I)                   At the start thediode is forward biased, this allows the capacitor to charge to the input peak.

However, when the input starts to decrease below its peak, the capacitor holds itscharge and the diode becomes reverse biased. At this point, the capacitorstarts discharging through the resistor at a set time constant. How much the capacitorwill discharge will depend on the time constant. The capacitor will dischargemore when the time constant is smaller, vice versa the capacitor will dischargeless if the time consent is larger. Then the diode becomes forward biased sothe capacitor charges back to peak of input.

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Here are a graphical illustrationto justify what I said:                 The expected output voltage-time waveform is like asine wave. For the input voltage the wave goes up and peaks to a certain pointand then drops down with the same amplitude with a negative value below the xaxis. On the other hand, for the output voltage the wave will always bepositive. Therefor, it would never go below the x axis. The wave would have thesame time period and will bounce back up to the same value when it reaches to thex axis.

Here is a graph I produced by simulating the circuit on task 3 to showthis: II)                 To find theforward voltage for the diode 1N4001 I figured out the best way would be tofind a data sheet for the following diode. I did a quick search on google whichprovided me with a data sheet with diode characteristics graph. Since I knowthat we are using 1.5 amp, I used diode characteristics graph to find out theforward voltage for this diode by making a horizontal line on the graph at 1.

5amps and it give me around 0.95 volts. Therefore, the forward voltage for theIN4001 diode is 0.

95 volts.   Modifying my results using a constant-voltage diodemodel I can justify my finding for the forward voltage of 0.95 volts bysimulation on the software Multisim.  I used the circuit that is provided on task 3 andsimulated it. I put 2 cursers on the peak value of the red and green graph. Forthe peak value of the red graph I got 14.9608 and for the peak value of thegreen graph I got 14.

0120. I can the do 14.9608 – 14.

0120 which gives me 0.9488or 0.95 when rounded up. Therefor, the graph justifies my finding of 0.95 voltsfor forward voltage.hhh