SECTION - C (a) Differentiate between electrical resistance and resistivity of a conductor. (b) Two metallic rods, each of length {\L}, area of crosssection A_1 and A_2 having resistivity \rho_1 and \rho_2 are connected in parallel across a d.c. battery. Obtain the expression for the effective resistivity of this combination.

CBSE Class 12 Physics 2020 Delhi Set 1 Paper - Question 28

Question : 28 of 37

Marks: +1, -0

SECTION - C

(a) Differentiate between electrical resistance and resistivity of a conductor.
(b) Two metallic rods, each of length

{\L}

, area of crosssection

A_1

and

A_2

having resistivity

ρ_1

and

ρ_2

are connected in parallel across a d.c. battery. Obtain the expression for the effective resistivity of this combination.

Solution:

(a)

Electrical resistance	Resistivity
It is the property of material due to which it opposes the flow of electricity through the conductor	The resistivity is defined as the resistance of a material of 1 metre length and 1 square metre area of cross section.
Unit: Ohm	Unit: Ohm-Meter
Symbol: R	Symbol: $\ρ$
Length, cross-section area of conductor and temperature	Depends on temperature and material of the conductor

\;\;\text " (b) For resistor "\; R_1

\;\;\text " Resistivity "\;=ρ_1

\;\;\text " Length "\;=L

\;\;\text " Area "\;=A_1

\; ∴ R_1=ρ _1× \;{L}/{A_1}

\;\text " For resistor "\; {\R}_2\;

\;\text " Resistivity "\;\;=ρ_2

\;\text " Length "\;\;= {\L}

\;\text " Area "\;\;= {\A}_2

∴ \;\; {\R}_2\;=ρ_2 × \;{L}/{ {\A}_2}

For the equivalent resistor

\;\text " Resistivity "\;\;=ρ

\;\text " Length "\;\;= {\L}

\;\text " Area "\;\;= {\A}={\A}_1+ {\A}_2

∴ \;\; {\R}\;=ρ × \;{ {\L}}/{ {\A}}

Since the resistors are connected parallel
Equivalent resistance

=R=\;{R_1 R_2}/{R_1+R_2}

Or,

\;\; ρ × \;{L}/{A}=\;{\;{ρ_1 L}/{A_1} × {ρ_2 L}/{A_2}}/{{ρ_1 L}/{A_1}+\;{ρ_2 L}/{A_2}}

∴ ρ=

Effective resistivity

=\;{ρ_1 ρ_2 (A_1+A_2) }/{ρ_1 A_2+ρ_2 A_1}

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