00:01
So, given a network of capacitors and the network can be drawn like this.
00:16
So yeah, these are the, so these are the points, nodes a and b, these are the capacitors and the outer capacitors are c1 and the two inner one are c2 and each capacitance c1 in the network between the points a and b has a value 6 .6 microfarads and the capacitors c2 has a value 4 .4 microfarads.
01:07
The potential across, applied between the points a and b is vab that is 470 volts.
01:19
It has been asked to find the equivalent capacitance of the network between the points a and b and in part b it has been asked to compute the charge on each of the three capacitors nearest to a and b when vab is equal to 470 volts.
01:52
So the nearest capacitors are c1, c2 and c1 here and in part c taking vab equal to 470 volts it has been asked to find the value of vcd.
02:04
So these are the two nodes c and d here.
02:07
So what is the voltage across c and d? this, these three capacitors are in series, so if we draw the equivalent circuit taking this three capacitor combination as a equivalent, replacing it with an equivalent capacitance, we can redraw the circuit as this.
02:31
So these are the capacitors and this is the equivalent of the three capacitors.
02:43
These are the points c and d, these are the nodes a and b.
02:48
So it is a and b, these are c1, c1, c and d and this is c2, this is c2, this is c1, c1 and this is finally the equivalent capacitance c prime where 1 by c prime is equals to 1 by c1 plus 1 by c1 plus 1 by c1, there is 3 by c1.
03:05
So c prime would be c1 by 3 that is 6 .6 microfarad divided by 3 that is 2 .2 microfarads.
03:13
Now if you see this c2 and c prime are in parallel, so the equivalent circuit would be this.
03:26
So across c and d the equivalent capacitance is c2 plus c prime and let it be c double prime.
03:38
So these are the capacitance, this is c1, c1, this is c double prime, this is c1, this is c1, this is c2.
03:45
So c double prime will be c2 plus c prime that is c2 is 4 .4 microfarads and c prime just now calculated as 2 .2 microfarads, so it comes out to be 6 .6 microfarad which is nothing but c1.
04:02
Now if again this is the series combination of three capacitance, same capacitances, so the equivalent circuit would be this.
04:15
So yeah, this is, these are the capacitors, so these are the nodes a and b and this is c1, this is c2, this is c1 and this is c dash dash dash.
04:31
So this equivalent capacitance is a series combination of c1, c double prime and c1.
04:40
So c triple prime would be c1 by 3 that is 2 .2 microfarads.
04:46
Now again this c2 and c triple prime, they are in parallel combinations.
04:51
So the equivalent capacitance would be across a and b would be c1, c4 prime and the c4 prime is c2 plus c triple prime that is 4 .4 microfarads plus 2 .2 microfarad that is 6 .6 microfarad is nothing but c1.
05:17
We can write it to be as c1.
05:19
Now finally these three are in series combination.
05:23
So the equivalent capacitance across a and b would be, let it be c equivalent, so across a and b would be, let it be c equivalent and the c equivalent would be 1 by c1 plus 1 by c1 plus 1 by c1 and that is c1 by 3, c equivalent is c1 by 3, so it is 2 .2 microfarads.
05:52
So this is the, thus the equivalent capacitance of the network, the equivalent capacitance of the network between the points a and b is 2 .2 microfarads.
06:11
Coming to part b where it has been asked to compute the charge on each of the three capacitors nearest to a and b when vab is equals to 470 volts.
06:25
Now the total charge across the capacitor network in, across all the capacitors in the network, let it be qt and that it would be equal to c equivalent times vab and it comes out to be, it comes out to be 1 .034 into 10 to the power minus 3 coulombs that is 1034 microcoulombs or roughly 1000 microcoulombs.
06:57
Now since c1, the capacitors c1, c1 and c4 prime, this combination, so it is, these are the capacitors which are, so this is c1, c1 and c4 prime.
07:20
They are in series, so the charge, charges on these capacitors will be same and will be equal to the total charge of the capacitor network that is qt...