Arrester with 1600 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 level of 1600 kV. It behaves as a resistance $R_a$ terminating the line.

Data

• Incident voltage, $V_i = 4200$ kV
• Arrester voltage (BIL rating / protective level), $V_a = 1600$ 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

$$\begin{aligned} V_r &= V_a - V_i = 1600 - 4200 = -2600\ \text{kV} \\ I_r &= -\frac{V_r}{Z} = \frac{2600}{300} = 8.67\ \text{kA} \end{aligned}$$

ii) Arrester discharge current and resistance

$$\begin{aligned} I_a &= I_i + I_r = \frac{2V_i - V_a}{Z} \\ &= \frac{8400 - 1600}{300} = 22.67\ \text{kA} \end{aligned}$$ $$R_a = \frac{V_a}{I_a} = \frac{1600}{22.667} = 70.59\ \Omega$$

Check

$$\beta = \frac{R_a - Z}{R_a + Z} = \frac{70.59 - 300}{370.59} = -0.619, \quad \beta V_i = -0.619 \times 4200 = -2600\ \text{kV}$$

Answer: $V_r = -2600$ kV, $I_r = 8.67$ kA, $I_a = 22.67$ kA, $R_a = 70.59\ \Omega$.