second rod = p2.

Connected in parallel.

Find: Effective resistance of the combination.

Solution:

Formula for electrical resistance is R = ρ*L/A. So resistance is directly proportional to length and inversely proportional to cross-sectional area of the conductor.

Since the connection is in parallel, we know that total resistance R = 1/R1 + 1/R2.

 = 1/(ρ1 + ρ2) * (A1 + A2)/L

1/R1 + 1/R2 = (A1 + A2) / L(ρ1 + ρ2)

(R2+R1) / R1R2 = (A1+A2) / L(ρ1+ρ2)

So inverse of the above, we get:

R1R2/(R1+R2) = (ρ1+ρ2)L / (A1+A2)

Effective resistance, R(eq) = R1R2/(R1+R2) = (ρ1+ρ2)L / (A1+A2)