Closing resistance Rs to limit overvoltage to 1.8 pu

When the breaker closes, the line acts as a series R–L–C circuit excited by a step voltage $E$ (closing at the peak of the source voltage, 1 pu). With no trapped charge, the capacitor voltage is

$$v_C(t) = E\left[1 - e^{-\alpha t}\frac{\sqrt{\alpha^2+\omega^2}}{\omega}\cos(\omega t - \phi)\right], \quad \alpha = \frac{R}{2L},\ \omega = \sqrt{\frac{1}{LC} - \alpha^2}$$

Setting $dv_C/dt = 0$ gives $\sin\omega t = 0$, so the first (largest) peak is at $t = \pi/\omega$:

$$V_{max} = E\left(1 + e^{-\alpha\pi/\omega}\right)$$

Data: $L = 0.5$ H, $C = 5\ \mu$F. Total resistance $R = 0.12 + 10 + R_s$ (source + line + switching resistor in series during closing).

$$\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.5 \times 5\times10^{-6}}} = 632.46\ \text{rad/s}$$

Without $R_s$ (for reference)

$R = 10.12\ \Omega$, so $\alpha = 10.12/(2\times0.5) = 10.12\ \text{s}^{-1}$ and $\omega = 632.37$ rad/s.

$$V_{max} = 1 + e^{-10.12\pi/632.37} = 1 + 0.951 = 1.951\ \text{pu}$$

Required $R_s$ for 1.8 pu

$$\begin{aligned} 1 + e^{-\alpha\pi/\omega} &= 1.8 \\ e^{-\alpha\pi/\omega} &= 0.8 \\ \frac{\alpha}{\omega} &= \frac{\ln(1/0.8)}{\pi} = \frac{0.22314}{\pi} = 0.07103 \end{aligned}$$

Using $\omega^2 = \omega_0^2 - \alpha^2$:

$$\begin{aligned} \omega &= \frac{\omega_0}{\sqrt{1 + (\alpha/\omega)^2}} = \frac{632.46}{\sqrt{1 + 0.07103^2}} = 630.87\ \text{rad/s} \\ \alpha &= 0.07103 \times 630.87 = 44.81\ \text{s}^{-1} \\ R &= 2L\alpha = 2 \times 0.5 \times 44.81 = 44.81\ \Omega \\ R_s &= 44.81 - 10 - 0.12 = 34.69\ \Omega \end{aligned}$$

Check: with $R = 44.81\ \Omega$, $e^{-44.81\pi/630.87} = 0.800$, so $V_{max} = 1.80$ pu ✓

 Vs ~--0.12--+--CB--+--10 ohm--0.5 H--+-- Vo
             |      |                 |
             +-Rs-/-+               5 uF
                                      |
 -------------------------------------+-- gnd

The resistor is in circuit only for the first few milliseconds (the pre-insertion time). The main contacts then short it out, which produces a second, smaller transient.

Answer: $R_s \approx$ 34.7 Ω (total circuit resistance 44.81 Ω), assuming no trapped charge and closing at the voltage peak.