This picture appears in the original question. It shows that when the sinusoidal voltage is at zero (black waveform), the current in the inductor (red waveform) is at a peak (positive or negative). It also shows that when the voltage is at a peak, the current in the inductor is zero: -
So, the most appropriate time to disconnect an inductive load from a sinusoidal voltage supply is when the current is small and that only happens when the applied voltage is at a peak. So you open circuit at that point and naturally, there will be very little stored energy in the inductor that could create a spark and erode the switching contact.
The most appropriate time to connect an inductor to a sinusoidal voltage supply is when the voltage is at a peak because the developing current will naturally begin at the right point and there will be little or no inrush current peak. If you connect when the voltage passes through zero, the current peak (for the first cycle and diminishing thereafter due to losses) will be twice the steady-state maximum and, of course this is to be avoided to prevent excessive current flow in the circuit and possible core saturation problems.
Regarding what the op says in comments on another "answer", these misconceptions must be properly addressed: -
For me, this answer make sense. During initial conditions, (inductor
current =0), if we apply VAC=0, then the current on the inductor will
be 0A. But, using this way, there will be a short time, where the Load
(inductor) will be 'stopped'. (No voltage/current in the Load) –
Kotik_o
- "During initial conditions" - there are no initial conditions in the proposed circuit and this exacerbates the problem if you are trying to locate "the best moment" for switching.
- "if we apply VAC=0, then the current on the inductor will
be 0A. But, using this way, there will be a short time, where the Load (inductor) will be 'stopped'." - No that will not happen as you believe - the inductor current will immediately start to rise or fall to a value dictated by the formula \$V = L\frac{di}{dt}\$ and will potentially rise to a peak value that is twice the normal running peak current.
