Question

A conductor of Resistance 20 Ω has a length ‘L’, thickness ‘d’ and Resistivity ‘ρ’. Now this conductor is cut into 4 equal parts. a) How should these parts be connected such that their equivalent resistance becomes 12.5 Ω? b) What will be new resistivity of this new combination? Why?

Answer 1

Answer:

YOUR ANSWER

Explanation:

i Resistivity will not change as it depends on the nature of the material of the conductor. ii The length of each part becomes L/4. ρ A constant. R=ρL/A. Resistance of each part = Rpart= ρL/4/A = R/4.a  In parallel the 1/Reqv = 1/Reqv +1/Reqv +1/Reqv +1/Reqv = 4/Reqv = 16R →Reqv = R/16 Ωb In series the Reqv= R/4+ R/4 +R/4+R/4=R Ωiii P=V2 /R. If Reqv is less power consumed will be more. In the given case  Reqv is lesser in the parallel and power consumed will be more

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Answer 2

Given that, A conductor of resistance 20 Ω has a length L , thickness d and Resistivity ρ.

Also, The conductor is divided into 4 equal parts, So the resistance of each part will be 5 Ω because the length, thickness and resistivity of each part would be the same.

a)

First, We have 4 resistors of resistance 5 Ω ,

Also, If we combine 2 resistors in series we will get equivalent resistance as 10 Ω but we need 2.5 Ω more, So we will combine the other two resistors in parallel, where the resultant resistant will be 2.5 Ω

So, The total resistance of the circuit

will be 12.5 Ω.

b)

No, The resistivity of the conductor won't change, because resistivity depends on the nature of material. here, The nature of material remains the same hence the resistivity will also be the same.