A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R, the ratio R/R is

Options

A. `(1)/(25)`
B. `(1)/(5)`
C. 5
D. 25

Answer

B

Explanation

Let length of the wire is L and cross-sectional area is A.
Initially, value of its resistance is R is
`R = rho ((L)/(A)) " "…..(i)`
Now, it is cut into 5 equal parts, so effective length of each part is now L/5.
Now , effective resistance of each part , r is
` r = rho ((L)/(5A)) " ".....(ii)`
Now , these parts are connected in parallel , their effective resistance R will be
`(1)/(R) = (1)/(r) + (1)/(r) + (1)/(r) + ( 1)/(r) + (1)/(r) + (1)/(r)`
`rArr R = (r)/(5)" ".......(iii)`
From Eqs. (ii) and (iii) , we get
`R. = (rho (L)/(5A)) xx (1)/(5) = (1)/(25) (rho(L)/(A))`
`rArr " " R = (1)/(25) (R)` [from Eq.(i)]
`rArr (R)/(R) = 25`