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RLC Series AC Circuits

Module by: OpenStax College. E-mail the author

Summary:

  • Calculate the impedance, phase angle, resonant frequency, power, power factor, voltage, and/or current in a RLC series circuit.
  • Draw the circuit diagram for an RLC series circuit.
  • Explain the significance of the resonant frequency.

Impedance

When alone in an AC circuit, inductors, capacitors, and resistors all impede current. How do they behave when all three occur together? Interestingly, their individual resistances in ohms do not simply add. Because inductors and capacitors behave in opposite ways, they partially to totally cancel each other’s effect. Figure 1 shows an RLC series circuit with an AC voltage source, the behavior of which is the subject of this section. The crux of the analysis of an RLC circuit is the frequency dependence of XLXL size 12{X rSub { size 8{L} } } {} and XCXC size 12{X rSub { size 8{C} } } {}, and the effect they have on the phase of voltage versus current (established in the preceding section). These give rise to the frequency dependence of the circuit, with important “resonance” features that are the basis of many applications, such as radio tuners.

Figure 1: An RLC series circuit with an AC voltage source.
The figure describes an R LC series circuit. It shows a resistor R connected in series with an inductor L, connected to a capacitor C in series to an A C source V. The voltage of the A C source is given by V equals V zero sine two pi f t. The voltage across R is V R, across L is V L and across C is V C.

The combined effect of resistance RR size 12{R} {}, inductive reactance XLXL size 12{X rSub { size 8{L} } } {}, and capacitive reactance XCXC size 12{X rSub { size 8{C} } } {} is defined to be impedance, an AC analogue to resistance in a DC circuit. Current, voltage, and impedance in an RLC circuit are related by an AC version of Ohm’s law:

I 0 = V 0 Z or I rms = V rms Z . I 0 = V 0 Z or I rms = V rms Z . size 12{I rSub { size 8{0} } = { {V rSub { size 8{0} } } over {Z} } " or "I rSub { size 8{ ital "rms"} } = { {V rSub { size 8{ ital "rms"} } } over {Z} } "." } {}
(1)

Here I0I0 size 12{I rSub { size 8{0} } } {} is the peak current, V0V0 size 12{V rSub { size 8{0} } } {} the peak source voltage, and Z Z is 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 for ZZ size 12{Z} {} in terms of R R , XLXL size 12{X rSub { size 8{L} } } {}, and XCXC size 12{X rSub { size 8{C} } } {}, we will now examine how the voltages across the various components are related to the source voltage. Those voltages are labeled VRVR size 12{V rSub { size 8{R} } } {}, VLVL size 12{V rSub { size 8{L} } } {}, and VCVC size 12{V rSub { size 8{C} } } {} in Figure 1.

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 in RR size 12{R} {}, LL size 12{L} {}, and CC size 12{C} {} are equal and in phase. But we know from the preceding section that the voltage across the inductor VLVL size 12{V rSub { size 8{L} } } {} leads the current by one-fourth of a cycle, the voltage across the capacitor VCVC size 12{V rSub { size 8{C} } } {} follows the current by one-fourth of a cycle, and the voltage across the resistor VRVR size 12{V rSub { size 8{R} } } {} is exactly in phase with the current. Figure 2 shows these relationships in one graph, as well as showing the total voltage around the circuit V=VR+VL+VCV=VR+VL+VC size 12{V=V rSub { size 8{R} } +V rSub { size 8{L} } +V rSub { size 8{C} } } {}, where all four voltages are the instantaneous values. According to Kirchhoff’s loop rule, the total voltage around the circuit V V is also the voltage of the source.

You can see from Figure 2 that while VRVR size 12{V rSub { size 8{R} } } {} is in phase with the current, VLVL size 12{V rSub { size 8{L} } } {} leads by 90º 90º , and VCVC size 12{V rSub { size 8{C} } } {} follows by 90º 90º . Thus VLVL size 12{V rSub { size 8{L} } } {} and VCVC size 12{V rSub { size 8{C} } } {} are 180º 180º out 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 voltage V0V0 size 12{V rSub { size 8{0} } } {} of the source does not equal the sum of the peak voltages across RR size 12{R} {}, LL size 12{L} {}, and CC size 12{C} {}. The actual relationship is

