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.

Question

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

henry2, · Accepted Answer

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].