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
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].

Related Questions

A coil takes a current of 15A and dissipates 2800W when connected with the supply of 200V ,50Hz when... A circuit consists of a resistance of 20Ω, an inductance of 0.05H, connected in series. A single-p... A voltmeter and frequency meter are connected across the supply of a circuit give the following read... A coil with a resistance of 25ohms and inductance of 0,19H is connected in series with a capacitor o...