Question

A wire of resistance ‘R’ is cut into five equal parts. Find the ratio of the net resistance in series and parallel when these five parts are connected .

Answer in a proper way . i dont need the type of answer given before​

Answer 1

Answer :

The ratio of the net resistance in series and parallel when these five parts are connected = 25

Explanation :

A wire of resistance ‘R’ is cut into five equal parts.

we know,

Resistance of the wire is directly proportional to the length of the wire

R ∝ l

Since, the wire is cut into five equal parts,

Resistance of each piece of wire, R' = R/5 Ω

When connected in series,

equivalent resistance,

Rₛ = R₁ + R₂ + R₃ + R₄ + R₅

Rₛ = R' + R' + R' + R' + R'

 Rₛ = 5R'

 Rₛ = 5(R/5)

 Rₛ = R

Net resistance when connected in series = R

When connected in parallel,

equivalent resistance,

1/Rₚ = 1/R₁ + 1/R₂ + 1/R₃ + 1/R₄ + 1/R₅

1/Rₚ = 1/R' + 1/R' + 1/R' + 1/R' + 1/R'

1/Rₚ = 5/R'

Rₚ = R'/5

Rₚ = (R/5)/5

Rₚ = R/25

Net resistance when connected in parallel = R/25

Net resistance in series/Net resistance in parallel = ?

Rₛ/Rₚ = R/(R/25)

Rₛ/Rₚ = 25

The ratio of the net resistance in series and parallel when these five parts are connected = 25