Lumped line energisation with trapped charge

The line is represented by a series $R$–$L$–$C$ circuit energised by a step $E = 343$ kV. Given $R = 12.4\ \Omega$, $L = 0.4$ H, $C = 4\ \mu$F.

(i) Attenuation factor

$$a = \frac{R}{2L} = \frac{12.4}{2 \times 0.4} = 15.5\ \text{s}^{-1}$$

(ii) Natural angular frequency and frequency

Undamped:

$$\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.4 \times 4\times10^{-6}}} = 790.57\ \text{rad/s}$$

Damped:

$$\omega = \sqrt{\omega_0^2 - a^2} = \sqrt{790.57^2 - 15.5^2} = 790.42\ \text{rad/s}$$ $$f = \frac{\omega}{2\pi} = \frac{790.42}{2\pi} = 125.8\ \text{Hz}$$

(Since $R \ll 2\sqrt{L/C} = 632.5\ \Omega$, the circuit is oscillatory.)

(iii) Peak voltage across C

The capacitor voltage reaches its first peak at $t = \pi/\omega = 3.97$ ms:

$$V_{max} = E + (E - V_0)e^{-a\pi/\omega}$$

Damping factor:

$$e^{-a\pi/\omega} = e^{-15.5\pi/790.42} = e^{-0.0616} = 0.9403$$

(a) No trapped charge ($V_0 = 0$):

$$V_{max} = 343(1 + 0.9403) = 665.5\ \text{kV} \;(1.94\ \text{pu})$$

(b) Trapped charge of 343 kV. The worst (and usual textbook) case is reclosing when the trapped charge is of opposite polarity to the source, $V_0 = -343$ kV:

$$V_{max} = 343 + (343 + 343)(0.9403) = 343 + 645.0 = 988.0\ \text{kV} \;(2.88\ \text{pu})$$

If the trapped charge had the same polarity ($V_0 = +343$ kV), $E - V_0 = 0$ and there is no oscillation: $V_{max} = 343$ kV.

Answer: $a = 15.5\ \text{s}^{-1}$; $\omega = 790.4$ rad/s, $f = 125.8$ Hz; $V_{max} = 665.5$ kV with no trapped charge and $988.0$ kV with 343 kV trapped charge of opposite polarity.