Basic Electrical and Electronics Engineering: Unit I: Electrical Circuits
RL Series Circuit
Basic Circuit diagram, Phasor diagram, Formula with Solved Example Problems | Electrical Circuits
Apparent power, S = VI,Real power, P = VI cos ϕ,Reactive power Q = VI sin ϕ
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
VR
= Voltage drop across R = IR
VL
- Voltage drop across L = IXL
I
- Current flowing in RL series circuit.
Phasor
diagram for RL series circuit is shown in fig 1.32
Impedance
Z = R + j XL
Z
= V / I
Phase
angle ϕ = tan-1
﴾ XL/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, XL = 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 ﴾XL/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
Basic Electrical and Electronics Engineering: Unit I: Electrical Circuits : Tag: : Basic Circuit diagram, Phasor diagram, Formula with Solved Example Problems | Electrical Circuits - RL Series Circuit
Basic Electrical and Electronics Engineering: Unit I: Electrical Circuits
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