An electric lamp of 100Omega, 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?

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

1.	An electric lamp of 100Omega, 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?
Answer» Solution :Here. Resistance of an electric LAMP `R_1 = 100 Omega` Resistance of a toaster `R_2 = 50 Omega` Resistance of a water filter `R_3 =500 Omega` Equivalent resistance `R_p` of THREE resistors `R_1,R_2 and R_3` connected in parallel is given by, `1/R_P = 1/R_1 + 1/R_2 + 1/R_3` `:.1/R_P = 1/100 + 1/50 + 1/500` `= (5 + 10 + 1)/500` `=16/500` `:. R_p = 500/16` `=125/4` `= 31.25 Omega` Total current through the circuit (i.e.. through all the three APPLIANCES) `I = V/R_p` `=220/(31.25)` `= 7.04 A` Since the electric IRON connected to the same SOURCE (Le., 220 V) takes as much current as taken by all the three appliances (Le., I = 7.04 A), its resistance must be equal to `R_p`. So. the resistance of the electric iron = `31.25 Omega` and the current through the electric iron = 7.04 A