Engineering Why is the frequency different for circuits with the same RC?

[jaus tail]

Homework Statement

Solution in Book#1

Solution in Google:
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2. Homework Equations
I don't know which solution is correct. I am struggling to remember this formula.

The Attempt at a Solution

This was a solved example so I didn't attempt.

 

[Baluncore]

It will depend on what load current or voltage is deemed to signal the end of each half cycle.
Capacitors total = 0.2uF.
H bridge current is fixed at 10A. Load = 10 ohm.
Maximum Vload = R * I = 100V.
H-bridge voltage alternates between –100 and +100, so each transition must reverse cap voltage by 200V.

Ignoring the load current and using all the current to charge and discharge the capacitors.
C = Q / V; C = I * t / V; t = C * V / I; t = 0u2F * 200V / 10A = 4u0 sec per transition.
One cycle will take 2 * 4u0 = 8u0 sec = 125kHz.

 

[jaus tail]

I guess the In2 formula is for circuit turn off time and not for max frequency. Thanks.

 

[Baluncore]

Those capacitors are expensive when it comes to current at high switching frequencies.

It appears that this circuit would be more applicable to situations switching at power line frequencies of 50 or 60Hz, than at many tens or even hundreds of kHz. For higher frequencies, this circuit has higher currents charging and discharging the capacitors than it does through the load. After a transition, peak capacitor current can be twice the maximum load current. Total is always 10 amp but the capacitors partly discharge through the load.

Here is an idealised switching circuit simulated by LTspice. Shown clocked at 125kHz (T=8usec), and at 50kHz (T=20usec).
The three traces show total capacitor current, load current and the constant current source.

Driven at 125kHz = 8usec cycle.

Driven at 50 kHz = 20usec cycle.

 

[jaus tail]

Both circuits have same RC. So why is frequency different if we use your first formula in post2?