Dielectric power loss under AC and DC

Under AC, the loss in a dielectric is the total dielectric loss given by the loss tangent: $P = V^2\omega C\tan\delta$. Under DC, after charging, only the conduction (leakage) current flows, so $P = V^2/R$ with $R$ from the volume resistivity.

Given

• $A = 5$ m², $d = 5$ cm $= 0.05$ m, $\varepsilon_r = 3.5$, $\tan\delta = 0.02$
• $\rho = 6\times10^{12}$ Ω·cm $= 6\times10^{10}$ Ω·m
• $\varepsilon_0 = 8.854\times10^{-12}$ F/m

Capacitance of the specimen

$$\begin{aligned} C &= \frac{\varepsilon_0\varepsilon_r A}{d} = \frac{8.854\times10^{-12}\times3.5\times5}{0.05} \\ &= 3.099\times10^{-9}\ \text{F} = 3.099\ \text{nF} \end{aligned}$$

(i) 60 kV rms, 50 Hz AC

$$\begin{aligned} P_{ac} &= V^2\,\omega C\tan\delta \\ &= (60\times10^3)^2\times(2\pi\times50)\times3.099\times10^{-9}\times0.02 \\ &= 3.6\times10^{9}\times314.16\times3.099\times10^{-9}\times0.02 \\ &= 70.1\ \text{W} \end{aligned}$$

(ii) 60 kV DC

Volume resistance of the specimen:

$$\begin{aligned} R &= \frac{\rho\,d}{A} = \frac{6\times10^{10}\times0.05}{5} = 6\times10^{8}\ \Omega \\ P_{dc} &= \frac{V^2}{R} = \frac{(60\times10^3)^2}{6\times10^8} = 6.0\ \text{W} \end{aligned}$$

Results

Answer: (i) $P_{ac} \approx$ 70.1 W; (ii) $P_{dc} =$ 6.0 W.

The AC loss is about 11.7 times the DC loss because under AC the polarization (dipole and interfacial) losses add to the conduction loss. This is why thermal breakdown voltage of a dielectric is lower under AC than under DC.