Question

A piece of wire having a resistance R is cut into five equal parts. (i) How will the resistance of each part of the wire compare with the original resistance? (ii) If the five parts of the wire are placed in parallel, how will the resistance of the combination compare with the resistance of the original wire? What w ill be ratio of resistance in series to that of parallel?

Answer 1

i) R = R₀/ 5 , ii) Req = R₀ / 25, iii) R_series / R_parallel = 25

Step-by-step explanation:

i) The resistivity of a metal is constant, therefore resistance is

R = ρ L / A

where we can see that cutting the resistance in 5 parts each part has

R = R₀/ 5

where R ose the original resistance.

ii) The equivalent resinification of a parallel resistance is

1 / Req = ∑ 1 / Ri

1 / Req = 5 1/ (R₀ / 5)

1 / Req = 25 / R₀

Req = R₀ / 25

to buy these values ​​we find their relationship

R₀ /
`R_(paralel)` = 25

iii) the series resistance is

Req = 5 Ri

q = 5 (Ro / 5)

Req = Ro

the relationship between resistance in being and parallel is

R_serie / R_parallel = R₀ / (R₀ / 25)

R_series / R_parallel = 25