In this homework forum it is required that you show your own work first.
When you show your own work first it not only gets the ball rolling it also tells us a lot about how much flexibility is allowed in choosing components and design procedures. Without knowing that we could end up steering you in a direction which was never intended by your instructor and thus could spell failure of the class for you.
Just in case you havent started yet...
For question 1, you first need to derive the equation for the voltage at the left side of the diac, then for the triac turn on and turn off time. You can then start to calculate the power in the bulb.
However, due to the nature of the variability of some of the components used in the circuit there are some questionable items that come up as to what we are allowed to use as approximations to the real life components and this information you can only get from the previous course work or by direct instruction from the teacher.
One question is are you allowed to use a linear resistance in place of the bulb or do you have to look up the typical non linear resistance with voltage or current? This makes a very big difference.
Another question is are you allowed to use zero gate voltage turn on and zero triac conduction voltage or do you have to use some typical data sheet values? This is a little important for knowing now to calculate the turn on and turn off times as well as the more accurate power dissipation in the bulb and triac (if needed).
As to the second question you posted, are there any similar assumptions we can use for that one too and can we use IC chips or just transistors or what? It will obviously require a boost circuit of some type but to know what type we can use we need to know what kind of components are allowed in the design.
I have no idea what you mean by that. In particular, we should know if we should use typical values for components or we can assume ideal values for components.
For example, can we use a gate voltage of 0v or do we need to use something more like 2v which is more typical for a triac.
Can we use 0v for the triac 'on' voltage or do we have to use something like 1.4v when on.
Can we use a linear resistance for the bulb.
Using 0v for both and a constant resistance for the bulb, all we have to do is the following:
1. Calculate the time domain solution for Vc the voltage across the capacitor.
2. Equate that to 32v, then solve for t in terms of the pot resistance R, giving us t1.
3. Use t1 as the turn on time for the triac waveform, being a sine wave that switches 'on' at time t1.
4. Using that waveform, calculate the power in the bulb resistance assuming a linear resistance. The result will be dependent on R and so the power will vary from maximum (500 watts) to some minimum.
5. If in step 4 any solutions come out invalid, that means the capacitor voltage has to be checked for validity also, which means check to see that it always reaches to 32v or above. If it does not reach to 32v for any setting of R (some max value) then that means zero output, so the graph will go out and then down and then reach 0 and then a horizontal line for those values of R that do not allow the cap voltage to reach up to 32v. This happens if R is too big for the particular cap value.
Note that we ignored a lot of things in the above. We also ignored the fact that the cap will discharge more when the triac turns on because the gate conducts for the whole time it is on. This would mean that a transient response is needed for the cap value which includes an exponential part rather than just the sinusoidal part. For the first attempt though, try it with just the sinusoidal part as that should produce an analytical solution.
So the basic circuit operation is as follows...
1. The cap charges up through the pot R, producing a voltage across it. The voltage is phase shifted and reduced in amplitude from the line voltage.
2. Eventually if the cap voltage reaches 32v the triac turns on. If it does not reach 32v then the triac stays off.
3. If the triac turns on it turns on somewhere in the sine cycle, which energizes the load resistance. If it does not turn on then the load gets no power (0 watts). Once zero watts is reached if it is reached then any higher pot resistance setting also produces zero watts.
The "idea" that you at least make an attempt to work the problem and show that attempt will garner you all the help you need.
Otherwise you will quickly grow frustrated with the lack of help you get in the form of us doing your homework for you.