Question

Question 1 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’, then the ratio R/R’ is : (a) 1 25 (b) 1 5 (c) 5 (d) 25​

Answer 1

When we cut the wire into 5 parts, each one has a length of l5 and all other parameters remain unchanged. So the resistance of each new segment isR5. Connecting them parallel gives a network as shown in fig1. So the ratio RR′=25 , which is the required answer.

Answer 2

Answer:

Answer is d) 25

Complete step by step answer:

We know the resistance of a wire is proportional to the length of the wire (∵R=ρlA)

Let’s say the total resistance of the wire before cutting wasR.

When we cut the wire into 5 parts, each one has a length of l5 and all other parameters remain unchanged.

So the resistance of each new segment isR5.

Connecting them parallel gives a network

Now since the question asks for The effective resistance of this combination, lets recollect that the effective resistance of a parallel combination of n resistances - R1,R2,R3, . . . Rn is given by: 1Reff=1R1+1R2+...+1Rn

So if R′ is the resistance of this new combination, then:

1R′=1R5+1R5+1R5+1R5+1R5

Simplifying this expression a bit gives us:

1R′=5R5

R′=R25

So the ratio RR′=25 , which is the required