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Response_to_AC

# Response_to_AC - EE201 Response of the basic elements to AC...

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EE201 Response of the basic elements to AC / Phasor Notation Denard Lynch Page 1 of 4 Oct 2009 Let’s go back and look at the voltage w.r.t. time in a R-L-C series circuit: Take, for now, as a given: i(t) = 2.48sin(377t)A The voltage is also sinusoidal, of the same frequency, and a magnitude of 170V peak , but unknown phase. First, find R, X L and X C : R = 47 Ω (given) X L = ω L = (377)(.1H) = 37.7 Ω X C = 1/ ω C = 1/{(377)(100X10 -6 F)} = 26.5 Ω Next, find the maximum voltages across the elements: V R = I M R = (2.48A)(47 Ω ) = 116.6V V L = I M X L = (2.48A)(37.7 Ω ) = 93.5V V C = I M X C = (2.48A)(26.5 Ω ) = 65.7V If we simply add these up we get 275.8V!! Clearly this would violate Kirchhoff’s voltage law as we know it! To help understand why, let’s plot these voltages: We can now see, as we noted earlier, that the various voltages all have differing phases with respect to each other. The problem we face is how to add sine waves of differing phases and frequencies? Here are three options to consider: 1) Graphically: Use the plotted waveforms and graphically add them to get the resultant waveform. Looking at the plot above, we can easily check the validity of this method by testing a few points. For example, if we look at values of the component waves at t=0: v R = 0V , v L = max = 93.5V and v C = min = -65.7V . If we add these we see the result in the ‘sum’ waveform: 27.8V . Also, if we look at the point in time when v R = v C = 0V , the ‘sum’ = v R and we can see that the curves cross at that point. We could

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Response_to_AC - EE201 Response of the basic elements to AC...

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