(III) Two resistors and two uncharged capacitors are arranged as shown in Fig. 62 . Then a poten

step 1 Okay, so we're doing chapter 26, problem 51. So we have two resistors and two uncharged capacito

(III) Two resistors and two uncharged capacitors are...

Key Concepts

Voltage Divider

A voltage divider is a fundamental circuit where two or more resistors in series divide an input voltage into smaller voltages at their junctions. This principle is used to determine the potential at intermediate points in a circuit, making it a key tool in circuit analysis.

RC Circuit Behavior

An RC circuit consists of resistors and capacitors and exhibits time-dependent behavior. When such circuits are subject to a voltage source, the capacitor charging process involves an initial transient period followed by a steady state where the capacitor behaves like an open circuit. This concept is crucial for analyzing the evolution of voltage and charge distribution in circuits over time.

Capacitor Charging and Discharging

Capacitors store electrical energy in the form of charge and voltage. They charge through resistors exponentially over time when connected to a voltage source, and when discharged they also release their stored energy exponentially. This behavior is described by time constants and is central to understanding the dynamics of circuits that include capacitors.

Charge Redistribution

When circuit configurations change, for example by closing a switch, the stored charge on capacitors may redistribute to reach a new equilibrium state. This concept involves the conservation of charge and energy, and is essential in predicting the final voltages and the amount of charge transferred in dynamic circuit systems.

Transcript

00:01 Okay, so we're doing chapter 26, problem 51.

00:03 So we have two resistors and two uncharged capacitors shown in this circuit, and a potential difference of 24 volts is applied across the combination as shown.

00:16 So part a asks, what is the potential at point a with the switch open? it says let v equals zero at the negative terminal of the source.

00:28 Okay, so now we can just apply.

00:32 A kirchav loops law to this small loop right here.

00:39 And we should see that the equation we get from that is v minus i .r1 minus i .r2 equals zero.

00:51 So from this we can solve for what i is.

00:54 And that's given by v over r1 plus r2.

00:58 So this is 24 volts over 8 .8 plus 4 .4 oms.

01:05 And that comes to a current of being 1 .818 amps.

01:15 So it says we're looking for the voltage at point a, and it says here is v equals zero.

01:24 So if we're looking for voltage at a, this is given by the same as the voltage drop across r1, which is given by the current times r1.

01:42 Sorry, i said r1 here, but this is totally r2.

01:47 So we're looking for the voltage cross r2 because we know that there to there is getting to zero, va to there is zero.

01:56 So this is r2.

01:58 So this is i times up to 2 or 4 .4 oms times 1 .818 amps.

02:05 And that comes out to being 8 volts.

02:10 Okay.

02:11 Let's move on to part b now.

02:14 So part b says what is the potential at point b when the switch is open? so point b.

02:22 So first we need to know that when the switch is open, these two capacitors are in series to each other.

02:30 So we know that ceq is one over c1 plus one over c2 inverse.

02:37 If we plug that in, then this becomes 0 .2057 microfitters.

02:46 So this is the equivalent capacitance.

02:51 So if we're looking for the voltage at point b, since c2 is connected up to the zero voltage point, this is the same as voltage across c2.

03:10 We can see because there's zero point to point b can follow along this track to where it only goes across this one capacitor.

03:21 So now the voltage across the capacitor is given by the charge on the capacitor over the capacitance.

03:33 Okay.

03:36 So let's go back to the equivalent capacitiveness of our circuit.

03:41 We can write the equivalent charge of the circuit as the voltage times the equivalent capacitance.

03:49 We already know the overall voltage drop of across the both capacitors is 24 volts.

03:56 And why is that? that's because we can travel this path, which shows just across both capacitors from the positive 24 volts point to the v equals 0 point.

04:07 So that means we can take this equation and write this as 24 volts times 0 .257 microperds.

04:16 And this comes out to being 4 .937 micropulops.

04:27 And the charge stored on the equivalent capacitance is the same as the charge stored on each capacitor.

04:34 The reason for that is because each capacitor has its own voltage drop such that vc1 plus vc2 equals v.

04:44 So if you plug that in over here, you'll see why that that is going to equal the case.

04:49 So we can go ahead and write this charge eq as the same as charge 1 or charge 2...

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