00:01
So for this problem, in part a, the time constant for a circuit is tau -1 times r -c -1.
00:11
Plutinant numbers with r equals 28 times 10 to the cube, and c1 equals 35 times 10 to negative 6f.
00:20
This gives us tau -1 equals 0 .98 seconds.
00:27
We can see from the figure that there are two peaks per cycle.
00:33
Therefore, the period of the voltage is t equals 1 over 120 seconds, which is 0 .0083 seconds.
00:51
So we observe that tau 1 is much larger than t.
00:55
Therefore, the voltage across the capacitor will be essentially constant during a cycle.
01:00
So the average voltage is the same as the peak voltage, v -average equals v -peak.
01:06
So the average current is basically constant...