00:01
So, in this question we are given with an input ac source and it is connected to a resistor by two diodes that this is d1 and here it is your load and it is again connected with another diode in reverse bias condition.
00:26
So, this is your d2 and we are taking the output voltage here across the load.
00:36
So, the input source is given as 10 sine 100 t so 100 pi t and what else it is given, it is given as your vt or drop across diode is 1.
00:59
So, we have to write the output voltage.
01:05
So, when we will consider the positive half cycle, so see here for positive half cycle, positive half cycle means current will flow in this direction.
01:17
So, diode d1 will be positive half cycle d1 will be forward bias and d2 will be reverse bias.
01:28
So, d2 will be open and there will be no voltage across d2.
01:34
So, we can find out at this case vout will be 9 volt since vin will be 10 volt and what we can write vout equal to vin minus voltage drop across diode.
01:55
So, this will be 10 minus 1 that will be 9.
01:59
So, vout will be 9 volt for positive half cycle.
02:06
Now for negative half cycle let's see what happens.
02:09
For negative half cycle, negative half cycle we see negative half cycle here d1 will be reverse bias, d2 will be forward bias but it has no role, it has no role in vout or in the output voltage.
02:32
So, eventually it has no work.
02:35
So, since d1 is reverse biased or diode 1 is reverse biased, so we will get output voltage as 0 volt, no output voltage will be obtained in this period.
02:49
Okay.
02:50
So, we can draw the waveform, input and output waveform like this.
02:57
So, for vin we can write vinput.
03:02
So, input waveform will be like this.
03:05
So, here it is omega t and you can write like this.
03:13
This is a sine wave.
03:15
So, the maximum peak voltage will be 10 volt.
03:19
Here it is minus 10 volt.
03:23
So, minus only indicates the direction that is negative half cycle.
03:28
Now the corresponding output voltage we can write as, for the first 1 volt or the first time difference which will be t1.
03:42
Here we can say it is t1.
03:45
So, from here you will be getting an ac voltage and here let me draw the dotted diagrams from here.
03:57
So, here it is, this time period is t1 and the peak voltage for output will be 9...