Question

Two uniform wires A and B are of same metal have equal masses. The radius of wire A is twice that of wire B. The total resistance of A and B when connected in parallel is

• A. $4\Omega$ when the resistance of wire A is $4.25\Omega$
• B. $5\Omega$ when the resistance of wire A is $4\Omega$
• C. $4\Omega$ when the resistance of wire B is $4.25\Omega$
• D. $5\Omega$ when the resistance of wire B is $4\Omega$

Answer 1

The correct option is A $4\Omega$ when the resistance of wire A is $4.25\Omega$

Given: Two uniform wires A and B are of same metal have equal masses. The radius of wire A is twice that of wire B.

To find the total resistance of A and B when connected in parallel

Solution:

We know, mass=density $\times$ volume

$⟹m=d\times \pi r_A^2{l}_A=d\times \pi r_B^2{l}_B⟹l_A={\left(\frac{r_B}{r_A}\right)}^2{l}_B={\left(\frac{r_B}{2r_B}\right)}^2{l}_B⟹l_A=\frac{l_B}{4}$

Now substituting this value in equation

$R=\rho \frac{l}{A}⟹R_A=\rho \frac{l_A}{\pi r_A^2},R_B=\rho \frac{l_B}{\pi r_B^2}⟹R_A=\rho \frac{l_A}{\pi r_A^2},R_B=\rho \frac{4l_A}{\pi {\left(\frac{r_A}{2}\right)}^2}=\rho \frac{16l_A}{\pi r_A^2}$

or, $R_B=16R_A$

Hence, $R_{total}=\frac{R_A{R}_B}{R_A+R_B}=\frac{R_A\times 16R_A}{R_A+16R_A}⟹R_{total}=\frac{4.25\times 16R_A}{17R_A}=4\Omega$

is the total resistance of A and B when connected in parallel.