V 0 = V 0R 2 + ( V 0L V 0C ) 2 , V 0 = V 0R 2 + ( V 0L V 0C ) 2 , size 12{V rSub { size 8{0} } = sqrt {V rSub { size 8{0R} } "" lSup { size 8{2} } + \( V rSub { size 8{0L} } - V rSub { size 8{0C} } \) rSup { size 8{2} } } ,} {}
(2)

where V0RV0R size 12{V rSub { size 8{0R} } } {}, V0LV0L size 12{V rSub { size 8{0L} } } {}, and V0CV0C size 12{V rSub { size 8{0C} } } {} are the peak voltages across RR size 12{R} {}, LL size 12{L} {}, and CC size 12{C} {}, respectively. Now, using Ohm’s law and definitions from Reactance, Inductive and Capacitive, we substitute V0=I0ZV0=I0Z size 12{V rSub { size 8{0} } =I rSub { size 8{0} } Z} {} into the above, as well as V0R=I0RV0R=I0R size 12{V rSub { size 8{0R} } =I rSub { size 8{0} } R} {}, V0L=I0XLV0L=I0XL size 12{V rSub { size 8{0L} } =I rSub { size 8{0} } X rSub { size 8{L} } } {}, and V0C=I0XCV0C=I0XC size 12{V rSub { size 8{0C} } =I rSub { size 8{0} } X rSub { size 8{C} } } {}, yielding

I0Z= I 0 2 R2 + ( I0XLI0XC)2=I0R2+(XLXC)2.I0Z= I 0 2 R2 + ( I0XLI0XC)2=I0R2+(XLXC)2. size 12{I rSub { size 8{0} } Z= sqrt {I rSub { size 8{0} rSup { size 8{2} } } R rSup { size 8{2} } + \( I rSub { size 8{0} } X rSub { size 8{L} } - I rSub { size 8{0} } X rSub { size 8{C} } \) rSup { size 8{2} } } =I rSub { size 8{0} } sqrt {R rSup { size 8{2} } + \( X rSub { size 8{L} } - X rSub { size 8{C} } \) rSup { size 8{2} } } } {}
(3)

I0I0 size 12{I rSub { size 8{0} } } {} cancels to yield an expression for Z Z :

Z=R2+(XLXC)2,Z=R2+(XLXC)2, size 12{Z= sqrt {R rSup { size 8{2} } + \( X rSub { size 8{L} } - X rSub { size 8{C} } \) rSup { size 8{2} } } } {}
(4)

which is the impedance of an RLC series AC circuit. For circuits without a resistor, take R = 0 R = 0 ; for those without an inductor, take XL=0XL=0 size 12{X rSub { size 8{L} } =0} {}; and for those without a capacitor, take XC=0XC=0 size 12{X rSub { size 8{C} } =0} {}.

Figure 2: This graph shows the relationships of the voltages in an RLC circuit 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.
The figure shows graphs showing the relationships of the voltages in an RLC circuit to the current. It has five graphs on the left and two graphs on the right. The first graph on the right is for current I versus time t. Current is plotted along Y axis and time is along X axis. The curve is a smooth progressive sine wave. The second graph is on the right is for voltage V R versus time t. Voltage V R is plotted along Y axis and time is along X axis. The curve is a smooth progressive sine wave. The third graph is on the right is for voltage V L versus time t. Voltage V L is plotted along Y axis and time is along X axis. The curve is a smooth progressive cosine wave. The fourth graph is on the right is for voltage V C versus time t. Voltage V C is plotted along Y axis and time t is along X axis. The curve is a smooth progressive cosine wave starting from negative Y axis. The fifth graph shows the voltage V verses time t for the R L C circuit. Voltage V is plotted along Y axis and time t is along X axis. The curve is a smooth progressive sine wave starting from a point near to origin on negative X axis. The first and the fifth graphs are again shown on the right and their amplitudes and phases compared. The current graph is shown to have a lesser amplitude.

Example 1: Calculating Impedance and Current

An RLC series circuit has a 40.0 Ω 40.0 Ω resistor, a 3.00 mH inductor, and a 5.00 μF 5.00 μF capacitor. (a) Find the circuit’s impedance at 60.0 Hz and 10.0 kHz, noting that these frequencies and the values for L L and C C are the same as in (Reference) and (Reference). (b) If the voltage source has Vrms=120VVrms=120V size 12{V rSub { size 8{"rms"} } ="120"`V} {}, what is IrmsIrms size 12{I rSub { size 8{"rms"} } } {} at each frequency?

Strategy

For each frequency, we use Z=R2+(XLXC)2Z=R2+(XLXC)2 size 12{Z= sqrt {R rSup { size 8{2} } + \( X rSub { size 8{L} } - X rSub { size 8{C} } \) rSup { size 8{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)

At 60.0 Hz, the values of the reactances were found in (Reference) to be XL=1.13ΩXL=1.13Ω size 12{X rSub { size 8{L} } =1 "." "13" %OMEGA } {} and in (Reference) to be XC=531 Ω XC=531 Ω size 12{X rSub { size 8{C} } ="531 " %OMEGA } {}. Entering these and the given 40.0 Ω 40.0 Ω for resistance into Z=R2+(XLXC)2Z=R2+(XLXC)2 size 12{Z= sqrt {R rSup { size 8{2} } + \( X rSub { size 8{L} } - X rSub { size 8{C} } \) rSup { size 8{2} } } } {} yields

