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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. - Science

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प्रश्न

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. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?

संख्यात्मक

उत्तर

Resistance of electric lamp, R1 = 100 Ω

Resistance of toaster, R2 = 50 Ω

Resistance of water filter, R3 = 500 Ω

Potential difference of the source, V = 220 V

These are connected in parallel, as shown in the following figure

Let R be the equivalent resistance of the circuit.

The net resistance in parallel is given by

`1/R = 1/R_1 + 1/R_2 + 1/R_3`

Here, R1 = 100 Ω, R2 = 50 Ω and R3 = 500 Ω

`1/R = 1/100 + 1/50 + 1/500`

`= (5 + 10 + 1)/500`

= `16/500`

`R=500/16`

R = 31.25 Ω

Now, using Ohm’s law V = IR, we have

I = V/R

`=(220 V)/(31.25 Ω)`

= 7.04 A

Hence, the resistance of electric iron is 31.25 Ω and current through it is 7.04 A.

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अध्याय 12: Electricity - Intext Questions [पृष्ठ २१६]

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एनसीईआरटी Science [English] Class 10
अध्याय 12 Electricity
Intext Questions | Q 2 | पृष्ठ २१६

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