An electric lamp of $100 Ω$ , a toaster of resistance, $50 Ω$ and a water filter of resistance $500 Ω$ are connected in parallel to a $220 V$ source. The resistance of an electric iron connected to the same source that takes as much current as all three appliances is________________

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31 Ω

35 Ω

40 Ω

50 Ω

answer is A.

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Detailed Solution

31 Ω

resistance of an electric iron connected to the same source that takes as much current as all three appliances.
It is given that

R_{1} = 100 Ω, R_{2} = 50 Ω

and

R_{3} = 500 Ω

..
It is known that when the resistors are connected in parallel, the reciprocal of the equivalent resistance is equal to the sum of reciprocals of individual resistances, which makes it the least of all the individual resistances. Mathematically if

R_{p}

, represents the equivalent resistance of the resistors

R_{1}, R_{2}, R_{3}

connected in parallel, then the equivalent resistance is given by

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

On substituting the given values, we get

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

\Rightarrow \frac{1}{R} = \frac{1}{100} + \frac{1}{50} + \frac{1}{500}

\Rightarrow \frac{1}{R} = \frac{5 + 10 + 1}{500} = \frac{16}{500}

\Rightarrow R = \frac{125}{4}

= 31 Ω

(approximately)

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