Z = R2+(XLXC)2 = (40.0Ω)2+(1.13Ω531Ω)2 = 531Ω at 60.0 Hz.Z = R2+(XLXC)2 = (40.0Ω)2+(1.13Ω531Ω)2 = 531Ω at 60.0 Hz.alignl { stack { size 12{Z= sqrt {R rSup { size 8{2} } + \( X rSub { size 8{L} } - X rSub { size 8{C} } \) rSup { size 8{2} } } } {} # " "= sqrt { \( "40" "." 0` %OMEGA \) rSup { size 8{2} } + \( 1 "." "13" %OMEGA - "531" %OMEGA \) rSup { size 8{2} } } {} # " "="531" %OMEGA " at 60" "." "0 Hz" {} } } {}
(5)

Similarly, at 10.0 kHz, XL=188ΩXL=188Ω size 12{X rSub { size 8{L} } ="188" %OMEGA } {} and XC=3.18ΩXC=3.18Ω size 12{X rSub { size 8{C} } =3 "." "18" %OMEGA } {}, so that

Z = (40.0Ω)2+(188Ω3.18Ω)2 = 190Ω at 10.0 kHz. Z = (40.0Ω)2+(188Ω3.18Ω)2 = 190Ω at 10.0 kHz.alignl { stack { size 12{Z= sqrt { \( "40" "." 0` %OMEGA \) rSup { size 8{2} } + \( "188" %OMEGA - 3 "." "18" %OMEGA \) rSup { size 8{2} } } } {} # " "="190" %OMEGA " at 10" "." "0 kHz" {} } } {}
(6)

Discussion for (a)

In both cases, the result is nearly the same as the largest value, and the impedance is definitely not the sum of the individual values. It is clear that XLXL size 12{X rSub { size 8{L} } } {} dominates at high frequency and XCXC size 12{X rSub { size 8{C} } } {} dominates at low frequency.

Solution for (b)

The current IrmsIrms size 12{I rSub { size 8{"rms"} } } {} can be found using the AC version of Ohm’s law in Equation Irms=Vrms/ZIrms=Vrms/Z size 12{I rSub { size 8{"rms"} } =V rSub { size 8{"rms"} } /Z} {}:

Irms=VrmsZ=120 V531 Ω=0.226 AIrms=VrmsZ=120 V531 Ω=0.226 A size 12{I rSub { size 8{"rms"} } = { {V rSub { size 8{"rms"} } } over {Z} } = { {"120"" V"} over {"531 " %OMEGA } } =0 "." "226"" A"} {} at 60.0 Hz

Finally, at 10.0 kHz, we find

Irms=VrmsZ=120 V190 Ω=0.633 AIrms=VrmsZ=120 V190 Ω=0.633 A size 12{I rSub { size 8{"rms"} } = { {V rSub { size 8{"rms"} } } over {Z} } = { {"120"" V"} over {"190 " %OMEGA } } =0 "." "633"" A"} {} at 10.0 kHz

Discussion for (a)

The current at 60.0 Hz is the same (to three digits) as found for the capacitor alone in (Reference). The capacitor dominates at low frequency. The current at 10.0 kHz is only slightly different from that found for the inductor alone in (Reference). The inductor dominates at high frequency.

Resonance in RLC Series AC Circuits

How does an RLC circuit behave as a function of the frequency of the driving voltage source? Combining Ohm’s law, Irms=Vrms/ZIrms=Vrms/Z size 12{I rSub { size 8{"rms"} } =V rSub { size 8{"rms"} } /Z} {}, and the expression for impedance Z Z from Z=R2+(XLXC)2Z=R2+(XLXC)2 size 12{Z= sqrt {R rSup { size 8{2} } + \( X rSub { size 8{L} } - X rSub { size 8{C} } \) rSup { size 8{2} } } } {} gives

Irms=VrmsR2+(XLXC)2.Irms=VrmsR2+(XLXC)2. size 12{I rSub { size 8{"rms"} } = { {V rSub { size 8{"rms"} } } over { sqrt {R rSup { size 8{2} } + \( X rSub { size 8{L} } - X rSub { size 8{C} } \) rSup { size 8{2} } } } } } {}
(7)

The reactances vary with frequency, with XLXL size 12{X rSub { size 8{L} } } {} large at high frequencies and XCXC size 12{X rSub { size 8{C} } } {} large at low frequencies, as we have seen in three previous examples. At some intermediate frequency f0f0 size 12{f rSub { size 8{0} } } {}, the reactances will be equal and cancel, giving Z=RZ=R size 12{Z=R} {} —this is a minimum value for impedance, and a maximum value for IrmsIrms size 12{I rSub { size 8{"rms"} } } {} results. We can get an expression for f0f0 size 12{f rSub { size 8{0} } } {} by taking

