A uniform metallic wire of resistance R is stretched uniformly until its radius decreases … — JAMB UTME

A uniform metallic wire of resistance $R$ is stretched uniformly until its radius decreases by $20\%$ . The stretched wire is subsequently cut into two equal halves, which are then connected in parallel across a constant potential difference $V$ . What is the ratio of the total electrical power dissipated in this parallel combination to the power dissipated by the original unstretched wire under the same potential difference?

1.64Answer
B
0.61
C
2.56
D
0.41

Answer

The ratio of the total power dissipated in the parallel combination to the original power is 1.64

When a wire of initial resistance

R

is stretched so that its radius decreases by

20\%

, its new radius is

0.8r

. Because volume is conserved (

A_1 L_1 = A_2 L_2

), reducing the cross-sectional area to

0.64A

causes the length to increase to

L/0.64

. Consequently, the resistance scales inversely with the fourth power of the radius:

R_{\text{stretched}} = R / (0.8)^4 = R / 0.4096 = 2.4414R

. Cutting this wire into two equal pieces gives two resistors of

1.2207R

each. Connecting them in parallel yields an equivalent resistance

R_{\text{eq}} = 1.2207R / 2 = 0.61035R

. Power at constant voltage is

P = V^2/R

, so the new power is

P_{\text{new}} = V^2 / (0.61035R) = 1.64 (V^2/R) = 1.64 P_{\text{orig}}

Step-by-Step Solution

Determine the new resistance of the wire after stretching

R_{\text{stretched}} = \frac{R}{(0.8)^4} = \frac{R}{0.4096} \approx 2.4414 R

Since mass and density remain constant, volume

V_{\text{vol}} = A \cdot L

is conserved. Decreasing radius to

r_2 = 0.8 r_1

reduces area to

A_2 = 0.64 A_1

and increases length to

L_2 = L_1 / 0.64

. Resistance

R = \rho L / A \propto 1/r^4

Calculate the equivalent resistance of the two equal halves connected in parallel

R_{\text{eq}} = \frac{1}{4} R_{\text{stretched}} = \frac{2.4414 R}{4} \approx 0.61035 R

Cutting the stretched wire in half gives two pieces each of resistance

R_{\text{half}} = R_{\text{stretched}} / 2

. Connecting two identical resistors in parallel yields an equivalent resistance

R_{\text{eq}} = R_{\text{half}} / 2 = R_{\text{stretched}} / 4

Calculate the ratio of power dissipated across a constant potential difference V

\frac{P_{\text{new}}}{P_{\text{orig}}} = \frac{V^2 / R_{\text{eq}}}{V^2 / R} = \frac{R}{R_{\text{eq}}} = \frac{1}{0.61035} \approx 1.64

Electrical power dissipated at constant voltage is given by

P = V^2 / R

, which means power is inversely proportional to equivalent resistance.

Key Concept

Dependence of electrical resistance on conductor geometry under volume conservation, and power dissipation in parallel circuits.

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic