Tower top potential for lightning stroke

When lightning strikes the tower top, the stroke current $I_s$ divides between all the paths that meet at the tower top. These paths are in parallel: the stroke channel itself (surge impedance $Z_s$), the tower footing resistance $R$, and the ground wires (surge impedance $Z_g$ each). So the tower-top potential is

$$V_t = I_s Z_{eq}, \qquad \frac{1}{Z_{eq}} = \frac{1}{Z_s} + \frac{1}{R} + \frac{n}{Z_g}$$

where $n$ is the number of ground wires. (Tower inductance and travel time in the tower are neglected.)

Data: $R = 40\ \Omega$, $Z_g = 500\ \Omega$ (each), $Z_s = 400\ \Omega$, $I_s = 50$ kA.

(i)(a) Considering all impedances (two ground wires)

$$\begin{aligned} \frac{1}{Z_{eq}} &= \frac{1}{400} + \frac{1}{40} + \frac{2}{500} \\ &= 0.0025 + 0.025 + 0.004 = 0.0315\ \text{S} \\ Z_{eq} &= 31.75\ \Omega \\ V_t &= 50 \times 31.75 = 1587.3\ \text{kV} \end{aligned}$$

(i)(b) Neglecting ground wires and stroke impedance

Only the footing resistance carries the current:

$$V_t = I_s R = 50 \times 40 = 2000\ \text{kV}$$

(i)(c) One ground wire and stroke impedance

$$\begin{aligned} \frac{1}{Z_{eq}} &= \frac{1}{400} + \frac{1}{40} + \frac{1}{500} = 0.0295\ \text{S} \\ Z_{eq} &= 33.90\ \Omega \\ V_t &= 50 \times 33.90 = 1694.9\ \text{kV} \end{aligned}$$

The ground wires and the stroke channel take part of the current, so they lower the tower-top potential. The footing resistance is the largest single factor.

Answer: (a) 1587.3 kV, (b) 2000 kV, (c) 1694.9 kV.