XL=XC.XL=XC. size 12{X rSub { size 8{L} } =X rSub { size 8{C} } } {}
(8)

Substituting the definitions of XLXL size 12{X rSub { size 8{L} } } {} and XCXC size 12{X rSub { size 8{C} } } {},

2πf0L=12πf0C.2πf0L=12πf0C. size 12{2πf rSub { size 8{0} } L= { {1} over {2πf rSub { size 8{0} } C} } } {}
(9)

Solving this expression for f0f0 size 12{f rSub { size 8{0} } } {} yields

f0=1LC,f0=1LC, size 12{f rSub { size 8{0} } = { {1} over {2π sqrt { ital "LC"} } } } {}
(10)

where f0f0 size 12{f rSub { size 8{0} } } {} is the resonant frequency of an RLC series circuit. This is also the natural frequency at which the circuit would oscillate if not driven by the voltage source. At f0f0 size 12{f rSub { size 8{0} } } {}, the effects of the inductor and capacitor cancel, so that Z=RZ=R size 12{Z=R} {}, and IrmsIrms size 12{I rSub { size 8{"rms"} } } {} is a maximum.

Resonance in AC circuits is analogous to mechanical resonance, where resonance is defined to be a forced oscillation—in this case, forced by the voltage source—at the natural frequency of the system. The receiver in a radio is an RLC circuit that oscillates best at its f0f0 size 12{f rSub { size 8{0} } } {}. A variable capacitor is often used to adjust f0f0 size 12{f rSub { size 8{0} } } {} to receive a desired frequency and to reject others. Figure 3 is a graph of current as a function of frequency, illustrating a resonant peak in IrmsIrms size 12{I rSub { size 8{"rms"} } } {} at f0f0 size 12{f rSub { size 8{0} } } {}. The two curves are for two different circuits, which differ only in the amount of resistance in them. The peak is lower and broader for the higher-resistance circuit. Thus the higher-resistance circuit does not resonate as strongly and would not be as selective in a radio receiver, for example.

Figure 3: A graph of current versus frequency for two RLC series circuits differing only in the amount of resistance. Both have a resonance at f0f0 size 12{f rSub { size 8{0} } } {}, but that for the higher resistance is lower and broader. The driving AC voltage source has a fixed amplitude V0V0 size 12{V rSub { size 8{0} } } {}.
The figure describes a graph of current I versus frequency f. Current I r m s is plotted along Y axis and frequency f is plotted along X axis. Two curves are shown. The upper curve is for small resistance and lower curve is for large resistance. Both the curves have a smooth rise and a fall. The peaks are marked for frequency f zero. The curve for smaller resistance has a higher value of peak than the curve for large resistance.

Example 2: Calculating Resonant Frequency and Current

For the same RLC series circuit having a 40.0 Ω 40.0 Ω resistor, a 3.00 mH inductor, and a 5.00 μF 5.00 μF capacitor: (a) Find the resonant frequency. (b) Calculate IrmsIrms size 12{I rSub { size 8{"rms"} } } {} at resonance if VrmsVrms size 12{V rSub { size 8{"rms"} } } {} is 120 V.

Strategy

The resonant frequency is found by using the expression in f0=1LCf0=1LC size 12{f rSub { size 8{0} } = { {1} over {2π sqrt { ital "LC"} } } } {}. The current at that frequency is the same as if the resistor alone were in the circuit.

Solution for (a)

Entering the given values for L L and C C into the expression given for f0f0 size 12{f rSub { size 8{0} } } {} in f0=1LCf0=1LC size 12{f rSub { size 8{0} } = { {1} over {2π sqrt { ital "LC"} } } } {} yields

f0 = 1LC = 1(3.00×103 H)(5.00×106 F)=1.30 kHz. f0 = 1LC = 1(3.00×103 H)(5.00×106 F)=1.30 kHz.alignl { stack { size 12{f rSub { size 8{0} } = { {1} over {2π sqrt { ital "LC"} } } } {} # " "= { {1} over {2π sqrt { \( 3 "." "00" times "10" rSup { size 8{ - 3} } " H" \) \( 5 "." "00" times "10" rSup { size 8{ - 6} } " F" \) } } } =1 "." "30"" kHz" {} } } {}
(11)

Discussion for (a)

We see that the resonant frequency is between 60.0 Hz and 10.0 kHz, the two frequencies chosen in earlier examples. This was to be expected, since the capacitor dominated at the low frequency and the inductor dominated at the high frequency. Their effects are the same at this intermediate frequency.

