23 Electromagnetic Induction, AC Circuits, and Electrical Technologies

# 191 23.12 RLC Series AC Circuits

- 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. [link] 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.

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:

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 ZZ 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 RR, 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 [link].

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. [link] 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 VV is also the voltage of the source.

You can see from [link] 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 90º90º, and VCVC size 12{V rSub { size 8{C} } } {} follows by 90º90º. Thus VLVL size 12{V rSub { size 8{L} } } {} and VCVC size 12{V rSub { size 8{C} } } {} are 180º180º 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

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

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

which is the impedance of an *RLC* series AC circuit. For circuits without a resistor, take R=0R=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} {}.

An *RLC *series circuit has a 40.0 Ω40.0 Ω resistor, a 3.00 mH inductor, and a

5.00 μF5.00 μF capacitor. (a) Find the circuit’s impedance at 60.0 Hz and 10.0 kHz, noting that these frequencies and the values for

LL and

CC

are the same as in [link] and [link]. (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+(XL−XC)2Z=R2+(XL−XC)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 [link] to be XL=1.13ΩXL=1.13Ω size 12{X rSub { size 8{L} } =1 “.” “13” %OMEGA } {} and in [link] to be XC=531 ΩXC=531 Ω size 12{X rSub { size 8{C} } =”531 ” %OMEGA } {}. Entering these and the given 40.0 Ω40.0 Ω for resistance into Z=R2+(XL−XC)2Z=R2+(XL−XC)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

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” {}

} } {}

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

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” {}

} } {}

**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 [link]. The capacitor dominates at low frequency. The current at 10.0 kHz is only slightly different from that found for the inductor alone in [link]. 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 ZZ from Z=R2+(XL−XC)2Z=R2+(XL−XC)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

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

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

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

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. [link] 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.

For the same *RLC* series circuit having a 40.0 Ω40.0 Ω resistor, a 3.00 mH inductor, and a 5.00 μF5.00 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=12πLCf0=12πLC 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 LL and CC into the expression given for f0f0 size 12{f rSub { size 8{0} } } {} in f0=12πLCf0=12πLC size 12{f rSub { size 8{0} } = { {1} over {2π sqrt { ital “LC”} } } } {} yields

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” {}

} } {}

**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,

**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 [link], 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

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

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} {}.

For the same *RLC* series circuit having a 40.0 Ω40.0 Ω resistor, a 3.00 mH inductor, a

5.00 μF5.00 μF capacitor, and a voltage source with a VrmsVrms 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

We know Z= 531 ΩZ= 531 Ω from [link], so that

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

**Discussion for (a)**

The phase angle is close to

90º90º, consistent with the fact that the capacitor dominates the circuit at this low frequency (a pure *RC* circuit has its voltage and current 90º90º out of phase).

**Strategy and Solution for (b)**

The average power at 60.0 Hz is

IrmsIrms size 12{I rSub { size 8{“rms”} } } {} was found to be 0.226 A in [link]. Entering the known values gives

**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 [link]. 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 [link]. 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.

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. [link] shows the analogy between an *LC* circuit and a mass on a spring.

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 ZZ, the combined effect of resistors, inductors, and capacitors, defined by the AC version of Ohm’s law:
I0=V0Z or Irms=VrmsZ,I0=V0Z or Irms=VrmsZ, 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} } ,} {}
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+(XL−XC)2Z=R2+(XL−XC)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=12πLC.f0=12πLC. size 12{f rSub { size 8{0} } = { {1} over {2π sqrt { ital “LC”} } } } {}
- 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 fromcosϕ=RZ,cosϕ=RZ, size 12{“cos”ϕ= { {R} over {Z} } } {} - ϕ=0ºϕ=0º 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 byPave=IrmsVrmscosϕ,Pave=IrmsVrmscosϕ, size 12{P rSub { size 8{“ave”} } =I rSub { size 8{“rms”} } V rSub { size 8{“rms”} } “cos”ϕ} {}cosϕcosϕ size 12{“cos”ϕ} {} is called the power factor, which ranges from 0 to 1.

# Conceptual Questions

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

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

An *RL* circuit consists of a

40.0 Ω40.0 Ω resistor and a

3.00 mH inductor. (a) Find its impedance ZZ at 60.0 Hz and 10.0 kHz. (b) Compare these values of ZZ with those found in [link] in which there was also a capacitor.

(a) 40.02 Ω40.02 Ω at 60.0 Hz, 193 Ω193 Ω at 10.0 kHz

(b) At 60 Hz, with a capacitor, Z=531 ΩZ=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 ΩZ=190 Ω, about the same as without the capacitor. The capacitor has a smaller effect at high frequencies.

An *RC* circuit consists of a

40.0 Ω40.0 Ω resistor and a

5.00 μF5.00 μF capacitor. (a) Find its impedance at 60.0 Hz and 10.0 kHz. (b) Compare these values of ZZ with those found in [link], in which there was also an inductor.

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 [link] in which there was also a resistor.

(a) 529 Ω529 Ω at 60.0 Hz, 185 Ω185 Ω at 10.0 kHz

(b) These values are close to those obtained in [link] 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.

What is the resonant frequency of a 0.500 mH inductor connected to a

40.0 μF40.0 μF capacitor?

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 μH1.00 μH inductor connected to a variable capacitor. What range of capacitance is needed?

9.30 nF to 101 nF

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?

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

3.17 pF

What inductance do you need to produce a resonant frequency of 60.0 Hz, when using a 2.00 μF2.00 μF capacitor?

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?

(a) 1.31 μH1.31 μH

(b) 1.66 pF

An *RLC* series circuit has a

2.50 Ω2.50 Ω resistor, a

100 μH100 μH inductor, and an

80.0 μF80.0 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?

An *RLC* series circuit has a 1.00 kΩ1.00 kΩ resistor, a 150 μH150 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?

(a) 12.8 kΩ12.8 kΩ

(b) 1.31 kΩ1.31 kΩ

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

(d) 82.2 kHz

(e) 0.408 A

An *RLC* series circuit has a

2.50 Ω2.50 Ω resistor, a

100 μH100 μH inductor, and an

80.0 μF80.0 μF capacitor. (a) Find the power factor at f=120 Hzf=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.

An *RLC* series circuit has a

1.00 kΩ1.00 kΩ resistor, a

150 μH150 μH inductor, and a 25.0 nF capacitor. (a) Find the power factor at f=7.50 Hzf=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.

(a) 0.159

(b) 80.9º80.9º

(c) 26.4 W

(d) 166 W

An *RLC* series circuit has a

200 Ω200 Ω

resistor and a 25.0 mH inductor. At 8000 Hz, the phase angle is 45.0º45.0º. (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?

Referring to [link], find the average power at 10.0 kHz.

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+(XL−XC)2Z=R2+(XL−XC)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=12πLCf0=12πLC 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} {}