A wire of resistance $R_{1}$ is cut into five equal pieces. These five pieces of wire are then connected in parallel.
If the resultant of this combination be $R_{2}$ then the ratio $\frac{R_{1}}{R_{2}}$ is:
(a) $\frac{1}{25}$ (b) $\frac{1}{5}$ (c) 5 (d) 25

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Solution

We know that the resistance of a piece of wire is proportional to its length.

As the wire of resistance $R_{1}$ is cut into 5 equal parts, the resistance of each part, $R^{'} = \frac{R_{1}}{5}$

Now, these five parts are connected in parallel.

So, $\frac{1}{R_{2}} = \frac{5}{R^{'}} = \frac{25}{R_{1}}$

$\Rightarrow R_{2} = \frac{R_{1}}{25}$

$∴ \frac{R_{1}}{R_{2}} = 25$

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Similar questions

Q. A wire of resistance

R_{1}

is cut into

5

equal pieces. The

5

pieces of wire are then connected in parallel. If the resultant resistance of this combination is

R_{2}

. Then, find

\frac{R_{1}}{R_{2}}

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 $\frac{R}{R^{'}}$ is:

(a) $\frac{1}{25}$
(b) $\frac{1}{5}$
(c) 5
(d) 25

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A wire of resistance R1 is cut into five equal pieces. These five pieces of wire are then connected in parallel. If the resultant of this combination be R2 then the ratio R1R2 is: (a) 125 (b) 15 (c) 5 (d) 25

We know that the resistance of a piece of wire is proportional to its length. As the wire of resistance R1 is cut into 5 equal parts, the resistance of each part, R′=R15 Now, these five parts are connected in parallel. So, 1R2=5R′=25R1 ⇒R2=R125 ∴R1R2=25

A wire of resistance $R_{1}$ is cut into five equal pieces. These five pieces of wire are then connected in parallel.
If the resultant of this combination be $R_{2}$ then the ratio $\frac{R_{1}}{R_{2}}$ is:
(a) $\frac{1}{25}$ (b) $\frac{1}{5}$ (c) 5 (d) 25