Solution for (b)

The current is given by Ohm’s law. At resonance, the two reactances are equal and cancel, so that the impedance equals the resistance alone. Thus,

Irms=VrmsZ=120 V40.0 Ω=3.00 A.Irms=VrmsZ=120 V40.0 Ω=3.00 A. size 12{I rSub { size 8{"rms"} } = { {V rSub { size 8{"rms"} } } over {Z} } = { {"120"" V"} over {"40" "." "0 " %OMEGA } } =3 "." "00"" A"} {}
(12)

Discussion for (b)

At resonance, the current is greater than at the higher and lower frequencies considered for the same circuit in the preceding example.

Power in RLC Series AC Circuits

If current varies with frequency in an RLC circuit, then the power delivered to it also varies with frequency. But the average power is not simply current times voltage, as it is in purely resistive circuits. As was seen in Figure 2, voltage and current are out of phase in an RLC circuit. There is a phase angle ϕϕ size 12{ϕ} {} between the source voltage VV size 12{V} {} and the current II size 12{I} {}, which can be found from

cosϕ=RZ.cosϕ=RZ. size 12{"cos"ϕ= { {R} over {Z} } } {}
(13)

For example, at the resonant frequency or in a purely resistive circuit Z=RZ=R size 12{Z=R} {}, so that cosϕ=1cosϕ=1 size 12{"cos"ϕ=1} {}. This implies that ϕ=0ºϕ=0º size 12{ϕ=0 rSup { size 8{ circ } } } {} and that voltage and current are in phase, as expected for resistors. At other frequencies, average power is less than at resonance. This is both because voltage and current are out of phase and because IrmsIrms size 12{I rSub { size 8{"rms"} } } {} is lower. The fact that source voltage and current are out of phase affects the power delivered to the circuit. It can be shown that the average power is

P ave = I rms V rms cos ϕ , P ave = I rms V rms cos ϕ , size 12{P rSub { size 8{"ave"} } =I rSub { size 8{"rms"} } V rSub { size 8{"rms"} } "cos"ϕ} {}
(14)

Thus cosϕcosϕ size 12{"cos"ϕ} {} is called the power factor, which can range from 0 to 1. Power factors near 1 are desirable when designing an efficient motor, for example. At the resonant frequency, cosϕ=1cosϕ=1 size 12{"cos"ϕ=1} {}.

Example 3: Calculating the Power Factor and Power

For the same RLC series circuit having a 40.0 Ω 40.0 Ω resistor, a 3.00 mH inductor, a 5.00 μF 5.00 μF capacitor, and a voltage source with a V rms V rms of 120 V: (a) Calculate the power factor and phase angle for f=60.0Hzf=60.0Hz size 12{f="60" "." 0`"Hz"} {}. (b) What is the average power at 50.0 Hz? (c) Find the average power at the circuit’s resonant frequency.

Strategy and Solution for (a)

The power factor at 60.0 Hz is found from

cosϕ=RZ.cosϕ=RZ. size 12{"cos"ϕ= { {R} over {Z} } } {}
(15)

We know Z = 531 Ω Z = 531 Ω from Example 1, so that

cosϕ=40.0Ω531 Ω=0.0753 at 60.0 Hz.cosϕ=40.0Ω531 Ω=0.0753 at 60.0 Hz. size 12{"cos"Ø= { {"40" "." 0 %OMEGA } over {5"31 " %OMEGA } } =0 "." "0753"} {}
(16)

This small value indicates the voltage and current are significantly out of phase. In fact, the phase angle is

ϕ=cos10.0753=85.7º at 60.0 Hz.ϕ=cos10.0753=85.7º at 60.0 Hz. size 12{ϕ="cos" rSup { size 8{ - 1} } 0 "." "0753"="85" "." 7 rSup { size 8{ circ } } } {}
(17)

Discussion for (a)

The phase angle is close to 90º 90º , consistent with the fact that the capacitor dominates the circuit at this low frequency (a pure RC circuit has its voltage and current 90º 90º out of phase).

Strategy and Solution for (b)

The average power at 60.0 Hz is

Pave=IrmsVrmscosϕ.Pave=IrmsVrmscosϕ. size 12{P rSub { size 8{"ave"} } =I rSub { size 8{"rms"} } V rSub { size 8{"rms"} } "cos"ϕ} {}
(18)

IrmsIrms size 12{I rSub { size 8{"rms"} } } {} was found to be 0.226 A in Example 1. Entering the known values gives

Pave=(0.226 A)(120 V)(0.0753)=2.04 W at 60.0 Hz.Pave=(0.226 A)(120 V)(0.0753)=2.04 W at 60.0 Hz. size 12{P rSub { size 8{"ave"} } = \( 0 "." "226"" A" \) \( "120"" V" \) \( 0 "." "0753" \) =2 "." "04"" W"} {}
(19)

Strategy and Solution for (c)

At the resonant frequency, we know cosϕ=1cosϕ=1 size 12{"cos"ϕ=1} {}, and IrmsIrms size 12{I rSub { size 8{"rms"} } } {} was found to be 6.00 A in Example 2. Thus,

Pave=(3.00 A)(120 V)(1)=360 WPave=(3.00 A)(120 V)(1)=360 W size 12{P rSub { size 8{"ave"} } = \( 3 "." "00"" A" \) \( "120"" V" \) \( 1 \) ="350"" W"} {} at resonance (1.30 kHz)

Discussion

Both the current and the power factor are greater at resonance, producing significantly greater power than at higher and lower frequencies.

