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

i) Resistivity will remain same.

ii) a) $\sf{Requ=}$$\frac{R}{16}$

b)$\sf{Requ=R}$

iii)Parallel connection consume more power because it have low equivalent resistance.

$\bold\red{Given:}$

$\sf{Resistance =R}$

$\sf{Length =L}$

$\sf{Thickness=d}$

$\sf{Resistivity =ρ}$

Now when wire cuts in 4 equal parts

L'= L/4

• We know that resistivity is a material properties and does not depends on the physical dimensions.So even wire cuts in 4 equal but still resistivity will remain same.

$\sf{So~ the ~answer ~is,~ resistivity~ will~ be~ same.}$

$\sf{We~ know ~that}$

$\sf{R = ρ L/A}$

When L become 1/4 th then resistance for each part is R'

$\sf{R' = ρ L'/A }$

$\sf{L'= L/4}$

$\sf{R' = ρ L/4A}$

$\sf{R'= R/4}$

• It means that resistance will become one forth of initial resistance

a) Parallel connection:

We know that equivalent resistance given as:

$\frac{1}{R_{equ}=}$ $\frac{1}{R_1}+$+$\frac{1}{R_2}$+$\frac{1}{R_3}$+$\frac{1}{R_4}$

$\sf{R_1=R_2=R_3=R_4=}$$\frac{R}{4}$

$\frac{1}{R_{equ}=}$$\frac{4}{R}$+$\frac{4}{R}$+$\frac{4}{R}$+$\frac{4}{R}$

$\sf{R_{equ}=}$$\frac{R}{4}$

b)Series connection :

$\sf{R_1=R_2=R_3=R_4}$

$\sf{R_1=R_2=R_3=R_4=}$$\frac{R}{4}$

$\sf{R_{equ}=R}$

• We know that Power P given as:

$\sf{P=V.I =V²/R}$

$\sf{( V= IR)}$

.°. For given voltage parallel connection consume more power because it have low equivalent resistance.

Answer 2

${\fbox{\huge\sf{\pink{An}\purple{sW}\green{eR}}}}$

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.