Question

A metallic wire of resistance R is cut into ten parts of equal lengths . Two pieces are joined in parallel .What is the ratio of resistance of that combination with R

Answer 1

Answer: The resistance of a conductor is directly proportional to the length of the conductor. The resistance of the metallic wire, when it is cut into ten parts of equal length is

r=R/10

Two such pieces when joined in series, the equivalent resistance of these two parts

= r+r=2r=2R/10

series resistance of two parts=2xR/10=R/5,

5 such elements are connected in parallel .Therefore the total resistance R’ will be

1/R’=R/25

Answer 2

Answer :

• Ratio of resistance of that combination with R is 1 : 20

Explanation :

Let, initial length of wire be L , area of cross section be A , resistivity of wire be ρ and, given resistance of wire is R

then,

• R = ρ L / A ____equation (1)

Now,

when Wire is cut into 10 equal parts therefore,

new length of each part will be L/10 , area of cross section of each part will be A , resistivity will be ρ and, Let new resistance of each part be R₁

then,

• R₁ = ρ L / (10 A)
• R₁ = (1/10) ( ρ L / A )

Using equation (1)

• R₁ = R / 10 _____equation (2)

Now,

Two pieces among the 10 parts are connected in parallel

therefore, equivalent resistance of two parts will be

• 1 / R(eq) = 1/R₁  + 1/R₁
• 1 / R(eq) = 2 / R₁
• R(eq) = R₁ / 2

Using equation (2)

• R (eq) = ( R / 10 ) / 2
• R (eq) = R / 20  ______equation (3)

Now, we need to find the ratio of resistance of that combination with R

• R(eq) / R

Using equation (3)

• R(eq) / R = (R / 20) / R
• R (eq) / R = 1 / 20
• R(eq) : R = 1 : 20

Therefore,

Ratio of resistance of combination with R is 1 : 20 .