Question

(i) Consider a conductor of resistance ‘R’, length ‘L’, thickness‘d’ and resistivity ‘ρ’.

Now this conductor is cut into four equal parts. What will be the new resistivity of

each of these parts? Why?

(ii) Find the resistance if all of these parts are connected in:

(a) Parallel

(b) Series

(iii) Out of the combinations of resistors mentioned above in the previous part, for a

given voltage which combination will consume more power and why?​

Answer 1

Answer:

i.ρ

ii.

 a).${R_{euq}}=\dfrac{R}{16}$

   b).${R_{euq}}={R}$

iii). Parallel

Explanation:

Given that

Resistance = R

Length = R

Resistivity =ρ

Now this conductor is cut into four equal parts , it means that length will become one forth.

As we know that Resistivity is the property of the material and it does not depends on the dimension on the materials .So when length will become one forth but the Resistivity(ρ) will be remain for all four conductors.

We know that

R= ρL/A

When  conductor is cut into four equal parts

R'= ρL'/A

L'=L/4

So

R'=R/4

Parallel   :

We know that equivalent resistance for  Parallel  given as

$\dfrac{1}{R_{euq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+\dfrac{1}{R_4}$

$\dfrac{1}{R_{euq}}=\dfrac{4}{R}+\dfrac{4}{R}+\dfrac{4}{R}+\dfrac{4}{R}$

${R_{euq}}=\dfrac{R}{16}$

Series :

${R_{euq}}=R_1+R_2+R_3+R_4$

${R_{euq}}=\dfrac{R}{4}+\dfrac{R}{4}+\dfrac{R}{4}+\dfrac{R}{4}$

${R_{euq}}={R}$

We know that power supply given as

P=V²/R

If R is low the power consumption will be high.

So for parallel connection R is low,the power consumption for parallel connection will be more.

Answer 2

(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, Reqv = R/16 Ω

(b) In series, Reqv = R Ω

(iii) 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.