four capacitors are connected as shown in the figure below let c 180 f a circuit consists of fou

(a) First, we need to find the equivalent capacitance of the capacitors in series and parallel.

Four capacitors are connected as shown in the figure below...

Four capacitors are connected as shown in the figure below. (Let C = 18.0 µF.) A circuit consists of four capacitors. It begins at point a before the wire splits in two directions. On the upper split, there is a capacitor C followed by a 3.00 µF capacitor. On the lower split, there is a 6.00 µF capacitor. The two splits reconnect and are followed by a 20.0 µF capacitor, which is then followed by point b. (a) Find the equivalent capacitance between points a and b. µF (b) Calculate the charge on each capacitor, taking ΔVab = 20.0 V. 20.0 µF capacitor µC 6.00 µF capacitor µC 3.00 µF capacitor µC capacitor C µC

Transcript

00:01 All right the learners first draw the circuit diagram.

00:04 Here we have a capacitor.

00:06 With this capacitor we have one another capacitor and here with this node we have one another capacitor.

00:13 In parallel with this capacitor we have one another capacitor connected and this is connected here.

00:22 Here we have one node.

00:24 Let's label these circuits.

00:27 The circuit components.

00:29 This capacitor is c and it is of 18 microferrard this is of 3 microferrard this capacitor is of 20 micro ferrard and finally this is of 6 micro ferrard this node is labeled as node a and this node is leveled as node b first of all we are supposed to find the total capacitance of this circuit for date first of all we see that these two capacitors are in series all right then first we will combine we will find the equivalence equivalent capacitance of these two capacitors let's call that c equivalent 1 that is equal to 18 times 3 divided by sorry they are in series yes they are in series so series combination can be solved through the product divided by their sum micro all right this is equal to 18 time 3 is 54 18 plus 3 is 2154 divided by 21 will give you 2 .57 micro ferratt so instead of these two capacitors we have one capacitor there is 2 .57 then we will solve this with this capacitor they both are in parallel so their combination should be there's denote that with c 2 that is simply the addition of this one that is c equivalent 1 plus this micro ferrette capacitor because they are in parallel so the reculent capacitance will be added only this is 2 .57 plus this is 6 and both are in microferrarts so this is equal to 8 .57 microferrard so finally total total capacitance this is 2 is equal to this 1 2 3 their whole combination or their total capacitance is c equivalent to that is 8 .57 this is in series with 20 micro ferrc so it means that is simply equal to 20 times 8 .57 divided by their sum that is 20 plus 8 .57 and this is in micro ferrard so this will give us total capacitance that would be equal to 5 .97 micro ferrard so this is the total capacitance of this circuit so let me recall it first of all i solve for these two capacitors they were in series so their equivalent capacitance is this one then their series combination is in parallel with this 6 micro ferrard capacitor.

03:51 So then i take the series combination like this, that would be equal to 8 .51.

03:58 Then the combination of these three resistors is in series with this 20 micro ferrard capacitor.

04:05 So their series combination is solved by this...

Question

Please give Ace some feedback