Phase Angle and Phasors
In the previous lessons you learned that capacitors and inductors shift the phase of current relative to voltage — a capacitor makes current lead by 90°, an inductor makes voltage lead by 90°. When resistance and reactance both exist in a circuit, the actual phase shift is somewhere between 0° and 90°, depending on the relative magnitudes. Phasors are the graphical tool that makes these phase relationships visible, intuitive, and easy to calculate. A phasor is a rotating arrow whose length represents the amplitude of an AC quantity and whose angle represents its phase.
Phasor diagram for a series RLC circuit. VR is in phase with current I (pointing right). VL leads current by 90° (pointing up). VC lags current by 90° (pointing down). The total voltage VS is the vector sum, at angle θ from the current reference.
View LargerWhat Is a Phasor?
A phasor is a mathematical representation of a sinusoidal quantity — a voltage or current — as a rotating vector in a two-dimensional plane. It captures two essential properties of any AC quantity:
- Length (magnitude): proportional to the amplitude (usually RMS) of the sinusoidal quantity
- Angle: representing the phase of the quantity relative to some chosen reference
The word "phasor" is a portmanteau of "phase" and "vector." Unlike regular vectors, phasors rotate counterclockwise at the angular frequency ω = 2πf of the signal. But because all quantities in a linear circuit rotate at the same frequency, we can freeze the picture at any convenient instant and compare their relative angles — those relative angles (the phase differences) are what matter for circuit analysis.
Phasors are not the same as the physical sinusoidal waveforms. The phasor is a shorthand notation that captures the two pieces of information you need (amplitude and phase) in a compact graphical form. The actual time-domain waveform v(t) = Vpeak × sin(2πft + φ) can always be reconstructed from the phasor's length and angle.
Drawing Phasor Diagrams for Series RLC
For a series RLC circuit, the same current I flows through every component. This makes current the natural reference phasor — draw it along the positive horizontal axis (0° reference). Then:
- VR = I × R: The voltage across a resistor is always in phase with the current. Draw VR along the same horizontal axis as I (0°).
- VL = I × XL: The voltage across an inductor leads the current by 90°. Draw VL vertically upward (90°).
- VC = I × XC: The voltage across a capacitor lags the current by 90°. Draw VC vertically downward (−90°).
The total supply voltage VS is the phasor sum of VR, VL, and VC. Because VL and VC are opposite (180° apart), they partially cancel. The net reactive voltage is |VL − VC|, directed either upward (if VL > VC, inductive) or downward (if VC > VL, capacitive). Combined with VR at a right angle:
VS = √(VR² + (VL − VC)²)
Phase angle: θ = arctan((VL − VC) / VR)
Note that this is entirely consistent with the impedance triangle from M07E: since VR = IR, VL = IXL, VC = IXC, dividing all phasor lengths by I gives the impedance triangle. The voltage phasor diagram and the impedance triangle are the same shape — just scaled by the current.
A series circuit has R = 30 Ω, XL = 80 Ω, XC = 20 Ω. The current is 100 mA RMS. Draw the phasor diagram and find the supply voltage.
VR = I × R = 0.1 × 30 = 3.0 V (horizontal, at 0°)
VL = I × XL = 0.1 × 80 = 8.0 V (vertical up, at 90°)
VC = I × XC = 0.1 × 20 = 2.0 V (vertical down, at −90°)
Net vertical: VL − VC = 8.0 − 2.0 = 6.0 V (upward, inductive)
VS = √(3.0² + 6.0²) = √(9 + 36) = √45 = 6.71 V
θ = arctan(6.0 / 3.0) = arctan(2) = 63.4° (inductive, voltage leads current)
Adding Phasors Geometrically
Phasor addition is vector addition. To add two phasors geometrically:
- Draw the first phasor starting from the origin.
- Draw the second phasor starting from the tip of the first.
- The resultant phasor is the arrow from the origin to the tip of the second phasor.
For two perpendicular phasors (like VR and (VL − VC)), this becomes the Pythagorean theorem. For phasors at other angles, you need the general vector addition formula, but in most basic AC circuit analysis the perpendicular case covers all the essential situations.
Phasor subtraction works the same way — just reverse the direction of the phasor being subtracted before adding. This is how VL and VC combine: since they point in opposite directions, their net effect is their arithmetic difference.
What the Phase Angle Means Physically
The phase angle θ tells you how many degrees the total voltage leads the current (positive θ) or how many degrees the current leads the total voltage (negative θ).
| Circuit type | Phase angle θ | Who leads? | What it means |
|---|---|---|---|
| Purely resistive | 0° | In phase | No energy storage; all power dissipated as heat |
| Purely inductive | +90° | Voltage leads current | All energy stored in magnetic field; none dissipated |
| Purely capacitive | −90° | Current leads voltage | All energy stored in electric field; none dissipated |
| Inductive > Capacitive | 0° to +90° | Voltage leads current | Circuit appears inductive; some real power dissipated |
| Capacitive > Inductive | −90° to 0° | Current leads voltage | Circuit appears capacitive; some real power dissipated |
Understanding the sign of the phase angle is important when reading impedance specifications. An antenna described as having "impedance 50 − j30 Ω" is capacitive (negative imaginary part). An antenna with "impedance 50 + j20 Ω" is inductive. An antenna tuner must add the opposite reactance (in this case +j30 or −j20 respectively) to cancel the reactive component and achieve a purely resistive impedance at the transmitter's output.
