Surge voltage at transformer with capacitance

For a steep surge (a few microseconds), the transformer behaves like its equivalent capacitance to ground. The leakage inductance of 16 H has a reactance so high at surge frequencies that it carries almost no current during this time, so it is neglected. A capacitance terminating a line of surge impedance $Z$ charges through $Z$ from a source of $2V$ (Thevenin equivalent of the incident wave).

   Z = 400 ohm
 2V ~----/\/\/\----+---- V_C(t)
                   |
                 C = 0.002 uF
                   |
 ------------------+---- ground

Time constant

$$\tau = ZC = 400 \times 0.002\times10^{-6} = 0.8\times10^{-6}\ \text{s} = 0.8\ \mu\text{s}$$

Voltage at the transformer terminal

$$\begin{aligned} V_T(t) &= 2V\left(1 - e^{-t/ZC}\right) \\ &= 2 \times 1000\left(1 - e^{-t/0.8\,\mu s}\right)\ \text{kV} \\ &= 2000\left(1 - e^{-1.25\times10^{6}\,t}\right)\ \text{kV} \end{aligned}$$

At $t = 0$ the capacitor acts as a short circuit, so the voltage is zero. It rises exponentially and finally reaches 2000 kV, the open-circuit (doubled) value.

Reflected wave

$$V_r(t) = V_T(t) - V = 1000 - 2000e^{-t/0.8\,\mu s}\ \text{kV}$$

It starts at $-1000$ kV and becomes $+1000$ kV.

The capacitance does not reduce the final surge voltage. It only slopes the wavefront: the transformer voltage takes about $2.3\tau = 1.84\ \mu$s to reach 90% of the final value. This reduces the steepness and so the stress on the first few turns of the winding.

Answer: $V_T = 2000\,(1 - e^{-t/0.8\,\mu s})$ kV, with a final (maximum) value of 2000 kV.