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\(I=\frac{v}R =\frac{220}{31.25}=7.04A \)