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CBSE Class 12 Physics 2020 Delhi Set 1 Paper

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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 LL, area of crosssection A1A_1 and A2A_2 having resistivity ρ1\rho_1 and ρ2\rho_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\rho
 Length, cross-section area of conductor and temperature  Depends on temperature and material of the conductor
 (b) For resistor R1\text{ (b) For resistor } R_1
 Resistivity =ρ1\text{ Resistivity }=\rho_1
 Length =L\text{ Length }=L
 Area =A1\text{ Area }=A_1
∴R1=ρ1×LA1\therefore R_1=\rho_1\times \frac{L}{A_1}
 For resistor R2\text{ For resistor } R_2
 Resistivity =ρ2\text{ Resistivity }=\rho_2
 Length =L\text{ Length }=L
 Area =A2\text{ Area }=A_2
∴R2=ρ2×LA2\therefore R_2=\rho_2\times \frac{L}{A_2}
For the equivalent resistor
 Resistivity =ρ\text{ Resistivity }=\rho
 Length =L\text{ Length }=L
 Area =A=A1+A2\text{ Area }=A=A_1+A_2
∴R=ρ×LA\therefore R=\rho\times \frac{L}{A}
Since the resistors are connected parallel
Equivalent resistance =R=R1R2R1+R2=R=\frac{R_1 R_2}{R_1+R_2}
Or, ρ×LA=ρ1LA1×ρ2LA2ρ1LA1+ρ2LA2\rho \times \frac{L}{A}=\frac{ \frac{\rho_1 L}{A_1} \times \frac{\rho_2 L}{A_2} }{ \frac{\rho_1 L}{A_1} + \frac{\rho_2 L}{A_2} }
∴ρ=\therefore \rho = Effective resistivity =ρ1ρ2(A1+A2)ρ1A2+ρ2A1=\frac{\rho_1 \rho_2 (A_1+A_2)}{\rho_1 A_2+\rho_2 A_1}
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