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 ?

Explanation

Resistance of electric lamp, `R_(1)=100Omega`
Resistance of toaster, `R_(2)=50 Omega`
Resistance of water fiter, `R_(3)=500 Omega`
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.
`1/R=1/R_(1)+1/R_(2)+1/R_(3)=1/100+1/50+1/500`
According to Ohm's law,
V=IR
`I=V/R`
Where,
Current flowing through the circuit = I
`I=220/(500/16)=(220//16)/50=7.04 A`
7.04 A of current is drawn by all the three given appliaces.
Therefore, current frawn by an electric iron connected to the same source of potential 220 V = 7.04 A
Let R' be the resistance of the electric iron, According to Ohm's law,
V=IR'
`R'=V/I=220/7.04=31.35 Omega`
Therefore, the resistance of the electric iron is `31.25 Omega ` and the current flowing through is 7.04 A.