You set up the circuit shown in Fig. 26.20, where C = 5.00 ×10^-6 F. At time t = 0, you close th

You set up the circuit shown in Fig. 26.20, where C = 5.00...

You set up the circuit shown in Fig. 26.20, where $C =$ 5.00 $\times$ 10$^{-6}$ F. At time $t =$ 0, you close the switch and then measure the charge q on the capacitor as a function of the current i in the resistor. Your results are given in the table: (a) Graph $q$ as a function of $i$. Explain why the data points, when plotted this way, fall close to a straight line. Find the slope and y-intercept of the straight line that gives the best fit to the data. (b) Use your results from part (a) to calculate the resistance R of the resistor and the emf $\varepsilon$ of the battery. (c) At what time $t$ after the switch is closed is the voltage across the capacitor equal to 10.0 V? (d) When the voltage across the capacitor is 4.00 V, what is the voltage across the resistor?

You set up the circuit shown in Fig. 26.20, where $C =$ 5.00 $\times$ 10$^{-6}$ F. At time $t =$ 0, you close the switch and then measure the charge q on the capacitor as a function of the current i in the resistor. Your results are given in the table:

(a) Graph $q$ as a function of $i$. Explain why the data points, when plotted this way, fall close to a straight line. Find the slope and y-intercept of the straight line that gives the best fit to the data. (b) Use your results from part (a) to calculate the resistance R of the resistor and the emf $\varepsilon$ of the battery. (c) At what time $t$ after the switch is closed is the voltage across the capacitor equal to 10.0 V? (d) When the voltage across the capacitor is 4.00 V, what is the voltage across the resistor?

Transcript

00:01 In this question, we are setting up an rc circuit as shown in the figure, and then we're going to close the switch and investigate the charge versus current relationship for this circuit.

00:19 So why, when we have plotted our data, does it look like...

00:25 Actually, before we dive into why these data points look like this, let's go ahead and answer the part that asks about the slope and intercept.

00:36 Because i plotted this in a program called graphical analysis, and it gives me the slope and the intercept right here, but they are very small and hard to read.

00:57 So my slope is going to be negative 1 .248 microcoulombs for every milliampere, and my y -intercept is 79 .9 microcoulombs.

01:17 I don't like the way that nine looks.

01:25 All right, so now we're going to answer the question.

01:28 I'm going to type down there at the bottom underneath the images just because it will be easier to read than my handwriting, and i can go faster than writing by hand.

01:45 And so these data fall near a straight line because the current is the time rate of change of the charge.

02:06 The current is the greatest just after the switch is closed when the charge is zero, and as charge builds on the capacitor, it pushes back against the flow from battery and the current decreases.

02:51 And then the longer that charging goes on, the longer it takes to push the same amount of charge onto capacitor plates, resulting in a decreasing rate of change.

03:23 And when the capacitor is fully charged and at the same potential as the battery, the current drops to zero as no more charge can be moved.

03:51 So you know the charge and the current both rely on the same e to the negative t over rc, and that also plays into it.

04:10 But this is the best way to explain when we remove time as a variable when we set up these two quantities as a relationship with each other why this happens.

04:24 And so now use our results to calculate the resistance of the resistor and the emf of battery.

04:34 Well, i actually found it easier to do the emf of the battery first because we need to note here in part b that our resistance of the resistor can be found from the initial current equals the emf divided by the resistance.

05:03 So our resistance will be equal to the emf divided by i naught, and we're going to have to find the emf first.

05:19 And then we are also going to need to get i naught is of course the current right at start when charge is zero, which is going to be that x -intercept.

05:38 And we can actually use the slope and intercept from our linear fit to get a precise value for that.

05:48 So we're going to have to come back to current.

05:51 Let's go ahead and oh how about purple because i've already done a bunch of typing in so here's where we can look at the fact that our q final, really it's our q final, our final amount of charge is going to be equal to the capacitance times the emf of the battery.

06:16 And so we can find the emf of the battery with the final charge divided by the capacitance.

06:26 Well the final charge is just what happens, we're looking at when current has dropped to zero, and that's going to also be the same as our y -intercept.

06:41 And so that was 79 .9 micro coulombs, so i need to put it as 10 to the negative 6, divided by a capacitance of 5 times 10 to the 6 farads.

07:01 So that gives me an emf of 15 .98 volts.

07:06 And now here's the thing to remember.

07:08 One, this is the result of a linear fit to data.

07:13 I think it's more likely that it would be 16 volts.

07:19 And just practically speaking, we actually would say that the capacitor, you know, the charging capacitor approaches that and the current approaches zero.

07:37 What happens is the time for a noticeable change just becomes so very very long that it's not really measurable.

07:45 So we're just going to go ahead and let this emf be 16 volts.

07:50 So we're going to take that 16 volts and come up here to that resistance calculation.

07:57 But like i said, we are going to need to figure out what is that x -intercept.

08:08 So what is the current when charge is zero? so we'll take, i'll go a little bit lower down so i'm moving to the right, so we'll take that charge equals negative 1 .248 i plus 79 .9.

08:24 Again we're looking at when is charge zero, what was that value of the current.

08:32 So i'll plug in zero for charge and then rearranging i'm going to get that current i is 79 .9 divided by 1 .248.

08:44 And i get, so this is again, that's going to be my i naught...

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