A conducting wire of length 1.5\,\text{m} and initial resistance 8.0\,\Omega is stretched … — JAMB UTME

A conducting wire of length $1.5\,\text{m}$ and initial resistance $8.0\,\Omega$ is stretched uniformly until its length is doubled while its total volume remains unchanged. If a steady potential difference of $16\,\text{V}$ is applied across the ends of the stretched wire, what is the electric current flowing through it?

$0.50\,\text{A}$ Answer
B
$1.00\,\text{A}$
C
$2.00\,\text{A}$
D
$8.00\,\text{A}$

Answer

The electric current flowing through the stretched wire is

0.50\,\text{A}

When a cylindrical wire is stretched to double its length (

L_2 = 2L_1

) while maintaining constant volume, its cross-sectional area halves (

A_2 = A_1/2

). From the resistance formula

R = \rho L / A

, doubling length and halving area causes resistance to increase by a factor of four (

R_2 = 4 R_1 = 32.0\,\Omega

). By Ohm's law (

I = V/R

), applying

16\,\text{V}

across

32.0\,\Omega

yields a current of

0.50\,\text{A}

Step-by-Step Solution

Relate volume conservation to cross-sectional area.

Since volume

V = A_1 L_1 = A_2 L_2

remains constant when

L_2 = 2L_1

, the new cross-sectional area is

A_2 = A_1 / 2

Stretching a wire increases its length while proportionally reducing its area to preserve volume.

Calculate the new resistance of the stretched wire.

The new resistance is

R_2 = \rho \frac{L_2}{A_2} = \rho \frac{2L_1}{A_1 / 2} = 4 R_1 = 4 \times 8.0\,\Omega = 32.0\,\Omega

Resistance is directly proportional to length and inversely proportional to cross-sectional area.

Apply Ohm's law to calculate current.

I = \frac{V}{R_2} = \frac{16\,\text{V}}{32.0\,\Omega} = 0.50\,\text{A}

Electric current is calculated by dividing potential difference by total resistance.

Key Concept

Effect of wire stretching on electrical resistance under volume conservation

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