Question

Two metallic wires A and B are connected in series, wire A has length L and radius r

while B has length 2L and the radius 2r. If both the wires are of same material then the ratio of the total resistance of series combination to the resistance of the wire A

A. 3/4

B. 3/2

C. 6/2

D. 6/5
.

Answer 1

The correct answer is B) 3/2.

The total resistance of a series combination of wires A and B, with wire B having double the length and radius compared to wire A, is 3/2 times the resistance of wire A alone.

Step-by-step explanation:

The problem describes two metallic wires A and B connected in series, where wire A has length L and radius r, and wire B has length 2L and radius 2r.

Since the wires are made of the same material, their resistivities will be the same.

The resistance (R) of a wire is given by the formula R = ρL/A, where ρ is the resistivity, L is the length, and A is the cross-sectional area.

For wire A, the resistance will be R_A = ρL/πr^2.

For wire B, because the length is doubled and the radius is doubled, the resistance will be R_B = ρ(2L)/π(2r)^2 which simplifies to R_B = ρL/2πr^2.

Thus, the resistance of wire B is half that of wire A.

Since they are connected in series, the total resistance R_total is the sum of R_A and R_B, which yields

R_total = R_A + R_B = ρL/πr^2 + ρL/2πr^2

= (3/2)(ρL/πr^2).

Therefore, the ratio of the total resistance of the series combination to the resistance of the wire A is

R_total/R_A = (3/2)(ρL/πr^2) / (ρL/πr^2)

= 3/2.