A circuit consists of a copper wire of length 10 m and radius 1 mm.The wire is connected to a 10-V battery.An aluminum wire of length 5 m is connected to the same battery and dissipates the same amount of power.What is the radius of the aluminum wire?
A)0.91 mm
B)0.84 mm
C)1.12 mm
D)0.45 mm
E)0.65 mm

Question

Quizplus · Accepted Answer

The power dissipated in a wire is given by $P = \frac{V^2}{R}$, where $V$ is the voltage across the wire and $R$ is the resistance of the wire. The resistance of a wire is given by $R = \frac{\rho L}{A}$, where $\rho$ is the resistivity of the material, $L$ is the length of the wire, and $A$ is the cross-sectional area of the wire. For a wire with a circular cross-section, $A = \pi r^2$, where $r$ is the radius of the wire.Given that both wires dissipate the same amount of power when connected to the same battery, we can set up the equation based on their resistances being proportional to their power dissipation, considering their materials' resistivities ($\rho_{Cu}$ for copper and $\rho_{Al}$ for aluminum), lengths ($L_{Cu}$ and $L_{Al}$), and radii ($r_{Cu}$ and $r_{Al}$).Given:- $L_{Cu} = 10 m$, $r_{Cu} = 1 mm = 0.001 m$- $L_{Al} = 5 m$, $V = 10 V$- $\rho_{Cu} = 1.68 \times 10^{-8} \Omega \cdot m$, $\rho_{Al} = 2.82 \times 10^{-8} \Omega \cdot m$Since the power dissipated is the same, we have:$$\frac{V^2}{R_{Cu}} = \frac{V^2}{R_{Al}}$$Substituting $R = \frac{\rho L}{\pi r^2}$, we get:$$\frac{\rho_{Cu} L_{Cu}}{\pi r_{Cu}^2} = \frac{\rho_{Al} L_{Al}}{\pi r_{Al}^2}$$Solving for $r_{Al}$, we find:$$r_{Al} = r_{Cu} \sqrt{\frac{\rho_{Al} L_{Al}}{\rho_{Cu} L_{Cu}}}$$Plugging in the given values:$$r_{Al} = 0.001 m \sqrt{\frac{2.82 \times 10^{-8} \Omega \cdot m \cdot 5 m}{1.68 \times 10^{-8} \Omega \cdot m \cdot 10 m}}$$$$r_{Al} \approx 0.001 m \sqrt{\frac{2.82}{1.68} \cdot \frac{1}{2}}$$$$r_{Al} \approx 0.001 m \sqrt{0.839} \approx 0.00091 m = 0.91 mm$$Therefore, the radius of the aluminum wire is approximately 0.91 mm.

A Circuit Consists of a Copper Wire of Length 10