Phasor Real Part (Polar → Rect)
R = |Z| · cos(θ)
Quick answer
This calculator solves R = |Z| × cos(θ) for Real Part. With Magnitude 1.03 kΩ, Phase Angle 45 °, the result is 0.7283 kΩ. It solves in 2 directions, so you can enter the values you have and get the one you need.
Calculator
Formula
Description
Converting a phasor from polar to rectangular form, the real (in-phase) component is the magnitude times the cosine of the phase angle. This is the resistive part of an impedance or the in-phase part of a voltage or current, and it is what determines real power dissipation. It is the projection of the phasor onto the real axis.
Variables
- R — Real / in-phase component (Ω)
- |Z| — Phasor magnitude (Ω)
- θ — Phase angle (degrees)
Practical Notes
The imaginary part is X = |Z|·sin(θ). For power, the real component of current times voltage gives true (watt) power; the imaginary gives reactive (VAR) power.
Symbols and units
| Symbol | Meaning | Default unit | Accepted units |
|---|---|---|---|
| R | Real Part | Ω | mΩ, Ω, kΩ, MΩ |
| mag | Magnitude | Ω | mΩ, Ω, kΩ, MΩ |
| theta | Phase Angle | ° | ° |
Step by step
- 1. What you know: Magnitude 1.03 kΩ, Phase Angle 45 °
- 2. The relation: R = |Z| × cos(θ)
- 3. Substitute: mag = 1.03 kΩ, theta = 45 °
- 4. Result: Real Part = 0.7283 kΩ
Nearest standard component values
| Series | Nearest value | Error |
|---|---|---|
| E12 | 0.68 kΩ | 6.6% |
| E24 | 0.75 kΩ | 3% |
Worked examples
Minimum range
Given: Magnitude 2.2 Ω, Phase Angle 1.5 °
Result: Real Part = 2.199 Ω
Low range
Given: Magnitude 22 Ω, Phase Angle 15 °
Result: Real Part = 21.25 Ω
Typical
Given: Magnitude 1.03 kΩ, Phase Angle 45 °
Result: Real Part = 0.7283 kΩ
High range
Given: Magnitude 10.3 kΩ, Phase Angle 75 °
Result: Real Part = 2.666 kΩ
Maximum range
Given: Magnitude 103 kΩ, Phase Angle 750 °
Result: Real Part = 89.2 kΩ
Solving for each variable
Solve for Real Part
Magnitude 2.2 Ω, Phase Angle 1.5 ° → 2.199 Ω
Solve for Magnitude
Real Part 1 Ω, Phase Angle 1.5 ° → 1 Ω
How Real Part changes with Magnitude
| Magnitude (input) | Real Part (result) |
|---|---|
| 103 Ω | 72.83 Ω |
| 0.2575 kΩ | 0.1821 kΩ |
| 0.515 kΩ | 0.3642 kΩ |
| 0.7725 kΩ | 0.5462 kΩ |
| 1.03 kΩ | 0.7283 kΩ |
| 1.545 kΩ | 1.092 kΩ |
| 2.06 kΩ | 1.457 kΩ |
| 4.12 kΩ | 2.913 kΩ |
| 7.21 kΩ | 5.098 kΩ |
| 10.3 kΩ | 7.283 kΩ |
The same result in other units
| Unit | Value |
|---|---|
| mΩ | 728300 |
| Ω | 728.3 |
| kΩ | 0.7283 |
| MΩ | 0.0007283 |
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<p><a href="https://vbatpower.com/formulas/polar-to-rect-real/">Phasor Real Part (Polar → Rect)</a> Calculator by vbatpower.com</p>Frequently asked questions
How do you calculate Phasor Real Part (Polar → Rect)?
Use R = |Z| × cos(θ). With Magnitude 1.03 kΩ, Phase Angle 45 °, Real Part works out to 0.7283 kΩ.
What is the formula for Phasor Real Part (Polar → Rect)?
The formula is R = |Z| × cos(θ). Converting a phasor from polar to rectangular form, the real (in-phase) component is the magnitude times the cosine of the phase angle.
How do you find Magnitude when you know the other values?
Rearrange for Magnitude: with Real Part 1 Ω, Phase Angle 1.5 °, Magnitude is 1 Ω.
What units does Phasor Real Part (Polar → Rect) use?
Real Part is shown in mΩ, Ω, kΩ, MΩ. Values are held in SI base units internally, so you can mix prefixes without converting by hand.
Related Concepts
All AC Circuits formulas →Need more features?
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