Phasor Magnitude (Rect → Polar)
|Z| = √(R² + X²)
Quick answer
This calculator solves |Z| = √(R² + X²) for Magnitude. With Real Part 1.03 kΩ, Imaginary Part 2.21 kΩ, the result is 2.438 kΩ. It solves in 3 directions, so you can enter the values you have and get the one you need.
Calculator
Formula
Description
Converting a complex impedance from rectangular form (real resistance plus imaginary reactance) to polar form gives its magnitude as the Pythagorean sum of the two parts. The magnitude is what an AC ohmmeter reads—the ratio of voltage to current amplitudes—while the angle (computed separately) gives the phase. This is the everyday tool for combining resistive and reactive components.
Variables
- |Z| — Impedance magnitude (Ω)
- R — Resistance / real part (Ω)
- X — Reactance / imaginary part (Ω)
Practical Notes
Pair with the phasor angle θ = atan(X/R) for the full polar form. The same Pythagorean rule converts any rectangular phasor (voltage, current) to its amplitude.
Symbols and units
| Symbol | Meaning | Default unit | Accepted units |
|---|---|---|---|
| mag | Magnitude | Ω | mΩ, Ω, kΩ, MΩ |
| R | Real Part | Ω | mΩ, Ω, kΩ, MΩ |
| X | Imaginary Part | Ω | mΩ, Ω, kΩ, MΩ |
Step by step
- 1. What you know: Real Part 1.03 kΩ, Imaginary Part 2.21 kΩ
- 2. The relation: |Z| = √(R² + X²)
- 3. Substitute: R = 1.03 kΩ, X = 2.21 kΩ
- 4. Result: Magnitude = 2.438 kΩ
Nearest standard component values
| Series | Nearest value | Error |
|---|---|---|
| E12 | 2.2 kΩ | 9.8% |
| E24 | 2.4 kΩ | 1.6% |
Worked examples
Minimum range
Given: Real Part 2.2 Ω, Imaginary Part 4.7 Ω
Result: Magnitude = 5.189 Ω
Low range
Given: Real Part 22 Ω, Imaginary Part 47 Ω
Result: Magnitude = 51.89 Ω
Typical
Given: Real Part 1.03 kΩ, Imaginary Part 2.21 kΩ
Result: Magnitude = 2.438 kΩ
High range
Given: Real Part 10.3 kΩ, Imaginary Part 22.1 kΩ
Result: Magnitude = 24.38 kΩ
Maximum range
Given: Real Part 103 kΩ, Imaginary Part 0.221 MΩ
Result: Magnitude = 0.2438 MΩ
Solving for each variable
Solve for Magnitude
Real Part 2.2 Ω, Imaginary Part 4.7 Ω → 5.189 Ω
Solve for Real Part
Magnitude 10 Ω, Imaginary Part 2.2 Ω → 9.755 Ω
Solve for Imaginary Part
Magnitude 10 Ω, Real Part 4.7 Ω → 8.827 Ω
How Magnitude changes with Real Part
| Real Part (input) | Magnitude (result) |
|---|---|
| 0.227 kΩ | 4.845 kΩ |
| 0.5675 kΩ | 4.873 kΩ |
| 1.135 kΩ | 4.971 kΩ |
| 1.703 kΩ | 5.131 kΩ |
| 2.27 kΩ | 5.346 kΩ |
| 3.405 kΩ | 5.918 kΩ |
| 4.54 kΩ | 6.636 kΩ |
| 9.08 kΩ | 10.29 kΩ |
| 15.89 kΩ | 16.61 kΩ |
| 22.7 kΩ | 23.21 kΩ |
The same result in other units
| Unit | Value |
|---|---|
| mΩ | 2438000 |
| Ω | 2438 |
| kΩ | 2.438 |
| MΩ | 0.002438 |
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<p><a href="https://vbatpower.com/formulas/phasor-magnitude/">Phasor Magnitude (Rect → Polar)</a> Calculator by vbatpower.com</p>Frequently asked questions
How do you calculate Phasor Magnitude (Rect → Polar)?
Use |Z| = √(R² + X²). With Real Part 1.03 kΩ, Imaginary Part 2.21 kΩ, Magnitude works out to 2.438 kΩ.
What is the formula for Phasor Magnitude (Rect → Polar)?
The formula is |Z| = √(R² + X²). Converting a complex impedance from rectangular form (real resistance plus imaginary reactance) to polar form gives its magnitude as the Pythagorean sum of the two parts.
How do you find Real Part when you know the other values?
Rearrange for Real Part: with Magnitude 10 Ω, Imaginary Part 2.2 Ω, Real Part is 9.755 Ω.
How do you find Imaginary Part when you know the other values?
Rearrange for Imaginary Part: with Magnitude 10 Ω, Real Part 4.7 Ω, Imaginary Part is 8.827 Ω.
What units does Phasor Magnitude (Rect → Polar) use?
Magnitude 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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