Power delivered to an RLC series AC circuit is dissipated by the resistance alone. The inductor and capacitor have energy input and output but do not dissipate it out of the circuit. Rather they transfer energy back and forth to one another, with the resistor dissipating exactly what the voltage source puts into the circuit. This assumes no significant electromagnetic radiation from the inductor and capacitor, such as radio waves. Such radiation can happen and may even be desired, as we will see in the next chapter on electromagnetic radiation, but it can also be suppressed as is the case in this chapter. The circuit is analogous to the wheel of a car driven over a corrugated road as shown in Figure 4. The regularly spaced bumps in the road are analogous to the voltage source, driving the wheel up and down. The shock absorber is analogous to the resistance damping and limiting the amplitude of the oscillation. Energy within the system goes back and forth between kinetic (analogous to maximum current, and energy stored in an inductor) and potential energy stored in the car spring (analogous to no current, and energy stored in the electric field of a capacitor). The amplitude of the wheels’ motion is a maximum if the bumps in the road are hit at the resonant frequency.

Figure 4: The forced but damped motion of the wheel on the car spring is analogous to an RLC series AC circuit. The shock absorber damps the motion and dissipates energy, analogous to the resistance in an RLC circuit. The mass and spring determine the resonant frequency.
The figure describes the path of motion of a wheel of a car. The front wheel of a car is shown. A shock absorber attached to the wheel is also shown. The path of motion is shown as vertically up and down.

A pure LC circuit with negligible resistance oscillates at f0f0 size 12{f rSub { size 8{0} } } {}, the same resonant frequency as an RLC circuit. It can serve as a frequency standard or clock circuit—for example, in a digital wristwatch. With a very small resistance, only a very small energy input is necessary to maintain the oscillations. The circuit is analogous to a car with no shock absorbers. Once it starts oscillating, it continues at its natural frequency for some time. Figure 5 shows the analogy between an LC circuit and a mass on a spring.

Figure 5: An LC circuit is analogous to a mass oscillating on a spring with no friction and no driving force. Energy moves back and forth between the inductor and capacitor, just as it moves from kinetic to potential in the mass-spring system.
The figure describes four stages of an L C oscillation circuit compared to a mass oscillating on a spring. Part a of the figure shows a mass attached to a horizontal spring. The spring is attached to a fixed support on the left. The mass is at rest as shown by velocity v equals zero. The energy of the spring is shown as potential energy. This is compared with a circuit containing a capacitor C and inductor L connected together. The energy is shown as stored in the electric field E of the capacitor between the plates. One plate is shown to have a negative polarity and other plate is shown to have a positive polarity. Part b of the figure shows a mass attached to a horizontal spring which is attached to a fixed support on the left. The mass is shown to move horizontal toward the fixed support with velocity v. The energy here is stored as the kinetic energy of the spring. This is compared with a circuit containing a capacitor C and inductor L connected together. A current is shown in the circuit and energy is stored as magnetic field B in the inductor. Part c of the figure shows a mass attached to a horizontal spring which is attached to a fixed support on the left. The spring is shown as not stretched and the energy is shown as potential energy of the spring. The mass is show to have displaced toward left. This is compared with a circuit containing a capacitor C and inductor L connected together. The energy is shown as stored in the electric field E of the capacitor between the plates. One plate is shown to have a negative polarity and other plate is shown to have a positive polarity. But the polarities are reverse of the first case in part a. Part d of the figure shows a mass attached to a horizontal spring which is attached to a fixed support on the left. The mass is shown to move toward right with velocity v. the energy of the spring is kinetic energy. This is compared with a circuit containing a capacitor C and inductor L connected together. A current is shown in the circuit opposite to that in part b and energy is stored as magnetic field B in the inductor.

PhET Explorations: Circuit Construction Kit (AC+DC), Virtual Lab:

Build circuits with capacitors, inductors, resistors and AC or DC voltage sources, and inspect them using lab instruments such as voltmeters and ammeters.

