Asked by Kate

Three impedences are connected in series across a 200V,50Hz ac supply. The first impedance is a 10 ohm resistor , the second coil which has an inductive reactance of 15 ohms and resistance of 5 ohms and third comprises a 15 ohms resistor in series with a capacitor which has a capacitive reactance of 25 ohms.
Calculate the circuit current
Answered by henry2,
Z = 10 + (5+j15) + (15-j25) = 30-j10 = 31.6ohms[-18.4o].

I = E/Z = 200[0o]/31.6[-18.4] = 6.3A[18.4o].
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