In Fig. 25-29, a potential difference V=100 V is applied across a capacitor arrangement with ca

step 1 We have a capacitor network here composed of three capacitors. Capacitors C1 and C2 are parallel

In Fig. 25-29, a potential difference V=100 V is applied...

Key Concepts

Stored Energy in a Capacitor

The energy stored in a capacitor is given by the expression U=½CV². This formula shows that the energy increases with both higher capacitance and higher voltage. Understanding this concept is fundamental when analyzing circuits to determine how much energy is stored or can be discharged from individual capacitors or networks.

Charge Distribution in Capacitor Networks

In a network of capacitors, the distribution of charge depends on whether the capacitors are connected in series or parallel. In series, the charge on each capacitor is identical even though the voltage may vary. This relationship is used to calculate the individual charges when a total potential difference is applied to the network.

Capacitance

Capacitance is a measure of a capacitor's ability to store charge per unit voltage. It is defined by the ratio q/V, where q is the charge stored, and V is the potential difference across the capacitor. The unit of capacitance is the farad (F), and it determines how much energy can be stored in the capacitor for a given voltage.

Potential Difference (Voltage)

The potential difference, or voltage, across a capacitor is the work required per unit charge to move the charge between its plates. It plays a key role in determining both the charge stored on the capacitor (via the relation q=CV) and the energy stored (using U=½CV²).

Capacitors in Series and Parallel

Capacitor networks can be arranged in series, parallel, or a combination of both, which affects their equivalent capacitance. In series arrangements, capacitors share the same charge, whereas in parallel arrangements, they share the same voltage. The method of combining these arrangements is crucial for analyzing how the overall potential difference is distributed and how charge is stored within the circuit.

Transcript

00:01 We have a capacitor network here composed of three capacitors.

00:07 Capacitors c1 and c2 are parallel to each other, and their combination becomes in -series with c -3.

00:16 So if we further transform this circuit, we have here our voltage across our diagram here.

00:30 And then the combined parallel c -1 -c -2, so this will be c sub 1, 2, and then we still have our capacitor c3 here.

00:42 Okay.

00:43 So that we say that this is now in series.

00:49 So we are to determine the charges on the three capacitors, the voltages across each capacitor, and the electric potential energy stored in each capacitor.

01:03 Okay.

01:03 So since capacitor c1 and capacitor c2 are parallel to each other, you have the junction here where the q total or the total charge gets divided into two branches.

01:23 Some will go through capacitor 1 as charge 1 and then some will go as capacity charge in capacitor 2.

01:31 These two charges will meet up again at this junction so that the charge that actually passes through capacitor 3 which we will call q sub 3 is actually the total charge to begin with so if we get the total charge then we can determine charge 3 it doesn't have to be in order so we can solve for the unknown which has the most complete information to begin with.

02:10 So in this case, once we get the total charge, then immediately that will be charge three.

02:17 So to determine the total charge, okay, we just go back to the definition of capacitance, which is the ratio of charge to voltage.

02:27 So the charge total will just be equivalent to the equivalent capacitance times the total voltage here v across the network here.

02:41 So to determine the equivalent capacitance, let's have c1, c2 parallel.

02:48 So this is actually just the sum of the two capacitance.

02:53 So this becomes 10 plus 5.

02:55 So this becomes 15 .6.

02:57 Micro -farrad in series with c3 which is of course still 15 micro -farrad so for series capacitors which are equivalent if you have two capacitors in series and they are equivalent with each other the equivalent capacitance of that two equal capacitances will just be half of the individual capacitance so in other words, the equivalent capacitance here becomes half of 15, which is equal to 7 .50 micro -farrad.

03:38 You can still determine this by doing the long method, which you, where you get the sum of the reciprocal of the individual capacitances, and then you get the reciprocal of that, you'll still arrive at 7 .50.

03:56 So let's go back to our total charge here, the equivalent capacitance now.

04:02 This means that if we further transform the circuit here, then the equivalent would be you have our voltage across these two here.

04:14 And then we just have a single capacitance here, which we can label as c equivalent, which is equivalent to 7 .50.

04:22 That's the final transformation of our circuit.

04:25 So going back here, we just multiply 7 .50, write the prefix micro in powers of 10.

04:34 So this is now 10 to negative 6, and this is now in farad, times the total voltage of 100 volts.

04:42 So we can say that the total charge in our capacitor network is 7 .50.

04:55 Times 10 to the negative fourth and this is now in coulum therefore this is now equivalent to the charge through capacitor 3 so we are now we can now box this as one of our answers i hope the space is still enough so q3 is there so once we have q3 we can just determine voltage across capacitor 3 still using the fundamental definition of capacitance.

05:31 So therefore, we now say that the voltage across capacitor 3, based on this definition, is just the charge through capacitor 3 divided by the capacitance of capacitor 3.

05:45 So let's put in the values now.

05:47 We have 7 .50 times 10 to negative 4.

05:50 This is in coulom, divided by the capacitans.

05:54 Of c3, which is 15, and then micro -farrad, that's negative 6.

06:03 So we say here that the voltage across capacitor 3 is 50 volts.

06:10 Okay, let's box our final answer now.

06:14 50 volts for b3.

06:16 We are done with this.

06:18 Okay.

06:18 And then finally, for the chart, the electric potential energy that is stored in capacitor 3 you can just choose among the 3 here okay so let's say i will choose the first one so that the electric potential energy stored in capacitor 3 is just one half and then the charge of capacitor 3 and the voltage across capacitor 3 let's put in the values now so q3 you just need to be very careful in plugging in the values.

06:57 You have to be very mindful.

06:59 Voltage 3 is 50 volts so that the stored energy in capacitor 3, take note that capacitor 3 has the maximum capacitance.

07:09 If you look at the relationship here, the greater the capacitance, the higher the electric potential energy that is stored in that capacitor.

07:19 So let's find out as we go along.

07:23 This is around 18 .8 millijoules.

07:30 Take note of the prefix milly, that is actually 10 to the raise to negative 3 joules.

07:37 Okay.

07:39 So we are all done now with the variables involving capacitor 3.

07:45 We do the same process here for the remaining two capacitors.

07:53 Okay, so we can now go to, what do we have here? we can now solve, let's go to capacitor 1.

08:05 Now, remember, capacitor 1 is parallel to capacitor 2.

08:12 Recall the characteristics of parallel circuits.

08:16 So recalling parallel circuits, we can say, therefore, that the voltage across capacitor 1, is just equivalent to the voltage across capacitor 2 such that we can label this as voltage sub 1 2.

08:32 Okay, we already know the we already know capacitor 3 i mean we already know the charge 3 so to determine okay, let's use this diagram here in the second transformation of our diagram, we have here combined c1 -2 series with c3.

09:05 So if you recall, for a series circuit, we also call this as voltage divider.

09:12 So if the total voltage this applied to the circuit is 100 volts, then it gets divided into these two series capacitances.

09:21 Okay, since we already know the voltage across capacitor 3, then the voltage across the combined capacitor c1c2 can also be solved easily.

09:34 So we just say that total voltage v, it's just equivalent to the voltage across the combined 1 and 2 voltage plus voltage 3.

09:45 So we just have here 100 and then we transpose the voltage across the voltage across.

09:51 3 as 50.

09:53 They have negative 50 so that this becomes b1 2...

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