Arrester current and resistance (Z = 300 Ω)

When the surge reaches the arrester, the arrester conducts and holds the voltage at its protective level $V_p$. It acts as a shunt resistance $R_a$ at the end of the line, and the travelling-wave relations give the currents.

Data

• Incident surge, $V_w = 3000$ kV (crest)
• Protective level, $V_p = 1700$ kV
• Line surge impedance, $Z = 300\ \Omega$

(a) Current in the line before reaching the arrester

$$I_w = \frac{V_w}{Z} = \frac{3000}{300} = 10\ \text{kA}$$

(b) Current through the arrester

Reflected voltage: $V_r = V_p - V_w = 1700 - 3000 = -1300$ kV. Reflected current: $I_r = -V_r/Z = 4.33$ kA.

$$\begin{aligned} I_a &= I_w + I_r = \frac{2V_w - V_p}{Z} \\ &= \frac{6000 - 1700}{300} = 14.33\ \text{kA} \end{aligned}$$

(c) Arrester resistance

$$R_a = \frac{V_p}{I_a} = \frac{1700}{14.333} = 118.6\ \Omega$$

Reflection and refraction coefficients of voltage

$$\begin{aligned} \beta &= \frac{R_a - Z}{R_a + Z} = \frac{118.6 - 300}{418.6} = -0.433 \\ \alpha &= \frac{2R_a}{R_a + Z} = \frac{237.2}{418.6} = 0.567 \end{aligned}$$

Check: $\alpha = 1 + \beta = 0.567$, and $\alpha V_w = 0.567 \times 3000 = 1700$ kV $= V_p$.

The arrester current (14.33 kA) is larger than the line current (10 kA) because the negative reflected voltage wave carries a positive current. The arrester must be rated for this current.

Answer: (a) 10 kA, (b) 14.33 kA, (c) $R_a = 118.6\ \Omega$; $\beta = -0.433$, $\alpha = 0.567$.