3 5 points details my notes katzpse1 29p060 ask your teacher...

3. [-/5 Points] DETAILS MY NOTES KATZPSE1 29.P.060. ASK YOUR TEACHER PRAC The figure below shows a simple RC circuit with a 1.50-$\mu$F capacitor, a 2.80-M$\Omega$ resistor, a 9.00-V emf, and a switch. What are the following exactly 9.00 s after the switch is closed? S ε R (a) the charge on the capacitor $\mu$C (b) the current in the resistor $\mu$A (c) the rate at which the capacitor is storing energy $\mu$W (d) the rate at which the battery is delivering energy $\mu$W SUBMIT ANSWER

Transcript

00:01 So we have this circuit, we have a battery that is delivering current, power to charging a capacitor.

00:10 Okay, we have a current in a resistor.

00:14 And we want to find after 10 seconds that the switch is closed.

00:20 Okay, it means that the current is...

00:25 That the circuit is closed.

00:30 What is the charge on the capacitor? what is the current that flows through the resistor, at what rate the energy is stored in the capacitor, and at what rate the energy is delivered by the battery.

00:46 Okay, so it's actually pretty easy.

00:48 We just have to use formulas and use all the data that is given.

00:54 So for the charge and the capacitor, remember that the charge is the q.

01:00 The 1 minus e to the minus t rc okay remember that this is going to lose charge so for 10 seconds i know that is also c times e okay so the c times e is 1 minus e to the minus t rc and i just have to substitute all the data i have both so c is 1 times 10 to the minus 6 paradigm okay the potential difference is 10 volt okay then i have 1 minus e to the minus t remember t is 10 seconds here to the minus t divided by two times 10 to the six almost that is the value of the resistor and multiply by 1 times 10 to the minus 6 faraday.

02:08 Okay and when you make all these calculations you will find that the value of this is 9 .93 times 10 to the minus 6 faradai or you can just say that the charge after 10 seconds is 9 .9 .9.

02:30 93 micro -fadile.

02:33 Okay, so this is the answer for the first part, or for the a part.

02:39 To find the current.

02:41 So, b, we know that the current, okay, is the charge, the differential of charge in terms of time.

02:53 Okay, so if we make here, and then take the derivative of this respect to time, we will find that this is c .e.

03:05 Divided by r .c.

03:07 E to the minus t .r .c.

03:11 Okay.

03:12 So actually, you can see that this is consul each other out.

03:19 So you just have to, again, use all the values.

03:23 So 10 volts divided by the resistor value.

03:27 That is 2 times 10 to the 6 om...