Question

Two conductors, both having same length and same area of cross-section are joined in series. Their materials
are different and resistivity of one of them is p. If equivalent resistivity of their combination in 3p, resistivity
of the other conductor is :
(1) p
(2) 2p
(3) 4p
(4) 5p

Answer 1

Answer:

4

Explanation:

R' =R1 + R2

P'L'/A = P1L/A + P2L/A

3P(2L)/A = P1L/A + PL/A

6P = P1 + P

P1 = 5P

Answer 2

Answer:

RESISTIVITY BE 5ρ.

Explanation:

Given,

• Area = A
• Length=l
• When Conductors are connected in series:
• Area =2A
• Length =l
• Resistivity of the conductor after combination= 3ρ

RESISTANCE R= ρL/Α

${R_{eq} }=R_{1}}+{R_{2}}$

$\mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{1}+\mathrm{R}_{2} \\\\\frac{(3 \rho)(2 \mathrm{l})}{\mathrm{A}}=\frac{\rho \mathrm{l}}{\mathrm{A}}+\frac{\rho_{1} \mathrm{l}}{\mathrm{A}} \\\rho_{1}=5 \rho$