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A wire of resistance $20 Ω$ is cut into two identical parts and then connected in parallel. The equivalent resistance is [[1]] $Ω .$

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Detailed Solution

A wire of resistance

20 Ω

is cut into two identical parts and then connected in parallel. The equivalent resistance is

5 Ω .

Since the resistance of a wire is directly proportional to its length as length gets half therefore, the resistance of each identical part become

10 Ω .

As the resistances are connected in parallel thus, the equivalent resistance

R

, can be calculated by the equation, identical parts are connected in parallel, can be found as:

\frac{1}{R_{P}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}

Let

R_{1} = 10 Ω

R_{2} = 10 Ω

= > \frac{1}{R_{P}} = \frac{1}{10} + \frac{1}{10}

= > \frac{1}{R} = \frac{2}{10}

= > R = 5 Ω

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