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InterviewSolution
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An electric lamb 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 ans what is the current through it ? |
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Answer» Resistance of the electric lamp, `r_1 = 100 Omega` resistance of toaster, `r_1 = 50 Omega` Resistance of water filter, `r_3 = 500 Omega` Since `r_1,r_2 and r_3` are connected in parallel, 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)` or `R_p = (500)/(16) Omega = (125)/(4) Omega` Current through the three appliances,i.e., `I = (V)/(R_p) =(220 V)/((125//4)Omega) = 7.04 A` Since the electric iron connected to the same source `(i.e., 220 V)`, takes as much current as taken by all the three appliances,i.e., `I`, its resistance is equal to `R_p`,i.e., `(125//4) Omega = 31.25 Omega` Current through the electric iron, `I = 7.04 A`. |
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