This is what you call triac abuse; a triac is not made for this kind of application. Like Steve says, triacs remain switched on on DC loads. You may have zero crossings where they normally would switch off, but I'm not sure how the triac reacts to very fast switching, like in your 56k EIA-232 signal. The zero crossing may be too fast to switch it off.
I'd suggest an analog switch. Analog devices has zillions of them.
The values used will often be OK.
A larger than usual motor inductance may cause problems.
The snubber's job is to protect the switch contacts from inductive turnoff transients from the motor. Stopping the transient at source (across the motor) or at destination (across the contacts) both work. Arguably, having it at the switch is better as it deals with the energy that will do damage, as opposed to energy that may do damage, so it is more focused and it also then deals with other spikes that may happen along.
If you look at your circuit you'll note that in both cases the snubber connects from the motor-switch connection point to one leg of the mains. If the mains impedance is low at the spike frequency (-ies) then both are about equivalent.
The circuit current continues instantaneously at switch off. If it all flows through the snubber then it will pass through the 120 ohm resistor, so the voltage spike will initially will be \$V=IR = 10\mathrm{A} \times 120\mathrm{\Omega} = 1200\mathrm{V}\$. While that is a lot it is usually within the switch break capability (or else), and there are usually other impedances present which will also help to damp it.
The snubbing current will flow only until the capacitor charges to the driving voltage. If the motor inductance is large the capacitor may charge to a higher or much higher voltage.
The capacitor needs to be large enough to not be charged to the point where current decays through charging of the cap before the resistor dissipates the energy. To be sure that the component values present will do the job, you need to know motor inductance.
Energy in inductor is \$E=\frac{1}{2}LI^2\$
Capacitor will "ring" with an energy of \$E=\frac{1}{2}CV^2\$
The resistor needs to dissipate this energy.
Energy = \begin{align}\frac{1}{2}Li^2 &= \frac{1}{2}CV^2 \\
\\
\Rightarrow V &= \sqrt{\frac{Li^2}{C}}
\end{align}
Then there is some \$L/R\$ time constant as well and ...
You can start to calculate this (if you know L) or simulate it, but in most cases the values shown are OK for typical equipment.
Place a scope across the contacts. What peak V do you see (use a suitable probe!). Do the contacts spark? They shouldn't.
Note that increasing C improves snubbing action but also increases losses from the mains in normal operation. Note also that a capacitor across a mains switch may be frowned at in some contexts.
Added:
Dario said: One problem with placing the RS across the switch is that now you have some current in the circuit in the switched off mode. ...
User_long_gone responded: I'm absolutely certain that the 4-5 MILLIAMPS of current flowing through a 0.1 microfarad capacitor at 60 Hz will present no problem to a motor circuit. Wasteful of energy? It's less than 1/2 watt.
It's worth noting that
- The snubber across the motor may not bother the motor itself but may well severely bother anyone silly enough to think that the switch being off means that the circuit is "safe" or "dead". If the switch is in the phase/live lead the motor side of the switch may be near ground due to relative impedances. But there is no certainty that this will always be the way the connection is made - even if regulations say that it should be.
2 "Even" 1/2 a Watt of pointlessly wasted energy in an appliance is frowned on in modern scenarios.
Best Answer
One of the problema with RC snubbers is that they leak current. This doesn't really matter if the leakage current is much lower than the load requires to work, but otherwise it does. In the circuit diagram in the question there is a 10nF capacitor in series with the load. If we neglect the load and the series resistor, at 50Hz this capacitor acts as a
\$X_{capacitor}=\dfrac{1}{2\pi f C} = \dfrac{1}{2\pi × 50 × 10\cdot 10^{-9}} \approx 300\text{k}Ω\$
impedance which in series with your 5W bulb:
\$R_{load} = \dfrac{U^2}{P} = \dfrac{230^2}{5} \approx 10 \text{k}Ω\$
apparently allows enough current to make it glow. It actually is a bit more complicated than that, with phase angles and all, but we're talking order magnitude to get a feeling for what is happening. Therefore it is important to size the snubber for the attached load.
I did find rule of thumb design guideline in a book, but I have no practical experience with these guidelines:
Design guidelines:
Resistor:
Capacitor:
An unrelated practical example:
I once built a snubber network 22nF/630V and 100Ω/0.5W for switching a small PC amplifier rated at 240V(AC) and 70mA. That configuration did solve the problem I was facing back then even though it doesn't exactly follow the above design guidelines.