---
title: "Loss tangent and dielectric power loss"
subject: "Electrical Engineering Material"
chapter: 3
key: "semester_3/electrical_engineering_material:loss-tangent-power-loss"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/loss-tangent-power-loss
site: BCT Engineering
---

# Loss tangent and dielectric power loss

Electrical Engineering Material · Chapter 3: Dielectric Materials

- Asked in: 2081 Chaitra (new course), 4 marks; 2081 Baisakh, 6 marks; 2080 Baisakh, 6 marks; 2074 Chaitra, 6 marks; 2072 Chaitra, 6 marks (asked 5 times)
- Video: [Understanding Dielectric Losses Factor And Tangent](https://www.youtube.com/watch?v=z5sCffMUP30) — Technology Intensive Care Unit (ICU), Nigeria.
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/loss-tangent-power-loss
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-3

## Question

What is loss tangent? Show that power loss in a dielectric material per unit volume is the function of frequency of the applied field and loss tangent.

## Short answer

Loss tangent tan delta = epsilon_r'' / epsilon_r' quantifies dielectric loss. Power loss per unit volume is W_vol = omega epsilon_0 epsilon_r' E_rms^2 tan delta, showing it is proportional to angular frequency omega and tan delta.

## Answer

#### Loss tangent

In an ideal dielectric the current leads the voltage by exactly 90°, so no power is lost. In a real dielectric, polarization cannot follow an AC field instantly, so part of the energy is turned into heat. This is described by a **complex relative permittivity**

$$
\varepsilon_r = \varepsilon_r' - j\varepsilon_r''
$$

where $\varepsilon_r'$ is the ordinary (storage) part and $\varepsilon_r''$ is the loss part. The **loss tangent** is

$$
\tan\delta = \frac{\varepsilon_r''}{\varepsilon_r'}
$$

$\delta$ is the angle by which the current falls short of leading the voltage by 90°. Good insulators have $\tan\delta \sim 10^{-4}$ (polyethylene); lossy ones are 0.01–0.1.

```text
      I_total
        ^   /
  I_cap |  /
  (wC'V)| /  delta
        |/________> V
          I_loss (wC''V)
```

#### Power loss per unit volume

Consider a parallel-plate capacitor with area $A$, spacing $d$, filled with the dielectric, and an applied voltage $V = V_0 e^{j\omega t}$.

Capacitance with complex permittivity:

$$
C = \frac{\varepsilon_0(\varepsilon_r' - j\varepsilon_r'')A}{d}
$$

Current:

$$
I = j\omega C V = \frac{\omega\varepsilon_0 A}{d}\left(j\varepsilon_r' + \varepsilon_r''\right)V
$$

The term $\frac{\omega\varepsilon_0\varepsilon_r'' A}{d}V$ is **in phase** with $V$ and causes power loss; the $j\varepsilon_r'$ term is the lossless capacitive current. So the dielectric acts like a capacitor with a parallel conductance

$$
G_p = \frac{\omega\varepsilon_0\varepsilon_r'' A}{d}
$$

Power dissipated (using rms voltage $V_{rms}$):

$$
W = G_p V_{rms}^2 = \frac{\omega\varepsilon_0\varepsilon_r'' A}{d}V_{rms}^2
$$

Divide by the volume $Ad$ and use $E_{rms} = V_{rms}/d$:

$$
W_{vol} = \frac{W}{Ad} = \omega\varepsilon_0\varepsilon_r'' E_{rms}^2
$$

Since $\varepsilon_r'' = \varepsilon_r'\tan\delta$:

$$
\boxed{W_{vol} = \omega\, \varepsilon_0\, \varepsilon_r'\, E_{rms}^2 \tan\delta}
$$

#### Conclusion

The dielectric loss per unit volume is directly proportional to:
- the frequency $\omega = 2\pi f$ of the applied field,
- the loss tangent $\tan\delta$ (or loss factor $\varepsilon_r'\tan\delta$),
- the square of the field strength.

So insulation for high-frequency or high-voltage use (cables, RF capacitors) must have a very low $\tan\delta$. The same heating is used usefully in microwave ovens and dielectric heating of plastics and wood.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In a lossy dielectric, ε_r is
   A. ε' + ε''
   B. ε' - jε'' (storage and loss parts)
   C. ε' ε''
   D. only real
   Answer: B. The imaginary part describes loss. (Key point: Complex permittivity)

2. tan δ is
   A. ε''/ε'
   B. ε'/ε''
   C. ε'' ε'
   D. ε'' - ε'
   Answer: A. Ratio of loss to storage. (Key point: Loss tangent)

3. The loss comes from the current term that is
   A. 90° ahead of V
   B. 90° behind V
   C. Zero for real dielectrics
   D. In phase with V, ω ε0 ε'' (A/d) V
   Answer: D. In-phase current causes dissipation; the j ε' term is lossless. (Key point: Loss current)

4. Power loss per unit volume is
   A. ε0 ε' E tan δ / ω
   B. ω ε0 ε' E tan δ
   C. ω ε0 ε' E_rms^2 tan δ
   D. ω ε0 ε'^2 E^2
   Answer: C. Proportional to frequency, tan δ and E^2. (Key point: Power per volume)

## Related in Electrical Engineering Material

- [Power loss in a glass dielectric](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Aglass-dielectric-power-loss) — Electrical Engineering Material, chapter 3
- [Silver: states per unit volume vs atoms](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Asilver-states-per-volume) — Electrical Engineering Material, chapter 2, asked 2×
- [Si polarizability, local field and resonant frequency](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Asi-polarizability-resonant-frequency) — Electrical Engineering Material, chapter 3, asked 2×
- [Applications of dielectric materials](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Adielectric-applications) — Electrical Engineering Material, chapter 3

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- [Dielectric power loss under AC and DC](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Adielectric-power-loss-ac-dc) — High Voltage Engineering, chapter 6
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- [Volume of cycloid about tangent at vertex](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Avolume-cycloid-about-tangent-at-vertex) — Mathematics I, chapter 2