Wires A and B have resistivities rho_A and rh | vedclass

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(A) Given: $\rho_B = 2 \rho_A$ and $R_A = R_B = R$. Let the radius of wire $A$ be $r_A = r$. Since the diameter of $B$ is twice that of $A$,the radius

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(A) Given: $\rho_B = 2 \rho_A$ and $R_A = R_B = R$. Let the radius of wire $A$ be $r_A = r$. Since the diameter of $B$ is twice that of $A$,the radius of wire $B$ is $r_B = 2r$. The resistance of a wire is given by $R = \rho \frac{l}{A} = \rho \frac{l}{\pi r^2}$. Since $R_A = R_B$,we have: $\rho_A \frac{l_A}{\pi r_A^2} = \rho_B \frac{l_B}{\pi r_B^2}$ Substituting the given values: $\rho_A \frac{l_A}{\pi r^2} = (2 \rho_A) \frac{l_B}{\pi (2r)^2}$ $\rho_A \frac{l_A}{r^2} = 2 \rho_A \frac{l_B}{4r^2}$ $l_A = \frac{2}{4} l_B$ $l_A = \frac{1}{2} l_B$ Therefore,$\frac{l_B}{l_A} = 2$.

Wires $A$ and $B$ have resistivities $\rho_A$ and $\rho_B$,where $\rho_B = 2 \rho_A$,and have lengths $l_A$ and $l_B$. If the diameter of wire $B$ is twice that of $A$ and the two wires have the same resistance,then the ratio $\frac{l_B}{l_A}$ is:

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