Question

Consider the two cylindrical wires of lengths 41=3.8m and 1=2.2m are merged back to back as shown in the figure below to form a single wire. The wires have the radii of r1=1mm and r2=9mm and the current runs along the lengths of the wires. If the thin and the thick wires resistivities are $1=108x10-60.m and P2=337x10 60.m respectively, determine the resistance of the wire. Please express your answer in units of using one decimal places. l2 li 11 12

Answer 1

Answer- R= 157 $\Omega$

A1 = 7 *ram=7*(1 * 10-3)2 = 3.14 * 10-6m?

$A_2 = \pi *r_2^{2}m= \pi *(9*10^{-3})^{2}=2.83*10^{-5} m^{2}$

$R=R_1+R_2=\frac{\rho_1 *L_1}{A_1}+\frac{\rho_2 *L_2}{A_2}$

$R=\frac{108*10^{-6} *3.8}{3.14*10^{-6}}+\frac{337*10^{-6} *2.2}{2.83*10^{-5}}$

$R=156.9 \Omega$

$R=157 \Omega$

Answer 2

SOLUTION :

Resistance is directly proportional to length of wire and inversely proportional to cross sectional area of wire. The constant of proportionality (p) is termed as resistivity of wire with respect length and area.

So,

R1 

= p1 *  l1 / A1

= 108 * 10^(-6) * 3.8 / (pi * (1*10^(-3))^2)

= 130.6344 Ω 

R2 

= p2 *  l2 / A2

= 337 * 10^(-6) * 2.2 / (pi * (9*10^(-3))^2)

= 2.9135 Ω 

R1 an R2 are in series.

So,

Total resistance of the joint system, R

= R1 + R2

= 130.6344 + 2.9135

= 133.5479

= 133.5  Ω (ANSWER)