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AC Circuit Graphs

Patrick Ford
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Hey, guys, let's do an example about a C circuits current and voltage in in a see circuit or graft in the following figure. What are the functions that describe these values? Okay, so just remember that voltage as a function of time in a sea circuits, is going to be equal to some maximum voltage produced by the source times cosine of Omega T War Omega is the frequency of the source and I of t the current produced by the source is gonna be some i max, which is the maximum current produced by the source times cosine of omega T. So, in order to find these functions, all we need to do is find what the's maximum values are, right. And then what The angular frequency off this oscillation is, Once we know those three values, right, the angular frequency for both of these functions is gonna be the same. Once we know those three values, we can plug them into our functions and be done with it. Okay, Now, remember that the maximum voltage and the maximum current, according to these equations right above me are just the amplitude of these oscillations. So what's the amplitude of the voltage oscillation. 11 volts. This is the max. What's the amplitude of the current oscillation? It's 2.5 amps. This is actually negative. Imax. That's why this is negative. 2.5 amps because you're at the negative amplitude. The only question remaining is what's the angular frequency? Well, we're told that from this point up here to this point down here takes half. Sorry seconds, not half a second 0.5 seconds. Okay, well, this distance right here is half of a cycle off. Full cycle would be starting from the amplitude coming down to the negative amplitude and going back up to the positive amplitude where you started going from the positive to the negative. Amplitude is half of a cycle, and that takes one half of a period. So that time 10.5 seconds is actually half of the period. So if you say one half of the period is 10. seconds, then we can just multiply this to up to the other side, and we can say that the period is point. Sorry. Yes. 0.1 seconds. Okay. 0.5 times two is just one. Now we wanna find angular frequency from the period. Okay, remember that the angular frequency is divided is defined this two pi f, which is the same as two pi over t. So this is two pi over 0.1 seconds, which is gonna be 62.8 inverse seconds. Okay, so now we know all three of our values. We know that the angular frequency 62.8 seconds in verse. We know that the maximum voltage is 11 volts, and we know that the maximum current is 2.5 amps. So all we have to do is plug in those three values to the two equations above me and will say that the Matt sorry. The current as a function of time is going to be 11 volts, which is the maximum. Sorry, this is the voltage is a function of time. The voltage is a function of time. Is the maximum voltage which is 11 volts times the co sign of thank you Lor frequencies 62 8 seconds times time and the current as a function of time is the maximum current, which is 2.5 amps. right. The amplitude of oscillations, times cosine of once again, the angular frequency, which is 62.8 seconds times time. And these are our answers. All right, guys, Thanks for watching.