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.

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
+
(
I0XL−I0XC)2=I0R2+(XL−XC)2.I0Z=
I
0
2
R2
+
(
I0XL−I0XC)2=I0R2+(XL−XC)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+(XL−XC)2,Z=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} } } } {}

(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} {}.

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+(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 (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+(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

Z
=
R2+(XL−XC)2
=
(40.0Ω)2+(1.13Ω−531Ω)2
=
531Ω at 60.0 Hz.Z
=
R2+(XL−XC)2
=
(40.0Ω)2+(1.13Ω−531Ω)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+(188Ω−3.18Ω)2
=
190Ω at 10.0 kHz.
Z
=
(40.0Ω)2+(188Ω−3.18Ω)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.