91影视

We will first find the resistance of each wire using the formula: $$ R=\frac{\rho*L}{A} $$, where R is the resistance, 蟻 (rho) is the resistivity of copper, L is the length of the wire, and A is the cross-sectional area of the wire. We are given that the thicknesses of the wires are 1 mm and 2 mm, so the cross-sectional areas will be: $$ A_1 = \pi (0.5 \times 10^{-3})^2 $$ $$ A_2 = \pi (1 \times 10^{-3})^2 $$ Let's assume that both wires have the same mass m. We know that mass = volume 脳 density, and volume = area 脳 length, so: $$ m = A_1*L_1*\rho_c = A_2*L_2*\rho_c $$ Here, 蟻_c is the density of copper (which is a constant), so it cancels out: $$ \frac{A_1}{A_2} = \frac{L_2}{L_1} $$ Since $$ \frac{A_2}{A_1} = 4 $$ We conclude that $$ \frac{L_1}{L_2} = 4 $$ Now, we can calculate the resistance of both wires, using the given resistivity of copper 蟻, and substituting the cross-sectional areas and lengths into the resistance formula: $$ R_1 = \frac{\rho*L_1}{A_1} = 4 *\rho*\frac{L_1}{\pi*(0.5\times10^{-3})^2} $$ $$ R_2 = \frac{\rho*L_2}{A_2}= \rho*\frac{L_1}{\pi*(1 \times 10^{-3})^2} $$

Now, we will use the heat produced formula \( H=I^2*R*t \) for each wire: $$ H_1 = I^2*R_1*t = 4*I^2*\rho*L_1*\frac{t}{\pi*(0.5 \times 10^{-3})^2} $$ $$ H_2 = I^2*R_2*t = I^2*\rho*L_1*\frac{t}{\pi*(1 \times 10^{-3})^2} $$

Finally, we will find the ratio of heat produced in both wires: $$ \frac{H_1}{H_2} = \frac{4*I^2*\rho*L_1*\frac{t}{\pi*(0.5 \times 10^{-3})^2}}{I^2*\rho*L_1*\frac{t}{\pi*(1 \times 10^{-3})^2}} $$ Some terms get canceled out, and we have: $$ \frac{H_1}{H_2} = \frac{4}{1} $$ Thus, the heat produced in the wires is in the ratio 4:1, which corresponds to option (C).