It's easy to divide a sine wave (curve) into any number of slices so that area of all slices become equal. It's a problem of 'definite integral'. Area of a sine curve with amplitude 'a'(peak) between '0' and 'pi' is '2a'.
Now 2a/40 is the area what you want !!
I think this much hint is ample for the problem to be solved.
You can take help of any Calculas book where "area under curve" is described.
Don't hesitate to ask if you still cannot solve the problem .
AGAIN please let me know about the circuit you used for detecting the zero cross of the mains voltage
Now 2a/40 is the area what you want !!
I think this much hint is ample for the problem to be solved.
You can take help of any Calculas book where "area under curve" is described.
Don't hesitate to ask if you still cannot solve the problem .
AGAIN please let me know about the circuit you used for detecting the zero cross of the mains voltage
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