RL Series Circuit

RL SERIES CIRCUIT

Fig 1.31 shown the RL series circuit.

In the above circuit

R - Resistance in Ω

L - Inductance in Henry

V - Input voltage

V_R = Voltage drop across R = IR

V_L - Voltage drop across L = IX_L

I - Current flowing in RL series circuit.

Phasor diagram for RL series circuit is shown in fig 1.32

Impedance Z = R + j X_L

Z = V / I

Phase angle ϕ = tan^-1  ﴾ X_L/R﴿

Power factor = cos ϕ = R/Z

Apparent power, S = VI

Real power, P = VI cos ϕ

Reactive power Q = VI sin ϕ

Example: 15

A coil having a resistance of 7Ω and an inductance of 31.8 mH is connected to 230V, 50H supply. Calculated the circuit current, phase angle, power factor and power consumed.

Inductive reactance, X_L = 2π ƒ L = 2 × л × 50 × 31.8 × 10^-3

= 10 Ω

Impedance, Z = = =12.2 Ω

Current I  = = 18.85 Q

Phase angle ϕ = tan^-1 ﴾X_L/R﴿

= tan^-1 ﴾10/7﴿

= 55°

Power factor cos ϕ = cos 55

= 0.573

Power consumed P = VI cos ϕ

= 230 × 18.85 × 0.573

= 2484.24 W