Eight copper wires of length l and diameter d | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The resistance of a single wire of length $l$ and diameter $d$ is given by $r = \rho \frac{l}{A} = \rho \frac{l}{\pi (d/2)^2} = \rho \frac{4l}{\pi

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) The resistance of a single wire of length $l$ and diameter $d$ is given by $r = \rho \frac{l}{A} = \rho \frac{l}{\pi (d/2)^2} = \rho \frac{4l}{\pi d^2}$.When $8$ such wires are connected in parallel,the equivalent resistance $R$ is given by $R = \frac{r}{8} = \frac{1}{8} \left( \rho \frac{4l}{\pi d^2} \right) = \frac{\rho l}{2 \pi d^2}$.For a single wire of length $2l$ and diameter $d_1$ to have the same resistance $R$,we have $R = \rho \frac{2l}{\pi (d_1/2)^2} = \rho \frac{8l}{\pi d_1^2}$.Equating the two expressions for $R$: $\frac{\rho l}{2 \pi d^2} = \frac{8 \rho l}{\pi d_1^2}$.Simplifying,we get $\frac{1}{2 d^2} = \frac{8}{d_1^2}$,which implies $d_1^2 = 16 d^2$.Taking the square root,we find $d_1 = 4d$.

Eight copper wires of length $l$ and diameter $d$ are joined in parallel to form a single composite conductor of resistance $R$. If a single copper wire of length $2\,l$ has the same resistance $R$,then its diameter will be $.....d$.

Similar Questions

Find the equivalent resistance between points $A$ and $B$.

In the following star circuit diagram,the equivalent resistance between the points $A$ and $H$ will be ..............$r$.

$A$ wire of length $l$ and resistance $100 \Omega$ is divided into $10$ equal parts. The first $5$ parts are connected in series while the next $5$ parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is: (in $\Omega$)

In the network shown in the adjoining figure,each resistance is $1\,\Omega$. The effective resistance between $A$ and $B$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module