Two cylindrical wires, X and Y, are made of the same uniform conducting material. Wire X h… — JAMB UTME

Two cylindrical wires, X and Y, are made of the same uniform conducting material. Wire X has length $L$ , diameter $d$ , and an electrical resistance of $12\,\Omega$ . Wire Y has length $2L$ and diameter $2d$ . What is the resistance of wire Y?

A
$3\,\Omega$
$6\,\Omega$ Answer
C
$12\,\Omega$
D
$24\,\Omega$

Answer

The resistance of wire Y is

6\,\Omega

The resistance of a uniform conductor is given by

R = \frac{4\rho L}{\pi d^2}

. For wire X,

R_X = 12\,\Omega

. For wire Y with length

2L

and diameter

2d

, the new resistance becomes

R_Y = \frac{4\rho(2L)}{\pi (2d)^2} = \frac{8\rho L}{4\pi d^2} = \frac{1}{2}\left(\frac{4\rho L}{\pi d^2}\right) = \frac{1}{2} R_X = 6\,\Omega

Step-by-Step Solution

Express the resistance of a cylindrical conductor in terms of length

L

and diameter

d

R = \rho \frac{L}{A} = \rho \frac{L}{\pi (d/2)^2} = \frac{4\rho L}{\pi d^2}

The cross-sectional area

A

of a circular wire with diameter

d

is given by

A = \frac{\pi d^2}{4}

Write the resistance formula for wire X using its given value.

R_X = \frac{4\rho L}{\pi d^2} = 12\,\Omega

Wire X has length

L

and diameter

d

Substitute the parameters of wire Y (

L_Y = 2L

and

d_Y = 2d

) into the resistance formula.

R_Y = \frac{4\rho (2L)}{\pi (2d)^2} = \frac{8\rho L}{4\pi d^2} = \frac{2\rho L}{\pi d^2}

Doubling diameter increases the cross-sectional area by a factor of

2^2 = 4

Relate the resistance of wire Y to the resistance of wire X and calculate the final numerical value.

R_Y = \frac{1}{2} \left(\frac{4\rho L}{\pi d^2}\right) = \frac{1}{2} R_X = \frac{12\,\Omega}{2} = 6\,\Omega

Since

R_Y = \frac{1}{2} R_X

, halving

12\,\Omega

yields

6\,\Omega

Key Concept

Dependence of Electrical Resistance on Conductor Dimensions

Estimated Time:1m 30s

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