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Series AC Circuits with Resistors, Inductors, and Capacitors
Introduction | Series RL Circuit with AC Source | Series RC Circuit with AC Source | Series RLC Circuit with AC Source
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Introduction |
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We already learned how to apply the KVL to series DC circuits comprised of simple resistors. This week, we will advance our learning into the realm of AC systems comprised of resistors, inductors, and capacitors.
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Series RL Circuit with AC Source |
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First, let us consider the AC circuits involving the resistors and inductors in series. We will directly proceed to an analysis involving some real-life numbers. Consider the following example (see Figure 1).
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In the circuit above, we are using a resistor, R1 = 100 
, in series with an inductor L = 10 mH that has a self-resistance, RL = 25 . By the way, every inductor has some resistance, which is due to the wire with which it is fabricated.
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AC Circuit with R and L in Series Source radian frequency, The inductive reactance, j XL = j The complex impedance of the inductor = [25 + j 125.6] Total circuit impedance ZT, in complex notation = [100 + 25 + j 125.6]
Source current Notice from the above expression for the source current that it lags the applied voltage by 45o. We now can calculate the voltage drops, VR and VL.
We can add these two voltage drops to establish the KVL.
The power dissipated by the source = Vin-RMS x IS-RMS x cos(
Finally, power dissipated by the source |
Figure 1a shows the phasor diagram for the source voltage, current, and the voltage drops, VR and VL.
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Series RC Circuit with AC Source |
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Now, let us consider the AC circuits involving the resistors and capacitors in series. We will directly proceed to an analysis involving some real life numbers. Consider the following example (see Figure 2).
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In the circuit above, we are using a resistor, R1 = 100 in series with a capacitor C = 2 F.
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AC Circuit with R and C in Series Source radian frequency, The capacitive reactance, - j XC = - j 1 / [ = {1 / [12,560 x 2 x 10-6 ]} - 90o = - j 40 Total circuit impedance ZT, in complex notation = [100 - j 40 ] Source current Notice from the above expression for the source current that it leads the applied voltage by 21.8o. We now can calculate the voltage drops, VR and VL.
We can add these two voltage drops to establish the KVL.
The power dissipated by the source PD = Vin-RMS x IS-RMS x cos( In the above equation, notice that power factor PF = cos( Finally, power dissipated by the source |
Figure 2a shows the phasor diagram for the source voltage, current, and the voltage drops, VR and VC.
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Series RLC Circuit with AC Source |
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Finally, let us consider the AC circuits involving the resistors and capacitors and inductors in series. Consider the following example (see Figure 3). Notice that the component values are the same as those used in the previous examples.
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AC Circuit with R, L and C in Series Total circuit impedance ZT, in complex notation = [125 - j 40 + j 125]
Once again, the source current, Knowing the source current, the different voltage drops, VR, VC, and VL can be calculated as before. We have
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We can now expand the different voltages into complex format and apply the KVL to verify if the voltage law has been satisfied.
Next week, we will learn to analyze an AC circuit with the components connected in parallel.
Students can do a simulation on MultiSim to verify the above calculations. However, caution must be used as the simulator cannot give the answers in complex mode for the voltages and currents. Students are advised to find out what can be verified with the simulator.
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= 12,560 rads/sec.
90o = 12,560 x 10 x 10-3 90o = j 125.6 
[125 + j 125] 
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