Voltage rise at cable–overhead line junction

A surge travelling on a cable (low surge impedance) into an overhead line (high surge impedance) is partly reflected and the junction voltage rises above the incident value.

Surge impedances

Cable:

$$Z_1 = \sqrt{\frac{L_1}{C_1}} = \sqrt{\frac{0.3\times10^{-3}}{0.4\times10^{-6}}} = \sqrt{750} = 27.39\ \Omega$$

Overhead line:

$$Z_2 = \sqrt{\frac{L_2}{C_2}} = \sqrt{\frac{1.5\times10^{-3}}{0.012\times10^{-6}}} = \sqrt{125000} = 353.55\ \Omega$$

Transmitted (junction) voltage

Transmission coefficient:

$$\tau = \frac{2Z_2}{Z_1 + Z_2} = \frac{2 \times 353.55}{27.39 + 353.55} = 1.856$$ $$V_t = \tau V = 1.856 \times 15 = 27.84\ \text{kV}$$

Reflected voltage on the cable:

$$V_r = \frac{Z_2 - Z_1}{Z_1 + Z_2}V = \frac{326.17}{380.94} \times 15 = 12.84\ \text{kV}$$

Check: $V + V_r = 15 + 12.84 = 27.84$ kV $= V_t$.

Currents: incident $I = 15000/27.39 = 547.7$ A; transmitted into the line $I_t = 27843/353.55 = 78.75$ A.

Answer: the voltage at the junction rises to about 27.84 kV (a rise of 12.84 kV over the 15 kV surge).