Voltage at junction with shunt capacitor

A step wave $V$ on line $Z_1$ reaches junction A, where line $Z_2$ continues and a capacitor $C$ is connected to ground.

Derivation

Thevenin equivalent at A: the incident wave on $Z_1$ acts as a source $2V$ behind $Z_1$. Line $Z_2$ acts as a resistance $Z_2$ to ground. So the circuit is $2V$ in series with $Z_1$, feeding $C$ in parallel with $Z_2$.

       Z1               A
 2V ~--/\/\/--+---------+
              |         |
              C         Z2
              |         |
 -------------+---------+---- gnd

Converting $2V$, $Z_1$ and $Z_2$ to a Thevenin source seen by $C$:

$$V_{th} = \frac{2VZ_2}{Z_1 + Z_2}, \qquad R_{th} = \frac{Z_1Z_2}{Z_1 + Z_2}$$

The capacitor charges with time constant $\tau = R_{th}C$:

$$\boxed{V_A(t) = \frac{2Z_2}{Z_1 + Z_2}V\left(1 - e^{-t/\tau}\right), \quad \tau = \frac{Z_1Z_2}{Z_1 + Z_2}C}$$

At $t = 0$ the capacitor short-circuits A, so $V_A = 0$. As $t \to \infty$ it is open, and $V_A$ tends to the normal refracted value.

Numerical values

$Z_1 = 400\ \Omega$, $Z_2 = 380\ \Omega$, $C = 3$ nF, $V = 4000$ kV.

$$\begin{aligned} R_{th} &= \frac{400 \times 380}{780} = 194.87\ \Omega \\ \tau &= 194.87 \times 3\times10^{-9} = 0.5846\ \mu\text{s} \\ V_{final} &= \frac{2 \times 380}{780} \times 4000 = 3897.4\ \text{kV} \end{aligned}$$

Voltage at $t = 1\ \mu$s

$$\begin{aligned} \frac{t}{\tau} &= \frac{1}{0.5846} = 1.7105 \\ e^{-1.7105} &= 0.1808 \\ V_A(1\,\mu s) &= 3897.4(1 - 0.1808) = 3192.9\ \text{kV} \end{aligned}$$

The capacitor does not reduce the final voltage. It slows the rise of the wavefront: after 1 μs the voltage is only 82% of its final value.

Answer: $V_A(t) = 3897.4\left(1 - e^{-t/0.5846\,\mu s}\right)$ kV, so $V_A$ = 3192.9 kV at $t$ = 1 μs.