Question

a piece of wire having a resistance R is cut into five equal parts.
(1) how will the resistance of each part of the while compared with the original resistance?
(2)if the five pares of the wire or placed in parallel how will the resistance in the serious to the of parallel?​

Answer 1

Answer:

Wire is cut into 5 equal parts. Since all dimensions are same for pieces, Resistance of each wire is equal and has a value = R/5.

Resistance is proportional to length of wire and inversely proportional to area of cross-section. IF length is 1/5th, then Resistance is also 1/5 th of original value.

When these five are connected in parallel combination, then net resistance becomes 1/5th again. So R1 = R/5 * 1/5 = R/25

Ratio = R1/R = 1/25

Explanation:

hope it will help you

Answer 2

Answer:

(1) Now, the length of the wire L is cut into five equal pieces.

New length= L÷5

We know that R₁=ρ(L/A)...... (1), where R₁ is the resistance of the actual wire.(original resistance)

R₂=ρ{(L÷5)/A}, where R₂ is the resistance of each individual piece of wire.

R₂=ρ(L/5A)

R₂=1/5{ρ(L/A)}...............................(2)

Substitute (1) in (2).

R₂=1/5(R₁).

Thus, the resistance of each wire=1/5(original resistance).

(2) Let the Resistance of parallel wire be Rₐ.

Rₐ=1/R+1/R+1/R+1/R+1/R..............(the resistance of the wires are equal since the length and material is equal)

Rₐ=5/R.

We know that the resistance of each wire=1/5(original resistance).

Thus, Rₐ=5/1/5

Rₐ=5×4

Rₐ=25Ω

Let the Resistance of series wire be Rₓ.

Rₓ=R+R+R+R+R..............(the resistance of the wires are equal since the length and material is equal)

Rₓ=5R.

We know that the resistance of each wire=1/5(original resistance).

Thus, Rₓ=5×1/5

Rₓ=1Ω

According to me, i think u asked the ratio of series resistance to the parallel resistance.(Ur 2nd part of question is not clear.)

So, Rₓ:Rₐ.

1/25

The ratio is 1:25.

HOPE THIS HELPS :D