Solution

The electric arc occurs when the switch is opened.

When the circuit contains an inductance L, the current cannot be interrupted abruptly. When attempting to open the contacts, the energy stored in the magnetic field of the coil $(\frac{1}{2}LI^2)$forces the continuity of the current, generating a very high self-induced electromotive force $(-L\,di/dt)$. This voltage ionizes the air between the separating contacts and produces a spark or arc.

When closing the switch, on the other hand, the current starts from zero and grows gradually due to the same inductance, without voltage spikes or ionization of the gap between contacts.

If a capacitor is connected in parallel with the switch, when the contacts open, the current finds a low-impedance path through the capacitor. The energy from the inductor is transferred to the capacitor, charging it in an oscillatory manner instead of being dissipated in an arc. Thus, the voltage across the switch terminals rises in a controlled way and does not reach the breakdown voltage of air.

b) Capacitance required to limit the overvoltage

When the switch opens, the inductor L and the capacitor C form a resonant circuit. In the worst case, all the initial magnetic energy is converted into electrical energy in the capacitor. If the RMS current before opening is I, the stored energy is$W = \frac{1}{2}LI^2$

The maximum voltage across the capacitor terminals (and therefore across the switch) is$V_{\max} = Q_{\max}/C$
By conservation of energy in the LC exchange:

$\frac{1}{2}LI^2 = \frac{1}{2}CV_{\max}^2 \quad\Rightarrow\quad V_{\max} = I\sqrt{\frac{L}{C}}$

We want$V_{\max} \le N V_{\text{mains}}$where $V_{\text{mains}}$ is the supply voltage. In sinusoidal steady state,$I = V_{\text{mains}}/(2\pi\nu L$) (neglecting the resistive part). Substituting:

$N V_{\text{mains}} \ge \frac{V_{\text{mains}}}{2\pi\nu L}\sqrt{\frac{L}{C}} = \frac{V_{\text{mains}}}{2\pi\nu\sqrt{LC}}$

Solving for the capacitance:

$C \ge \frac{1}{(2\pi\nu N)^2 L}$

For the data$\nu = 50\ \text{Hz}, L = 0.1\ \text{H}, N = 10$

$C \ge \frac{1}{(2\pi\cdot 50\cdot 10)^2 \cdot 0.1} = \frac{1}{(1000\pi)^2 \cdot 0.1} \approx \frac{1}{9.87\times 10^6 \cdot 0.1} \approx 1.01\times 10^{-6}\ \text{F}$

$\boxed{C \approx 1\ \mu\text{F}}$