Arrester with 1000 kV BIL on 300 Ω line

When the travelling wave reaches the arrester at the line end, the arrester conducts and holds the voltage at its rated (protective) level of 1000 kV. The arrester acts as a resistance $R_a$, so part of the wave is reflected.

Data

• Incident voltage, $V_i = 4200$ kV
• Arrester voltage (BIL rating / protective level), $V_a = 1000$ kV
• Surge impedance, $Z = 300\ \Omega$

Incident current

$$I_i = \frac{V_i}{Z} = \frac{4200}{300} = 14\ \text{kA}$$

(i) Reflected voltage and current

Voltage at the arrester = incident + reflected:

$$\begin{aligned} V_r &= V_a - V_i = 1000 - 4200 = -3200\ \text{kV} \\ I_r &= -\frac{V_r}{Z} = \frac{3200}{300} = 10.67\ \text{kA} \end{aligned}$$

The reflected voltage is negative (it cancels most of the incident wave), and the reflected current is positive (it adds to the incident current).

(ii) Arrester discharge current and resistance

$$\begin{aligned} I_a &= I_i + I_r = \frac{2V_i - V_a}{Z} \\ &= \frac{2 \times 4200 - 1000}{300} = 24.67\ \text{kA} \end{aligned}$$ $$R_a = \frac{V_a}{I_a} = \frac{1000}{24.667} = 40.54\ \Omega$$

Check with reflection coefficient

$$\beta = \frac{R_a - Z}{R_a + Z} = \frac{40.54 - 300}{340.54} = -0.762, \quad \beta V_i = -0.762 \times 4200 = -3200\ \text{kV}$$

Answer: $V_r = -3200$ kV, $I_r = 10.67$ kA, $I_a = 24.67$ kA, $R_a = 40.54\ \Omega$.