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Consider a series RC circuit as in Figure below for which R =1.8 MQ; C-1.8 UF; and €=30.0 V. Find the maximum charge ( in units of uC) the capacitor can have ...

Question

Consider a series RC circuit as in Figure below for which R =1.8 MQ; C-1.8 UF; and €=30.0 V. Find the maximum charge ( in units of uC) the capacitor can have after a very long time of the switch S being closedSelect one: A. 70.20B. 54.00C.81.00D. 27.00OE. 108.00

Consider a series RC circuit as in Figure below for which R =1.8 MQ; C-1.8 UF; and €=30.0 V. Find the maximum charge ( in units of uC) the capacitor can have after a very long time of the switch S being closed Select one: A. 70.20 B. 54.00 C.81.00 D. 27.00 OE. 108.00



Answers

Consider a series $R C$ circuit as in Figure $P 18.35$ for which $R=1.00 \mathrm{M\Omega}, C=$ $5.00 \mu \mathrm{F},$ and $\mathcal{E}=30.0 \mathrm{V}.$ Find (a) the time constant of the circuit and (b) the maximum charge on the capacitor after the switch is thrown closed. (c) Find the current in the resistor 10.0 s after the switch is closed.

In this question, you are given this RC circuit. So the mm for the batteries. 30 votes. The resistor is one. Make our own. The capacitance is five. Micro ferret impact A. We want to determine the time constant. So the time constant Call this. I see. Okay, so this is 10 to 6 times five times 10 to the minus six. So this is five seconds. And then in practice be we want to find a maximum charge on the capacitor after the switch is closed. So cue Max is the capacitance times the Yemen for the battery eso It's five times 10 to the minus six times 30. So you have 150 micro column. And then, um, you want in practice? See, we want to find a current in the resistor 10 seconds after the switch is closed. So we'll be using the equation for current. Yes. During the charging process? Uh, yes, I goes to I know times e minus t or with how. Okay, so the current decays exponentially. Okay. And then, um, so we know that he goes to 10 seconds. Mhm. 20 goes to 10 seconds. I equals two. I know it's You are E minus 10 divided by five. So you have, uh, 30 votes. Divide by one. Make our own tend to the six hands into their money's, too. And then this gives 4.6 Micro NPR's. Okay, so the answer for this question, So in part, a C five seconds in part B is 150. Micro column in part C is 4.6 micro mbia, okay?

So this problem were given an RC circuit and we want to find out some of the characteristics of the circuit once we close this switch at point A. So first us to determine is the time constant on the time Constant Tau can simply be calculated by taking the resistance and multiplying it by the capacitance. So we're keeping up. The resistance is one My God home and the capacitance is five Michael fire ads. So we go ahead and multiply those two values together. We find that our time constant is five seconds. Now, in the second part of this problem we want to do is we're if I know what the maximum charge of this capacitor will be once we close the switch A. So, in order to find the maximum charge, we can do this by taking the capacitance and multiplying it by the IMF of the voltage source. And we know that the IMF is 30 volts, so we go ahead and multiply the capacitance type CMF. We'll find that the maximum charge is going to be 100 and 50. Michael, cool arms. What? And now finally, what we want to do is you want to find the current that is going through the resistor 10 seconds after the switch is closed at a So we know that T is equal to 10 seconds. And we know that the equation for the current can be written as such. So we have function of the current in terms of the time and that will be equal to the half divided by the resistance times e raise too negative t the time divided by the time constant. So we know the value of the IMF, the resistance we calculated the time constant and were given that T is equal to 10 seconds. So when we go ahead and just plug in, all the values and solve we get is a current of 4.6 micrograms.

So in this problem, we are given restaurants secret too. 1 may go home, which is equal toe one cross taking power six. Welcome, uh, Capitals nickel toe five. Micro fraud, which is equal to five multiplied by taking power minus six. Fraud and M f is equal to 30 words. And the time is it cool to incident. First of all, we will find RC physical too. Value far is 1.0 crossed and keep our six. Oh. Hm. And the cramp distances five multiply. I think about minus six. Fraud. When you would write this, we get time. Constant, which is a good to five seconds. This is the answer of the first part of the question. How now? Musical part. We have to find charge. As you know, their charges equal to cap pistols into IMF is both are given, so we will multiply both absences. Got 5.0 cross taking power minus six fraud. And I m f is 30 worlds. So when you will like this, we get charges. Go to 1 50 Micro code, Um, which is the answer of the second part of the question. In the third part, we have to find I after 10 seconds and formula Ford, this is equal toe MF divided by our into he in power minus t Do it by RC so we will substitute the values e m f 30 are is one month like I think about six of them in exponent minus tense against divided by 1.0 cross. Thank you. Power Six of them is Our embassy is five cross. Thank you. Power minus six column. So when we saw this, we get current after 10 seconds is equal toe four point 06 micro ampere, which is the answer of third part of the question.

C. circuit. We were told that the resistor has a resistance of 75 Times 10 to the three homes. The capacitor you have a capacitance of 25 Times 10 to the -6 parents. And the m for the whole package of the battery. He's given us 12 votes. And so we want to calculate the time constant of the circuit for part A. And part B. Want to calculate the charging the capacitor after one second. So to find the time constantly. Just use the equation for time constant which is how is equal to the resistance. Time to capacity it we know both of those values. So we just plug them in to get the value for the time constant. And to find the charge on the capacitor after one second. For part B. Who is the equation for the charge on the capacitor as it's storing charge which is Q. Charging the capacitor is equal to capital Q. The maximum charge that the capacity can store times one minus E. To the minus T. To verify the time constant all time. See? Yeah and here capital Q. The maximum charge that the capacity can store is equal to the capacitance times the voltage of the battery. And you just plug that in. And we can see that we'll have all the values after we calculate the time constant in part A. So we're just poking our values and we get the answer for The not to charge after 1/2. So A. We're looking at the time constant our time. See So we're just poking our values 75 10- 10. 3 times 25 Times 10 to the -6 We get Time constant is 1.88 seconds now. Part B. We're looking for the charge on the capacitor after one second. So we have t to put in one second and we said that equation is little cue the charging. The capacitor is equally capital Q. And the maximum charge which we said is capacitance C terms of altered at the battery. And this is multiplied by one E. To the minus T. Divided by our time. See. Mhm. So we can just plug in what we know. So capacitance 25 10 to -6 Voltage in the battery 12 and then one minus E. To the minus 1 40 Divided by Art EMC which we found as one Mhm. Mhm. So we get Q. Is equal to one point two for attempts 10 to the minus form problems of change. Yeah.


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