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?

Asked by Vishal kumar | 1 year ago |  214

##### Solution :-

The electric lamp, the toaster and the water filter connected in parallel to a 220 V source can be shown as using a circuit diagram as follows: The equivalent resistance of the resistors can be calculated as follows:

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

$$=\frac{5+10+1}{500}=\frac{16}{500}$$

$$R=\frac{500}{16} \Omega$$

Now, using Ohm's law, the current flowing across the circuit can be calculated as follows:

$$l=\frac{V}{R}=\frac{220}{\frac{500}{16}}$$

$$l=\frac{220\times16}{500}=7.04\,A$$

As the appliance are connected in parallel, the current drawn across all of them is 7.04 A. Hence, the current drawn by the electric iron connected in parallel to the same source is 7.04 A. We can find the resistance of the iron box using Ohm's law as follows:

$$R = \frac{V}{I}=\frac{220V}{7.04A}=31.25 \Omega$$

The resistance of the electric iron box is 31.25 Ω.

Answered by Shivani Kumari | 1 year ago

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