Anderson Video - Capacitor Circuit Example

Professor Anderson
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<font color="#ffffff">Hello class, Professor Anderson here. Let's take a look at an example circuit</font> <font color="#ffffff">problem, this is just a very simple circuit, AC circuit, with one capacitor in</font> <font color="#ffffff">there. Okay, and let's think about what happens here. Of course the current</font> <font color="#ffffff">sloshes back and forth in this AC circuit, so it charges up the capacitor</font> <font color="#ffffff">one way and then discharges it and charges it up the other way. And then we</font> <font color="#ffffff">are given some numbers in this problem, namely the frequency of this circuit is</font> <font color="#ffffff">15 kilohertz, and at 15 kilohertz we get the following. We get a peak current of</font> <font color="#ffffff">65 milliamps and a rms voltage of 6 volts, and the question in this problem</font> <font color="#ffffff">is what is the capacitance? Okay, what is the capacitor right there?</font> <font color="#ffffff">Alright, so how do we deal with this? Well, we go back to good old Ohm's law, right?</font> <font color="#ffffff">Ohm's law says the following,</font> <font color="#ffffff">it says the voltage is equal to the current times the load, the resistance,</font> <font color="#ffffff">but remember when we deal with these capacitors, we have complex impedance,</font> <font color="#ffffff">and so the equivalent is V equals I times Z,</font> <font color="#ffffff">where this Z is the complex impedance of the device. And to be technically correct,</font> <font color="#ffffff">if we're talking about RMS voltage, we need to put RMS right there, and then</font> <font color="#ffffff">we're talking about RMS current so we put an RMS right there.</font> <font color="#ffffff">Now hopefully what you remember is Z for a capacitor is</font> <font color="#ffffff">1 over Omega times C, right?</font> <font color="#ffffff">As the frequency goes up the impedance of the capacitor goes down,</font> <font color="#ffffff">that current can jump right across there.</font> <font color="#ffffff">As the frequency goes down, the impedance</font> <font color="#ffffff">goes up, and that makes sense if you try to drive DC current through the</font> <font color="#ffffff">capacitor, it's not going to go through, it charges up one plate and then it</font> <font color="#ffffff">stops moving. All right.</font> <font color="#ffffff">Let's see if we can solve this equation now,</font> <font color="#ffffff">Vrms equals Irms 1 over Omega C, for C. Well that's not too bad, right? We can</font> <font color="#ffffff">move C over the other side and what do we have left on the right side? We have</font> <font color="#ffffff">Irms, we have an Omega in the bottom and then we have Vrms in the bottom.</font> <font color="#ffffff">Okay, but we're not given Irms, we're given I naught, but of course the peak current is</font> <font color="#ffffff">related to Ir -- Irms by root 2.</font> <font color="#ffffff">Irms is just I naught over root 2.</font> <font color="#ffffff">We're not given Omega but we are given the frequency F,</font> <font color="#ffffff">but we remember that Omega is just 2 pi times F,</font> <font color="#ffffff">and then we have Vrms, which we are given.</font> <font color="#ffffff">So now it looks like we have everything we can --</font> <font color="#ffffff">everything we need to solve this problem so we can just plug them in and try it.</font> <font color="#ffffff">All right, so let's plug this stuff in.</font> <font color="#ffffff">We've got I naught, which is 65 times 10 to the minus 3 amps,</font> <font color="#ffffff">we're going to divide that by root 2,</font> <font color="#ffffff">and then all of that is over 2 pi times F</font> <font color="#ffffff">after we said was 15 kilohertz, so 15 times 10 to the 3,</font> <font color="#ffffff">and then we have Vrms, which is just 6 volts.</font> <font color="#ffffff">So now everything is in SI units, should work out just fine. If you punch</font> <font color="#ffffff">all these numbers into your calculator you should get something around</font> <font color="#ffffff">81.3 nanofarad. Remember Nano is ten to the minus nine, so this is 81 point three</font> <font color="#ffffff">times ten to the minus nine farads. All right.</font> <font color="#ffffff">That's how you attack problems like that, hopefully that's clear.</font> <font color="#ffffff">If there's any questions, come see me in office hours. Cheers!</font>
