HereI 0 is the peak current,V 0 the peak source voltage, andZis the impedance of the circuit. The units of impedance are ohms, and its effect on
the circuit is as you might expect: the greater the impedance, the smaller the current. To get an expression forZin terms ofR,XL, andXC, we
will now examine how the voltages across the various components are related to the source voltage. Those voltages are labeledVR,VL, andVC
inFigure 23.48.
Conservation of charge requires current to be the same in each part of the circuit at all times, so that we can say the currents inR,L, andCare
equal and in phase. But we know from the preceding section that the voltage across the inductorVLleads the current by one-fourth of a cycle, the
voltage across the capacitorVCfollows the current by one-fourth of a cycle, and the voltage across the resistorVRis exactly in phase with the
current.Figure 23.49shows these relationships in one graph, as well as showing the total voltage around the circuitV=VR+VL+VC, where all
four voltages are the instantaneous values. According to Kirchhoff’s loop rule, the total voltage around the circuitVis also the voltage of the source.
You can see fromFigure 23.49that whileVRis in phase with the current,VLleads by90º, andVCfollows by90º. ThusVLandVCare
180ºout of phase (crest to trough) and tend to cancel, although not completely unless they have the same magnitude. Since the peak voltages are
not aligned (not in phase), the peak voltageV 0 of the source doesnotequal the sum of the peak voltages acrossR,L, andC. The actual
relationship is
(23.64)V 0 = V 0 R^2 +(V 0 L−V 0 C)^2 ,
whereV 0 R,V 0 L, andV 0 Care the peak voltages acrossR,L, andC, respectively. Now, using Ohm’s law and definitions fromReactance,
Inductive and Capacitive, we substituteV 0 =I 0 Zinto the above, as well asV 0 R=I 0 R,V 0 L=I 0 XL, andV 0 C=I 0 XC, yielding
(23.65)
I 0 Z= I 02 R^2 + (I 0 XL−I 0 XC)^2 =I 0 R^2 + (XL−XC)^2.
I 0 cancels to yield an expression forZ:
(23.66)
Z= R^2 + (XL−XC)^2 ,
which is the impedance of anRLCseries AC circuit. For circuits without a resistor, takeR= 0; for those without an inductor, takeXL= 0; and for
those without a capacitor, takeXC= 0.
Figure 23.49This graph shows the relationships of the voltages in anRLCcircuit to the current. The voltages across the circuit elements add to equal the voltage of the
source, which is seen to be out of phase with the current.
Example 23.12 Calculating Impedance and Current
AnRLCseries circuit has a40.0 Ωresistor, a 3.00 mH inductor, and a5.00 μFcapacitor. (a) Find the circuit’s impedance at 60.0 Hz and 10.0
kHz, noting that these frequencies and the values forLandCare the same as inExample 23.10andExample 23.11. (b) If the voltage
source hasVrms= 120 V, what isIrmsat each frequency?
StrategyFor each frequency, we useZ= R^2 + (XL−XC)^2 to find the impedance and then Ohm’s law to find current. We can take advantage of the
results of the previous two examples rather than calculate the reactances again.
Solution for (a)CHAPTER 23 | ELECTROMAGNETIC INDUCTION, AC CIRCUITS, AND ELECTRICAL TECHNOLOGIES 845