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If a wire of resistance $20 Ω$ is bent to form a closed square, the resistance is ____ $Ω$ across a diagonal of the square.

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Detailed Solution

If a wire of resistance

20 Ω

is bent to form a closed square, the resistance is

5 Ω

across a diagonal of the square.
It is given that the resistance of the wire is

20 Ω

.
The number of sides in a square is

4 .

Therefore, the resistance of a side of the square is

\Rightarrow R = \frac{20}{4}

\Rightarrow R = 5 Ω

The resistance of a single side is

5 Ω

Now, side AB and BC are in a series combination. Also, side AD and DC are in series combination. We know that,
the equivalent resistance of resistances connected in a series combination is given by

R_{s} = R_{1} + R_{2}

Therefore, their equivalent resistance Rs would be

\Rightarrow R_{s} = 5 Ω + 5 Ω

\Rightarrow R_{s} = 10 Ω

Now, both 10

Ω

resistances are in parallel.
Also, the net resistance in parallel combination is given by

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

Therefore, the equivalent resistance across a diagonal is

\Rightarrow \frac{1}{R_{p}} = \frac{1}{10} + \frac{1}{10}

Ω

\Rightarrow \frac{1}{R_{p}} = \frac{2}{10} Ω

\Rightarrow \frac{1}{R_{p}} = \frac{1}{5} Ω

\Rightarrow R_{p} = 5 Ω

Hence, the equivalent resistance across a diagonal of the square is

5 Ω

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