### Chapter 3: Electricity

Q
##### Electricity

Question:

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

We know,

$R = \frac{\rho L}{A}$

It is given in the question that wire is divided into five equal parts, so after division, the area of cross-section(A) and resistivity(ρ) of the wire does not change only the length of wire will change.

Therefore, the resistance of wire $R \propto L$

After division, the length of each part becomes $\frac{L}{5}$.

So, the resistance of each part is $\frac{R}{5}$.

The equivalent resistance of five parts in parallel combination:

$\frac{1}{R_P} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} + \frac{1}{R_5}$

$\frac{1}{R'} = \frac{1}{\frac{R}{5}} + \frac{1}{\frac{R}{5}} + \frac{1}{\frac{R}{5}} + \frac{1}{\frac{R}{5}} + \frac{1}{\frac{R}{5}}$

$\frac{1}{R'} = \frac{5}{R} + \frac{5}{R} + \frac{5}{R} + \frac{5}{R} + \frac{5}{R}$

$\frac{1}{R'} = \frac{5 + 5 + 5 + 5 + 5}{R}$

$\frac{1}{R'} = \frac{25}{R}$

$\frac{R}{R'} = 25$

So, the correct option is (d).

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