CHAPTER 16 (1): RL CIRCUITS
Chapter 16: RL Circuits (Part 1)
Series RL Circuits
Focus on series RL circuits in Part 1 of this chapter.
Objectives
After completing Part 1, students will be able to:
Analyze a series RL circuit.
Determine AC impedance of a series RL circuit.
Understand phase relationship between voltages and current in a series RC circuit.
Draw impedance and phasor diagrams of a series RL circuit.
Main Topics Covered
Analysis of series RL circuits (Part 1)
Analysis of parallel RL circuits (Part 2)
Power and power factor in RL circuits (Part 2)
Revision
In AC circuits, the opposition to current flow is called impedance, measured in ohms (Ω).
Impedance of resistor is R (real) and the impedance of inductor is jXL (imaginary) where XL is the inductive reactance.
Impedances in Circuits
Impedances can exist in parallel and in series:
In series: ZT = Z1 + Z2 + … + Zn.
Note the complexity of mathematics involved.
Impedance of Series RL Circuit
Total impedance (Z) in a series RL circuit is the sum of the impedance of the resistor and the inductor.
Real part of Z is contributed by resistors (R).
Imaginary part is contributed by inductors (jXL).
Represented in polar form as R + jXL.
Impedance Diagram
An Impedance Diagram displays all the impedances in the circuit, expressed as:
Total impedance = R + jXL.
Voltages and Current in Series RL Circuit
Common electrical quantity (I) is referenced in phasors.
Supplied voltage relationships:
I leads VR by 0 degrees.
VL (voltage across the inductor) leads I by 90 degrees.
Circuit Laws Applied to Series RL Circuit
Ohm's Law: V = I * Z
VR is in phase with I.
VL leads I by 90 degrees.
Kirchhoff's Voltage Law: V = VR + VL.
Phasor Diagram of Series RL Circuit
Shows relationship, where impedance diagrams relate to phasor diagrams through circuit current.
Relationship Between Diagrams
Impedance and phasor diagrams are similar; multiplied by circuit current gives phasors.
The phase of circuit impedance matches the applied voltage.
Analysis Procedure
Calculate circuit impedance: Z = R + jXL
Calculate circuit current: I = Vs / Z
Calculate voltages VR and VL.
Draw phasor and impedance diagrams if necessary.
Example 16-1
A 10 kHz AC voltage source connected to:
R = 10 kΩ
L = 100 mH
Circuit current: Irms = 0.2 ∠0° mA.
Tasks: Draw phasor domain schematic, impedance diagram, phasor diagram and write time domain expressions.
Problem Solutions
(a) Calculate XL = 2πfL = 6.28 kΩ.
(b) Z = R + jXL = 10 + j6.28 kΩ, which gives Z = 11.8 ∠32.1° kΩ.
(c) Voltage calculations using the values of Z, yielding:
V1 = 2.36 ∠32.1° V,
V_R = 2 ∠0° V,
V_L = 1.256 ∠90° V.
(d) Time domain expression for current and voltage:
i(t) = 0.2√2 sin(20000πt) mA,
v(t) = 2.36√2 sin(20000πt + 32.1°) V.
Summary
The impedance in a series RL circuit leads to current lagging the source voltage by a phase angle (ϕ).