00:01
In the first case, total resistance equals to r plus rf where r is 4 kω so 4000 plus rf is given as 12 ω.
00:18
This is equals to 4012 ω.
00:23
The current in the circuit i equals to vps divided by r plus rf which is equals to 9 divided by 4012 ω.
00:41
So this is equals to 2 .24 into 10 to the power minus 3 ω.
00:50
So the diode voltage vd equals to vps minus r into i.
01:01
Now substituting the values, we will have 9 minus 4 into 10 to the power 3 into 2 .24 into 10 to the power minus 3.
01:16
So this will be in the unit of volt which is equals to 0 .04 volt.
01:25
The diode voltage vd equals to 0 .04 volt.
01:37
Now power dissipated across the diode is vd into i so which is equals to 0 .04 into i is 2 .24 into 10 to the power minus 3 watt.
02:00
So which is equals to 0 .089 milliwatt.
02:07
Now in case b, b gamma is 0 .7 volt and rf is 0 ω.
02:20
Now here i equals to vps, current in the circuit is vps divided by r so 9 divided by 4000 ampere so which is equals to 2 .25 into 10 to the power minus 3 ampere...