<font color="#ffffff">Hello class, Professor Anderson here. Let's take a look at an example circuit</font> <font color="#ffffff">problem, this is just a very simple circuit, AC circuit, with one capacitor in</font> <font color="#ffffff">there. Okay, and let's think about what happens here. Of course the current</font> <font color="#ffffff">sloshes back and forth in this AC circuit, so it charges up the capacitor</font> <font color="#ffffff">one way and then discharges it and charges it up the other way. And then we</font> <font color="#ffffff">are given some numbers in this problem, namely the frequency of this circuit is</font> <font color="#ffffff">15 kilohertz, and at 15 kilohertz we get the following. We get a peak current of</font> <font color="#ffffff">65 milliamps and a rms voltage of 6 volts, and the question in this problem</font> <font color="#ffffff">is what is the capacitance? Okay, what is the capacitor right there?</font> <font color="#ffffff">Alright, so how do we deal with this? Well, we go back to good old Ohm's law, right?</font> <font color="#ffffff">Ohm's law says the following,</font> <font color="#ffffff">it says the voltage is equal to the current times the load, the resistance,</font> <font color="#ffffff">but remember when we deal with these capacitors, we have complex impedance,</font> <font color="#ffffff">and so the equivalent is V equals I times Z,</font> <font color="#ffffff">where this Z is the complex impedance of the device. And to be technically correct,</font> <font color="#ffffff">if we're talking about RMS voltage, we need to put RMS right there, and then</font> <font color="#ffffff">we're talking about RMS current so we put an RMS right there.</font> <font color="#ffffff">Now hopefully what you remember is Z for a capacitor is</font> <font color="#ffffff">1 over Omega times C, right?</font> <font color="#ffffff">As the frequency goes up the impedance of the capacitor goes down,</font> <font color="#ffffff">that current can jump right across there.</font> <font color="#ffffff">As the frequency goes down, the impedance</font> <font color="#ffffff">goes up, and that makes sense if you try to drive DC current through the</font> <font color="#ffffff">capacitor, it's not going to go through, it charges up one plate and then it</font> <font color="#ffffff">stops moving. All right.</font> <font color="#ffffff">Let's see if we can solve this equation now,</font> <font color="#ffffff">Vrms equals Irms 1 over Omega C, for C. Well that's not too bad, right? We can</font> <font color="#ffffff">move C over the other side and what do we have left on the right side? We have</font> <font color="#ffffff">Irms, we have an Omega in the bottom and then we have Vrms in the bottom.</font> <font color="#ffffff">Okay, but we're not given Irms, we're given I naught, but of course the peak current is</font> <font color="#ffffff">related to Ir -- Irms by root 2.</font> <font color="#ffffff">Irms is just I naught over root 2.</font> <font color="#ffffff">We're not given Omega but we are given the frequency F,</font> <font color="#ffffff">but we remember that Omega is just 2 pi times F,</font> <font color="#ffffff">and then we have Vrms, which we are given.</font> <font color="#ffffff">So now it looks like we have everything we can --</font> <font color="#ffffff">everything we need to solve this problem so we can just plug them in and try it.</font> <font color="#ffffff">All right, so let's plug this stuff in.</font> <font color="#ffffff">We've got I naught, which is 65 times 10 to the minus 3 amps,</font> <font color="#ffffff">we're going to divide that by root 2,</font> <font color="#ffffff">and then all of that is over 2 pi times F</font> <font color="#ffffff">after we said was 15 kilohertz, so 15 times 10 to the 3,</font> <font color="#ffffff">and then we have Vrms, which is just 6 volts.</font> <font color="#ffffff">So now everything is in SI units, should work out just fine. If you punch</font> <font color="#ffffff">all these numbers into your calculator you should get something around</font> <font color="#ffffff">81.3 nanofarad. Remember Nano is ten to the minus nine, so this is 81 point three</font> <font color="#ffffff">times ten to the minus nine farads. All right.</font> <font color="#ffffff">That's how you attack problems like that, hopefully that's clear.</font> <font color="#ffffff">If there's any questions, come see me in office hours. Cheers!</font>