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 220V 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?

An electric lamp of 100 Omega, a toaster of resistance 50 Omega and a

1.	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 220V 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?
Answer» Solution :Here voltage of given voltage SOURCE V=220V As three gadgets of resistances `R_1=100 OMEGA,R_2=50 Omega and R_3=500 Omega` have been connected to parallel across the voltage source, hence their EQUIVALENT resistance `R_p` is given by `1/R_p=1/R_1+1/R_2+1/R_3=1/100+1/50+1/500=(5+10+1)/500=16/500` `R_p=500/16 Omega =31.25 Omega` `therefore` Resistances of electric iron, which draws as much current as all three appliances (lamp, toaster and WATER filter) taken together =`R=R_p=31.25 Omega` Current passing through electric iron `I=V/R=(220V)/(31.25 Omega)=7.04 A`