Question

A PIECE OF wire of resistance R us cut into 5 equal parts. these parts are then connected in parallel. if the equivalent resistance of this combination is R', then the ration R/R' is

Answer 1

Given that, a wire whose resistance is R ohm is cut into five pieces of equal parts.

We have to find the resistance of the combination. (means Req = R')

Assume a long cylindrical wire, which is cut into five equal pieces. Where the resistance of wire is R ohm, which is cut into five equal pieces then it's resistance = R/5 ohm

Also, given that all the five pieces of wire are connected in parallel. So,

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

R1 = R2 = R3 = R4 = R5 = R/5

Substitute value of R1, R2, R3, R4 and R5 in the above formula,

1/Rp = 1/(R/5) + 1/(R/5) + 1/(R/5) + 1/(R/5) + 1/(R/5)

1/Rp = 5/R + 5/R + 5/R + 5/R + 5/R

1/Rp = (5 + 5 + 5 + 5 + 5)/R

1/Rp = 25/R

Rp = R/25 ohm

Now,

R' = 1/R1 + 1/R2 + 1/R3 + 1/R4 + 1/R5

R' = 1/Rp

R' = Rp

R' = R/25

R'/R = 1/25

R/R' = 25/1

Therefore, the ratio of R:R' is 25:1.

Answer 2

GIVEN:

• Resistance of wire = R Ω
• Wire cut into 5 equal parts
• Parts are connected in parallel
• Equivalent resistance of this combination is R'

TO FIND:

• R : R'

SOLUTION:

Given , a wire of resistance R cut into 5 equal parts

Resistance of total wire → RΩ

Resistance when cut into 5 parts → R/5 Ω

Assuming , resistance of each part

1st part : R₁

2nd part : R₂

3rd part : R₃

4th part : R₄

5th part : R₅

Also given that all parts are in parallel so,

1/Rp = 1/R₁ + 1/R₂ + 1/R₃ + 1/R₄ + 1/R₅…… ( i )

R₁ = R₂ = R₃ = R₄ = R₅ = R/5

Now, substituting R₁ , R₂ , R₃ , R₄ , R₅ = R/5 in eq(i)

1/Rp = 1/(R/5) + 1/(R/5) + 1/(R/5) + 1/(R/5) + 1/(R/5)

1/Rp = 5 + 5 + 5 + 5 + 5/R = 5(5)/R

1/Rp = 25/R

Rp = R/25 Ω

Now,

R' = 1/R₁ + 1/R₂ + 1/R₃ + 1/R₄ + 1/R₅

R' = 1/Rp

R' = R/25

→ R : R' = 25 : 1 ( Answer )