Power Factor and Reactive Power
The phase angle between voltage and current has a direct impact on how efficiently power is delivered to a load. The power factor (PF) is defined as:
Power Factor = cos(θ)
Real Power P = Vrms × Irms × cos(θ) = S × cos(θ) (watts)
Apparent Power S = Vrms × Irms (volt-amperes, VA)
Reactive Power Q = Vrms × Irms × sin(θ) (volt-amperes reactive, VAR)
When θ = 0° (purely resistive load), cos(0°) = 1 and all the apparent power is real power — nothing wasted. When θ = 90° (purely reactive load), cos(90°) = 0 and no real power is delivered — all the current produces only magnetic or electric field energy that sloshes back and forth without being consumed.
An antenna presents an impedance of 50 + j40 Ω at an operating frequency. The transmitter delivers 10 W into this load. What is the real power actually radiated?
Z = √(50² + 40²) = √4100 = 64.03 Ω
θ = arctan(40/50) = 38.7°
Power factor = cos(38.7°) = 0.781
Wait — let me reconsider. If the transmitter is designed to drive 50 ohms and the antenna presents 64 ohms, the transmitter won't actually deliver full rated power. But taking the question at face value: if 10 W of apparent power were delivered, only PF × 10 = 7.81 W would be real power. The remaining 2.19 W would be reactive power sloshing back and forth.
More practically: An antenna tuner would be used to cancel the j40 Ω reactive component, presenting 50 + j0 Ω to the transmitter so it can deliver full power into a resistive load. The tuner does not add resistance — it adds −j40 Ω reactance (via a capacitor in series) to cancel the antenna's +j40 Ω inductive component.
Phase Angle in Ham Radio Practice
SWR and phase angle. A transmission line between your transmitter and antenna carries both forward and reflected waves. Reflections occur whenever the antenna impedance does not match the feedline characteristic impedance. The phase angle of the reflection coefficient — which determines the type and magnitude of impedance mismatch — is directly related to the reactive component of the antenna's impedance. A purely resistive mismatch has a reflection phase of 0° or 180°; a reactive mismatch has a phase somewhere in between.
Antenna analyzers. Modern antenna analyzers (such as the NanoVNA) measure both the magnitude and phase of the antenna's impedance. The display typically shows R (resistance) and X (reactance) separately, or equivalently |Z| (magnitude) and θ (phase angle). A reading of 50 Ω and 0° phase means a perfect 50-ohm resistive load. A reading of 73 Ω and −15° means a slightly capacitive impedance — a half-wave dipole that is slightly long for the frequency.
Phased antenna arrays. Some directional antenna systems use two or more elements fed with currents at controlled phase angles. A 90°-phased pair of vertical antennas produces a cardioid (heart-shaped) radiation pattern. Increasing or decreasing the phase difference changes the pattern's directionality and null direction. Understanding phase angles is essential for designing and adjusting phased arrays.
Frequently Asked Questions
Are phasors the same as vectors?
Phasors are a type of vector, but with one important distinction: phasors represent sinusoidal quantities at a specific frequency and are assumed to rotate at that frequency. In steady-state AC analysis, we "freeze" the rotation and only care about the relative angles between phasors. Physical vectors represent static or slowly changing quantities with direction in physical space. Mathematically, phasors use complex number arithmetic (multiplication and division rotate phasors as well as scale them), which is why complex impedance Z = R + jX is so useful.
Can the voltage across a capacitor or inductor be larger than the supply voltage?
Yes — this seems paradoxical but is entirely correct. In a series RLC circuit near resonance, the voltages across the inductor and capacitor can greatly exceed the supply voltage. They are large but nearly equal in magnitude and opposite in phase, so they almost cancel in the phasor sum. The supply voltage only needs to equal the voltage across R (which provides the resistive loss). This phenomenon — where reactive component voltages exceed the supply voltage — is called voltage magnification and is one of the reasons resonant circuits can build up large oscillating voltages. It is the principle behind Tesla coil operation and the step-up action of series resonant circuits.
Why does power factor matter in ham radio?
Power factor affects how efficiently power is transferred from your transmitter to the antenna. A power factor less than 1 means some of the current flowing in the circuit is not doing useful work — it is just charging and discharging reactive components without radiating any signal. This reactive current stresses the transmitter's output transistors and causes unnecessary heating. Antenna tuners correct the power factor by canceling the reactive component of the antenna's impedance, allowing all available power to be transferred as real (radiated) power.
Test Your Knowledge
Answer the questions below to check your understanding. Every answer can be found in the lesson above.