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

given

conductor ,resistance = R

length = l

thickness= d

resistivity = ρ

Solution

ρ= R × (area/length )

When the resistance is cut into 4 equal pieces, then the cross sectional area remains same before and after.

Hence resistivity becomes-

ρ

$= \frac{r}{4} \frac{(area)}{ \frac{l}{4} }$

hence resistivity remains same .

part 1

when all parts are connected in series.

in series

R equivalent = R1 + R2+ R3 +R4

then,

R equivalent = ρ(l/4 area) + ρ(l/4 area)+ρ(l/4 area)+ρ(l/4 area)

= R/4

part 2

when all are connected in parallel

$\frac{1}{r \: eq.} = \frac{1}{r1} + \frac{1}{r2} + \frac{1}{r3} + \frac{1}{r4}$

on solving you get

R equivalent = R/16

part 3

power = I²R

power loss is Directly proportional to resistance of circuit.

hence ,

resistance in series will consume more power

I hope it helps you

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