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# An electric lamp of 100 $$\Omega$$ , a toaster of resistance 50 $$\Omega$$, and a water filter of resistance 500 $$\Omega$$ 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 ?

Resistance of electric lamp, $$R_1 = 100 \Omega$$
Resistance of toaster, $$R_2 = 50 \Omega$$

Resistance of water filter, $$R_3 = 500 \Omega$$

Equivalent resistance R of the three appliances connected in parallel, is:

$$\frac{1}R=\frac{1}{R_1} +\frac{1}{R_2}+\frac{1}{R_3}$$

$$\frac{1}R= \frac{1}{100} +\frac{1}{50}+\frac{1}{500}=\frac{16}{500}$$

$$R=\frac{500}{16} \Omega =31.25 \Omega$$

Resistance of electric iron = Equivalent resistance of the three appliances connected in parallel = 31.25 $$\Omega$$

Applied voltage, V = 220 V

So,current$$I=\frac{v}R =\frac{220}{31.25}=7.04A$$