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

Explanation

(a)
(a) Shows the connection of all three appliances .
`(1)/(R_(p)) =(1)/(50)+(1)/(500) +(1)/(100) =(16)/(500)`
`R_(p)= (500)/(16) Omega = (125)/(4) Omega`
`I = (V)/(R) = (220xx4)/(125)= 7.04 A`
Now, since the current in electric iron is 7.04 A and is connected to the same source of 220 V. Hence the resistance of electron iron is
` R =(220)/(7.04) Omega = 31 .25 Omega`