Section Summary

  • The AC analogy to resistance is impedance Z Z , the combined effect of resistors, inductors, and capacitors, defined by the AC version of Ohm’s law:
    I 0 = V 0 Z or I rms = V rms Z , I 0 = V 0 Z or I rms = V rms Z , size 12{I rSub { size 8{0} } = { {V rSub { size 8{0} } } over {Z} } " or "I rSub { size 8{ ital "rms"} } = { {V rSub { size 8{ ital "rms"} } } over {Z} } ,} {}
    (20)
    where I0I0 size 12{I rSub { size 8{0} } } {} is the peak current and V0V0 size 12{V rSub { size 8{0} } } {} is the peak source voltage.
  • Impedance has units of ohms and is given by Z=R2+(XLXC)2Z=R2+(XLXC)2 size 12{Z= sqrt {R rSup { size 8{2} } + \( X rSub { size 8{L} } - X rSub { size 8{C} } \) rSup { size 8{2} } } } {}.
  • The resonant frequency f0f0 size 12{f rSub { size 8{0} } } {}, at which XL=XCXL=XC size 12{X rSub { size 8{L} } =X rSub { size 8{C} } } {}, is
    f0=1LC.f0=1LC. size 12{f rSub { size 8{0} } = { {1} over {2π sqrt { ital "LC"} } } } {}
    (21)
  • In an AC circuit, there is a phase angle ϕϕ size 12{ϕ} {} between source voltage VV size 12{V} {} and the current II size 12{I} {}, which can be found from
    cosϕ=RZ,cosϕ=RZ, size 12{"cos"ϕ= { {R} over {Z} } } {}
    (22)
  • ϕ=ϕ= size 12{ϕ=0 rSup { size 8{ circ } } } {} for a purely resistive circuit or an RLC circuit at resonance.
  • The average power delivered to an RLC circuit is affected by the phase angle and is given by
    Pave=IrmsVrmscosϕ,Pave=IrmsVrmscosϕ, size 12{P rSub { size 8{"ave"} } =I rSub { size 8{"rms"} } V rSub { size 8{"rms"} } "cos"ϕ} {}
    (23)
    cosϕcosϕ size 12{"cos"ϕ} {} is called the power factor, which ranges from 0 to 1.

Conceptual Questions

Exercise 1

Does the resonant frequency of an AC circuit depend on the peak voltage of the AC source? Explain why or why not.

Exercise 2

Suppose you have a motor with a power factor significantly less than 1. Explain why it would be better to improve the power factor as a method of improving the motor’s output, rather than to increase the voltage input.

Problems & Exercises

Exercise 1

An RL circuit consists of a 40.0 Ω 40.0 Ω resistor and a 3.00 mH inductor. (a) Find its impedance Z Z at 60.0 Hz and 10.0 kHz. (b) Compare these values of Z Z with those found in Example 1 in which there was also a capacitor.

Solution

(a) 40.02 Ω 40.02 Ω at 60.0 Hz, 193 Ω 193 Ω at 10.0 kHz

(b) At 60 Hz, with a capacitor, Z=531 Ω Z=531 Ω , over 13 times as high as without the capacitor. The capacitor makes a large difference at low frequencies. At 10 kHz, with a capacitor Z=190 Ω Z=190 Ω , about the same as without the capacitor. The capacitor has a smaller effect at high frequencies.

Exercise 2

An RC circuit consists of a 40.0 Ω 40.0 Ω resistor and a 5.00 μF 5.00 μF capacitor. (a) Find its impedance at 60.0 Hz and 10.0 kHz. (b) Compare these values of Z Z with those found in Example 1, in which there was also an inductor.

Exercise 3

An LC circuit consists of a 3.00mH3.00mH size 12{3 "." "00" μH} {} inductor and a 5.00μF5.00μF size 12{5 "." "00" μF} {} capacitor. (a) Find its impedance at 60.0 Hz and 10.0 kHz. (b) Compare these values of ZZ size 12{Z} {} with those found in Example 1 in which there was also a resistor.

Solution

(a) 529 Ω 529 Ω at 60.0 Hz, 185 Ω 185 Ω at 10.0 kHz

(b) These values are close to those obtained in Example 1 because at low frequency the capacitor dominates and at high frequency the inductor dominates. So in both cases the resistor makes little contribution to the total impedance.

Exercise 4

What is the resonant frequency of a 0.500 mH inductor connected to a 40.0 μF 40.0 μF capacitor?

Exercise 5

To receive AM radio, you want an RLC circuit that can be made to resonate at any frequency between 500 and 1650 kHz. This is accomplished with a fixed 1.00 μH 1.00 μH inductor connected to a variable capacitor. What range of capacitance is needed?

Solution

9.30 nF to 101 nF

Exercise 6

Suppose you have a supply of inductors ranging from 1.00 nH to 10.0 H, and capacitors ranging from 1.00 pF to 0.100 F. What is the range of resonant frequencies that can be achieved from combinations of a single inductor and a single capacitor?

