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The net resistance of the given circuit is ____ $Ω$ .

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Detailed Solution

The net resistance of the given circuit is

2.5 Ω

.
From the circuit, it can be said that the resistors

3 Ω

and

2 Ω

are in a series arrangement. This series arrangement is in parallel combination with the

5 Ω

.
We know that the net resistance of two resistances in series is given by

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

Therefore, the equivalent resistance of the resistors

3 Ω

and

2 Ω

is given by

\Rightarrow R_{S} = 3 + 2

{\Rightarrow R}_{S} = 5 Ω

Also, the net resistance in parallel combination is given by

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

Also, resistor

5 Ω

and

R_{S}

are in parallel arrangement and therefore their equivalent resistance is given by

\Rightarrow \frac{1}{R_{P}} = \frac{1}{5} + \frac{1}{5}

\Rightarrow \frac{1}{R_{P}} = \frac{2}{5}

{\Rightarrow R}_{P} = 2.5 Ω

The equivalent resistance of the given circuit is

2.5 Ω

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