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

i) New resistivity:

The conductor is divided into four equal parts and there is no change in length or thickness of the conductor.

The resistivity depends on the nature and size of the conductor. Thus, the resistivity remains the same.

ii) Combination:

Length of each part = L/4

Resistance of each part = (ρL/4)/A

Now, the resistance of the each part = R/4 = R1 = R2 = R3 = R4

(a) Parallel:

1/Reqv = 1/R1 + 1/R2 + 1/R3 + 1/R4

1/Reqv = 4/R + 4/R + 4/R + 4/R

1/Reqv = 16/R

∴ Reqv = R/16 Ω

(b) Series:

Reqv = R1 + R2 + R3 + R4

Reqv = R/4 + R/4 + R/4 + R/4

∴ Reqv = R Ω

iii) Power consumption:

The power and resistance relationship is given as:

P = V²/R

∴ P ∝ 1/R

Since, the resistance in parallel combination is less. Thus, the power consumed in parallel combination is more.

Answer 2

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