what is the charge q on the capacitor immediately after the switch is moved to position b? This is a topic that many people are looking for. amritsang.org is a channel providing useful information about learning, life, digital marketing and online courses …. it will help you have an overview and solid multi-faceted knowledge . Today, amritsang.org would like to introduce to you RC Circuits – Problem 9 – Energy dissipated during discharge. Following along are instructions in the video below:
All right heres problem. 9. This is more complicated problem here it says the the switch in the circuit diagram is initially in position.
A. So you should see should be a charging circuit. Right because youve got a battery a resistor and a capacitor and its there for a long time ok.
So thats a keyword. We know what that means that means that the voltage across this 200. Micro farad capacitor.
Is 50 right all right. And then its switched right you move. The switch over to position b.
Once you move it over to position b. It now becomes a discharging because you got a fully charged capacitor and it discharges through this 5500 ohm resistor. However you only do that for one second and then you return it to position a so.
The question is while the switch is in position b. So. The question is while the switch is in position b.
How much energy is dissipated in this 550. Ohm resistor. So we start with a whole bunch of energy.
Right because weve charged the capacitor. Using this 50 volt battery. And then it starts to once youve moved the switch over it starts to discharge so after a second the voltage is no longer 50.
If the voltage is no longer 50 the potential energy stored in this capacitor is no longer at its maximum. So lets go ahead and calculate what the initial energy is right you know if you want to know how much energy is being dissipated you simply have to calculate how much energy you had initially initial vs. How much potential energy you had at the end so lets start with the initial case.
The potential energy. Remember in a capacitor is one half cv squared. Theres three forms of this equation.
However in this case. I know the voltage and i know the capacitance. I dont really know the charge.
I could calculate it but its more straightforward simply to use this equation. So this is simply one half the value of the capacitance is 200 micro times. 10 to the minus 6.
And the voltage. And dont forget the square. Its 50 squared all right lets call this initial.
Here if i punch everything in the. Calculator i get 025. Joule.
Thats the initial energy stored in this capacitor before i move the switch all right and now we want to know a final final is after one second so after one second. What is the voltage across once i know the voltage across. I can then calculate the final potential energy.
So once im in position b. I. Start discharging so the voltage as a function of time is my initial voltage multiplied by this exponential decay now be careful.
Which resistance would you use here you clearly use the 5500 ohm resistor. Because this is the circuit here in the second part okay. So lets calculate what this product of rc.
Is here. Rc is simply 550 500. Multiplied by 200 times 10 to the minus.
6 this gives me an rc time constant of 11. Seconds. This makes the calculation a little bit easier okay so at the end the voltage.
After one second simply going to be my initial. Voltage which is 50 and exponential of minus. 1 divided by 11.
My time constant you substitute everything in there. I got twenty point one volts. So if the capacitance is 200.
Micro farad and the voltage is now twenty point. One that means my final potential energy is its no longer. 025.
Joule. You have to calculate it so i get 1 2. 200.
Times. 10 to the minus 6. To get the 200 micro farad.
And again. Now. I have twenty point one squared.
So my final potential energy. Is point zero four joule therefore. The energy dissipated in this 5500.
Ohm resister im simply going to be 025 004. And thats a pretty straightforward. .
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