Surge from overhead line into cable and transformer

A travelling wave meeting a change of surge impedance is partly reflected and partly transmitted. The transmitted voltage is

$$V_t = \frac{2 Z_2}{Z_1 + Z_2} V$$

Voltage entering the cable

$Z_1 = Z_0 = 400\ \Omega$ (overhead line), $Z_2 = Z_c = 100\ \Omega$ (cable), $V = 1000$ kV.

$$\begin{aligned} V_{cable} &= \frac{2 \times 100}{400 + 100} \times 1000 \\ &= 0.4 \times 1000 = 400\ \text{kV} \end{aligned}$$

Reflected voltage on the line $= 400 - 1000 = -600$ kV. The low cable impedance greatly reduces the surge entering the cable.

Voltage at the transformer

The transformer has practically infinite surge impedance (bushing capacitance neglected), so the cable end behaves as an open circuit. The reflection coefficient is +1 and the voltage doubles:

$$V_{tr} = 2 \times V_{cable} = 2 \times 400 = 800\ \text{kV}$$

 OH line 400 ohm     cable 100 ohm     transformer
 1000 kV --->  J1 ---- 400 kV ---> J2   (open: 800 kV)

Note: the reflected 400 kV wave from the transformer travels back to J1, where it is partly reflected again. Over successive reflections the transformer voltage settles towards $2 \times 1000 = 2000$ kV (the open-end value for the original line). A short cable therefore only protects the transformer during the first few microseconds, unless an arrester is also used.

Answer: voltage in the cable = 400 kV; transformer voltage (first arrival) = 800 kV.