Exercise 7

What capacitance do you need to produce a resonant frequency of 1.00 GHz, when using an 8.00 nH inductor?

Solution

3.17 pF

Exercise 8

What inductance do you need to produce a resonant frequency of 60.0 Hz, when using a 2.00 μF 2.00 μF capacitor?

Exercise 9

The lowest frequency in the FM radio band is 88.0 MHz. (a) What inductance is needed to produce this resonant frequency if it is connected to a 2.50 pF capacitor? (b) The capacitor is variable, to allow the resonant frequency to be adjusted to as high as 108 MHz. What must the capacitance be at this frequency?

Solution

(a) 1.31 μH 1.31 μH

(b) 1.66 pF

Exercise 10

An RLC series circuit has a 2.50 Ω 2.50 Ω resistor, a 100 μH 100 μH inductor, and an 80.0 μF 80.0 μF capacitor.(a) Find the circuit’s impedance at 120 Hz. (b) Find the circuit’s impedance at 5.00 kHz. (c) If the voltage source has Vrms=5.60VVrms=5.60V size 12{V rSub { size 8{"rms"} } =5 "." "60"`V} {}, what is IrmsIrms size 12{I rSub { size 8{"rms"} } } {} at each frequency? (d) What is the resonant frequency of the circuit? (e) What is IrmsIrms size 12{I rSub { size 8{"rms"} } } {} at resonance?

Exercise 11

An RLC series circuit has a 1.00 kΩ 1.00 kΩ resistor, a 150 μH 150 μH inductor, and a 25.0 nF capacitor. (a) Find the circuit’s impedance at 500 Hz. (b) Find the circuit’s impedance at 7.50 kHz. (c) If the voltage source has Vrms=408VVrms=408V size 12{V rSub { size 8{"rms"} } ="408"`V} {}, what is IrmsIrms size 12{I rSub { size 8{"rms"} } } {} at each frequency? (d) What is the resonant frequency of the circuit? (e) What is IrmsIrms size 12{I rSub { size 8{"rms"} } } {} at resonance?

Solution

(a) 12.8 kΩ 12.8 kΩ

(b) 1.31 kΩ 1.31 kΩ

(c) 31.9 mA at 500 Hz, 312 mA at 7.50 kHz

(d) 82.2 kHz

(e) 0.408 A

Exercise 12

An RLC series circuit has a 2.50 Ω 2.50 Ω resistor, a 100 μH 100 μH inductor, and an 80.0 μF 80.0 μF capacitor. (a) Find the power factor at f = 120 Hz f = 120 Hz . (b) What is the phase angle at 120 Hz? (c) What is the average power at 120 Hz? (d) Find the average power at the circuit’s resonant frequency.

Exercise 13

An RLC series circuit has a 1.00 kΩ 1.00 kΩ resistor, a 150 μH 150 μH inductor, and a 25.0 nF capacitor. (a) Find the power factor at f = 7.50 Hz f = 7.50 Hz . (b) What is the phase angle at this frequency? (c) What is the average power at this frequency? (d) Find the average power at the circuit’s resonant frequency.

Solution

(a) 0.159

(b) 80.9º 80.9º

(c) 26.4 W

(d) 166 W

Exercise 14

An RLC series circuit has a 200 Ω 200 Ω resistor and a 25.0 mH inductor. At 8000 Hz, the phase angle is 45.0º 45.0º . (a) What is the impedance? (b) Find the circuit’s capacitance. (c) If Vrms=408VVrms=408V size 12{V rSub { size 8{"rms"} } ="408"`V} {} is applied, what is the average power supplied?

Exercise 15

Referring to Example 3, find the average power at 10.0 kHz.

Solution

16.0 W

Glossary

impedance:
the AC analogue to resistance in a DC circuit; it is the combined effect of resistance, inductive reactance, and capacitive reactance in the form Z=R2+(XLXC)2Z=R2+(XLXC)2 size 12{Z= sqrt {R rSup { size 8{2} } + \( X rSub { size 8{L} } - X rSub { size 8{C} } \) rSup { size 8{2} } } } {}
resonant frequency:
the frequency at which the impedance in a circuit is at a minimum, and also the frequency at which the circuit would oscillate if not driven by a voltage source; calculated by f0=1LCf0=1LC size 12{f rSub { size 8{0} } = { {1} over {2π sqrt { ital "LC"} } } } {}
phase angle:
denoted by ϕϕ size 12{ϕ} {}, the amount by which the voltage and current are out of phase with each other in a circuit
power factor:
the amount by which the power delivered in the circuit is less than the theoretical maximum of the circuit due to voltage and current being out of phase; calculated by cosϕcosϕ size 12{"cos"ϕ} {}

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