Question

Two metallic wires A and B are connected in second wire A has length l and radius r, while wire B has length 2l and radius 2r. Find the ratio of total resistance of series combination and the resistance of wire A, if both wires are of same material?​

Answer 1

Wire A

Length = l

Radius =r

Area =πr2

R1= rho x l /π r2

Wire B

Length = 2l

Radius = 2r

Area = π4r2.

R2= rho x 2l/π4r2=rho x l /πx 2r2.

total resistance = 1/Rp= 1/R1+1/R2.= πr2/rho x l+2 πr2/ rho x l.=3 π0r2/rho x l.

Rp= rho x l/ 3 π x r2.

ratio = Rp/ R1= rho /3 pie r2divided by rho x l / pie x r2.

Rp / R1= 1 : 3.

Answer 2

Answer:

3 : 1

Explanation:

Given that

A has length L and radius r

B has length 2l and radius 2r

Now, area of cross section of A would be πr² and that of B would be π(2r)² = 4πr²

Now, since the wires are of same material, their resistivity would be same.

So, we have :-

$\sf R = \rho \frac{l}{\pi r^2}$

and, $\sf R' = \rho \frac{2l}{4\pi r^2}$

Where, R is the resistance of A and R' is the resistance of B.

Now, ratio of R to R' =

$\sf \frac{R}{R'} = \rho\frac{l}{\pi r^2} \times \frac{4\pi r^2}{\rho 2l} \\ \\ \implies \frac{R}{R'} = \frac{4}{2} \\ \\ \implies \frac{R}{R'} = 2 \\ \\ \implies R = 2R'$

Now, Since the wires are in series, their resistance would be added.

→$\sf R_e = R + R' \\ \\ R_e = R + 2R \\ \\ R_e = 3R$

(Since, R' = 2R)

So, total resistance of series combination = 3R and that of wire A = R

So their ratio = 3R : R = 3 : 1