Phasor Imaginary Part (Polar → Rect)
X = |Z| · sin(θ)
Quick answer
This calculator solves X = |Z| × sin(θ) for Imaginary 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
The imaginary (quadrature) component of a phasor in rectangular form is its magnitude times the sine of the phase angle. This is the reactive part of an impedance or the out-of-phase part of a voltage or current. It stores and returns energy each cycle rather than dissipating it, and it determines the reactive (VAR) power in a circuit.
Variables
- X — Imaginary / quadrature component (Ω)
- |Z| — Phasor magnitude (Ω)
- θ — Phase angle (degrees)
Practical Notes
Positive X is inductive, negative is capacitive. Combine with the real part R = |Z|·cos(θ) to write the impedance as R + jX.
Symbols and units
| Symbol | Meaning | Default unit | Accepted units |
|---|---|---|---|
| X | Imaginary 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: X = |Z| × sin(θ)
- 3. Substitute: mag = 1.03 kΩ, theta = 45 °
- 4. Result: Imaginary 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: Imaginary Part = 57.59 mΩ
Low range
Given: Magnitude 22 Ω, Phase Angle 15 °
Result: Imaginary Part = 5.694 Ω
Typical
Given: Magnitude 1.03 kΩ, Phase Angle 45 °
Result: Imaginary Part = 0.7283 kΩ
High range
Given: Magnitude 10.3 kΩ, Phase Angle 75 °
Result: Imaginary Part = 9.949 kΩ
Maximum range
Given: Magnitude 103 kΩ, Phase Angle 750 °
Result: Imaginary Part = 51.5 kΩ
Solving for each variable
Solve for Imaginary Part
Magnitude 2.2 Ω, Phase Angle 1.5 ° → 57.59 mΩ
Solve for Magnitude
Imaginary Part 1 Ω, Phase Angle 1.5 ° → 38.2 Ω
How Imaginary Part changes with Magnitude
| Magnitude (input) | Imaginary Part (result) |
|---|---|
| 0.227 kΩ | 0.2244 kΩ |
| 0.5675 kΩ | 0.5611 kΩ |
| 1.135 kΩ | 1.122 kΩ |
| 1.703 kΩ | 1.683 kΩ |
| 2.27 kΩ | 2.244 kΩ |
| 3.405 kΩ | 3.367 kΩ |
| 4.54 kΩ | 4.489 kΩ |
| 9.08 kΩ | 8.978 kΩ |
| 15.89 kΩ | 15.71 kΩ |
| 22.7 kΩ | 22.44 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-imag/">Phasor Imaginary Part (Polar → Rect)</a> Calculator by vbatpower.com</p>Frequently asked questions
How do you calculate Phasor Imaginary Part (Polar → Rect)?
Use X = |Z| × sin(θ). With Magnitude 1.03 kΩ, Phase Angle 45 °, Imaginary Part works out to 0.7283 kΩ.
What is the formula for Phasor Imaginary Part (Polar → Rect)?
The formula is X = |Z| × sin(θ). The imaginary (quadrature) component of a phasor in rectangular form is its magnitude times the sine of the phase angle.
How do you find Magnitude when you know the other values?
Rearrange for Magnitude: with Imaginary Part 1 Ω, Phase Angle 1.5 °, Magnitude is 38.2 Ω.
What units does Phasor Imaginary Part (Polar → Rect) use?
Imaginary 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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