00:01
Determine the equivalent capacitance of the circuit shown in figure 24 through 27.
00:06
So based on the figure, c2 and c3 are in series which is in parallel with c1.
00:12
Let us first compute the capacitors in series.
00:19
So c s would be equal to c2 times c3 over c2 plus c3.
00:27
And then we can find the equivalent capacitance by computing the parallel capacitance.
00:32
And that would be equal to c1 plus c2, c3 over c2 plus c3.
01:02
And per part b, given that c1 is equal to c2, which is equal to 2 times c3, and that's 24 muf and v is equal to 35, since c2 and c3 are ensues, which is a parallel with c1, and the voltage of c1 should be the same with that of c2 and c3.
01:32
So with this, we can compute the charge using the voltage q1 equals c1v, which would be 24 mu f, so mu is times 10 to the negative 6, and that would be times 35, which would be 8 .4 times 10 to the negative 4th, or 8 .4 mu 840 mu c we move it twice.
02:24
Okay, so next let us find the series equivalent of the capacitors...