Consider the arrangement of the four initially uncharged...

Consider the arrangement of the four initially uncharged capacitors shown in Fig. Capacitors A, B, C, and D have capacitances 5.4F, 4.3F, 3.2F, and 2.1F, respectively. Suppose that a battery applies a potential difference of 3000 V across the circuit, which is then disconnected from the battery. What is the potential difference across each capacitor?

Transcript

00:01 In this circuit, we need to find out the potential difference across each capacitor.

00:07 So first we will find their total capacitance.

00:11 Capacetor b and d are in parallel.

00:15 So their total capacitance, capacitance equivalent capacitance of b and b will be capacitance of b plus capacitors of capacitor d.

00:26 So this means it is 4 .3 plus 2 .3 plus 2.

00:30 2 .1 means 6 .4 micro ferret.

00:38 Now this we will redraw the circuit again.

00:44 We will redraw the circuit and this will be a series combination of three capacitors.

00:54 Here is capacitor and connected to a battery.

01:01 So here we have a capacitor a of 5 .4 micro ferrard and here we have capacitor bd, the parallel combination of bd, it is 6 .4 microferr and here we have capacitor c which is 3 .2 microferral.

01:30 So this is again the three capacitors in parallel.

01:34 So the total capacitance of the circuit 1 upon c equivalent will be 1 upon 5 .4 plus 1 upon 6 .4 plus 1 upon 3 .2.

01:48 We are using the formula 1 upon c1 plus 1 upon c2 plus 1 upon c3 is equal to 1 upon c equivalent.

01:58 So if you solve this, if you solve this, the equivalent capacitance comes out to be 1 .53 microferral.

02:10 So what is the charge? total charge in the network.

02:16 Total charge in the network cube is equal to c equivalent into the potential.

02:24 Now c equivalent is 1 .53 into 10 to minus 6, 1 .53 micro ferret multiplied by the voltage.

02:33 Voltage is 3 ,000 volts.

02:36 So it means the total charge is 4 .59 into 10 raise to minus 3 kula now if you see this total charge will pass through capacitor a so charge passing through capacitor a will be same like q because capacitor a capacitor bd and capacitor c are in series so same charge will pass and see same way the charge passing through capacitor c will also be q, right? so, potential at a, b, a will be equal to q upon ca, right? this is very simple.

03:18 So q is the total charge is 4 .59 into 10 -2 minus 3 upon total capacitance.

03:30 Total capacitance is 1 .53 micro -ferat, means 10 -6 ferrat.

03:38 So we, we, we, we, we have to, we, we a comes out to be 850 volts.

03:46 Similarly, we can find vc also.

03:49 Vc will be q upon capricetance of c...

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