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A parallel combination of three resistors takes a current of $7.5 A$ form a $30 V$ supply, It the two resistors are $10 Ω$ and $12 Ω$ find which is the third one?

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4 Ω

15 Ω

12 Ω

22 Ω

answer is B.

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Detailed Solution

A parallel combination of three resistors takes a current of

7.5 A

form a

30 V

supply, It the two resistors are

10 Ω

and

12 Ω

then the third resistance is

15 Ω .

Given,

Voltage (V) = 30 V; current (I) = 7.5 A

The resistance of a conductor (R) is,

R = \frac{30 V}{7.5 A}

R = 4 Ω

The resistance of the three resistors equals to

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

R_{3}

represents the resistance of the third resistor, then

\frac{1}{R} = \frac{1}{10} + \frac{1}{12} + \frac{1}{R_{3}}

so,

\frac{1}{4} = \frac{1}{10} + \frac{1}{12} + \frac{1}{R_{3}}

\frac{1}{R_{3}} = \frac{1}{4} - \frac{1}{10} - \frac{1}{12}

R = \frac{15}{60} - \frac{6}{60} - \frac{5}{60}

R = \frac{4}{60}

Ω

Hence,

R_{3} = \frac{60}{4}

R_{3} = 15 Ω

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