How does an increase in operating frequency result in decrease in the size of an inverter circuit?
The largest single factor is usually inductor size. If you eg double frequency you can generally halve inductance (as the impedance of a pure inductor is proportional to frequency). In practice a number of factors apply so that it's not a directly linear relationship, but good enough.
If you need a peak current of say 1A then the time taken to ramp up from 0 to 1A is related mainly to inductance and applied voltage. If the inductor is say 10 x smaller the current ramps at ~ 10x the rate. The discharge time is also similarly speeded up and the overall cycle is faster so operating frequency is higher. You can look at this as the smaller inductor causing higher frequency operation or of the higher frequency allowing smaller inductors.
If the text mentioned thyristors in that context it is probably either an old one or is dealing with extremely high power levels. Nowadays, for most purposes inverters would usually use either MOSFETs or IGBTs. The very largest inverters may still use Thyratron valves - such as the many MegaWatt units used for DC to AC power conversion for DC submarine cables.
In typical portable modern applications an inverter which may have been operated at 100 kHz or less 10+ years ago is now liable to operate at 500 kHz to 2 MHz and a few operate at higher again. At 1 MHz+ and power levels of say a few Watts the inductor size may be 10%-20% of the size at 100 kHz and the inductor may still dominate the overall size.
Note that current carrying capacity ~ proportional to wire area but inductance is proportional to turns squared. This does not mean though that core size changes only with sqrt of frequency as you have issues of core cross section, core path length, winding window size and more to add to the fun.
The use of higher frequency requires smaller capacitors, physically smaller inductors / transformers and their cores, and therefore reduces overall size of a design.
- Capacitive reactance \$X_C = \dfrac{1}{2\pi fC}\$, so for any given reactance desired (filtering etc) a higher frequency
fallows a lower capacitanceC. - Inductive reactance \$X_L = 2 \pi fL\$ so again, for any given reactance, a higher frequency
fallows for a smaller inductanceL.
On the other hand, depending on the purpose intended, a high frequency inverter might not suit the purpose: For domestic power inverters, an output at least approximately close to the mains frequency is required for most equipment.
The way some sinewave inverters address this, is by operating at a far higher frequency, kilohertz to megahertz, and generating the sine waveform via PWM. Thus, the bulk of the power transmission occurs at the higher frequency, with a final stage low-pass filter to get rid of the higher harmonics from the PWM signal, and leave behind a smooth sine wave at the desired 50 / 60 Hz.