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Three resistors are connected in parallel such that their total resistance is $3 Ω$ . If the resistance of the two resistors is $10 Ω$ and $30 Ω$ respectively, the resistance of the third resistor is [[1]] $Ω$ .

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Detailed Solution

The resistance of the third resistor is

5 Ω

.
Given,
the total resistance

R = 3 Ω

R_{1} = 10 Ω

R_{2} = 30 Ω

R_{3} = ?

Here,

R_{1},

R_{2},

and

R_{3}

are resistors connected in parallel.
For the parallel arrangement of resistor equivalent resistance is
⇒

\frac{1}{R_{}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}

⇒

\frac{1}{3} = \frac{1}{10} + \frac{1}{30} + \frac{1}{R_{3}}

\Rightarrow

\frac{1}{3} = \frac{3 + 1}{30} + \frac{1}{R_{3}}

\Rightarrow

\frac{1}{3} = \frac{4}{30} + \frac{1}{R_{3}}

⇒

\frac{1}{R_{3}} = \frac{1}{3} - \frac{4}{30}

⇒

\frac{1}{R_{3}} = \frac{10 - 4}{30}

⇒

\frac{1}{R_{3}} = \frac{6}{30}

⇒

\frac{1}{R_{3}} = \frac{1}{5}

\Rightarrow R_{3} = 5 Ω

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