00:01
Here we are asked to find the current through the capacitor if the voltage across a 2 -tharid capacitor is shown by the graph here, where the y -axis is the voltage in volts and the x -axis is time in seconds.
00:19
Now the equation for the capacitor is q equals cv, where q is the charge across the capacitor, c is the capacitor, c is the capacitance, and v as the voltage.
00:29
If we get the time derivative of this equation, we'll get dq over dp is equal to a constant capacitance times dv over d .p.
00:45
Or simply, current is equal to capacitance times dv over d .p.
00:51
But if you remember your calculus, dv over dt, in this case, the function voltage over time, is simply the slope of.
01:02
Of the voltage v.
01:05
Now to get the current, we get the slope of the function as a piecewise.
01:11
So we can divide this graph as in regions of 0 seconds to 2 seconds, 2 seconds to 4 seconds, 4 seconds to 5 seconds, 5 seconds to 6 seconds, and 6 to 7.
01:30
Now getting the current for 0.
01:34
0 to 2 is equal to the capacitance 2 ferrets times the slope is the difference in the change in y divided by the difference in the change in x or delta y over delta x.
01:54
In this case, for 0 to 2 seconds, we have 10 volts minus 0 over 2 minus 0...