00:01
In the problem we have been given the diagram.
00:05
So here we have been given the diagram and there are two questions in the problem.
00:11
First question is a diagram shown over here.
00:14
There are two diodes and these are connected in the opposite direction with each other.
00:24
Therefore we can see that in the first half that is the positive half cycle of the ac signal one of the diode will be conducting while the other will be not conducting so it will be open circuit and in the second half that is negative ac signal we can see that one of the diode will be conducting and the other will be non conducting therefore we can represent or draw the same circuit over here therefore when positive cycle off the ac signal is taken into consideration, then diode d2 will be on and d1 will be off.
01:13
So here if d2 is on, then the circuit is drawn over here.
01:21
And in the problem it is given the direction across, the current direction across v0, that is the arrow.
01:31
Is shown in the problem itself.
01:37
Therefore this is the direction of the current.
01:39
Now we consider this is the direction of the current, therefore we'll apply the current to chops current law here in this node and incoming current will be equal to the outgoing current therefore we have 0 minus v0 upon 1 k that equals v0 minus v1 upon 1 plus v0 minus 0 upon 1 therefore further after solving it we obtained the v0 that is equal to v1 upon 3 now since the v0 is equal to v1 upon 3 that is the amplitude is lowered by 3 times the graph of the signal or we can see the v0 graph is a v1 upon 3 and this is the curve now when negative cycle of ac signal is taken into consideration, that is negative half cycle of the ac signal is taken into consideration, the da1 will be conducting and the id2 will be off.
02:44
So here this circuit is shown over here.
02:47
This can be drawn like this...