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An LCR Circuit at Its Natural Frequency

solved SeriesLcr

Q. A series LCR circuit with R = 20 ΩR=20Ω, L = 1.5 HL=1.5H and C = 35 μFC=35μF is connected to a variable-frequency 200 VV=200V ac supply. When the frequency of the supply equals the natural frequencyν=νr of the circuit, what is the average power transferred to the circuit in one complete cycle, in watts?

R = 20 ΩL = 1.5 HC = 3.5×10⁻⁵ FV = 200 Vν = ?I = ?Z = ?P = ?φ = ?
Inputs
R
20 Ω
L
1.5 H
C
35 μF
V
200 V
Outputs
powerP?

How it's solved

Solved with SeriesLcr

Given

  • R=20Ω
  • L=1.5H
  • C=35μF
  • V=200V
  • ν=νr from “equals the natural frequency”

Find P.

Working
  1. C=35μF=3.5×10−5F
  2. resonance of a series circuit(circuit)νr=12⋅π⋅L⋅C
  3. νr=12⋅3.14159265⋅1.5⋅3.5×10−5
  4. νr=21.965469Hz
  5. equals the natural frequencyν=νr
  6. ν=21.965469
  7. ν=21.965469Hz
  8. reactance of an inductor(circuit)XL=2⋅π⋅ν⋅L
  9. XL=2⋅3.14159265⋅21.965469⋅1.5
  10. XL=207.019668Ω
  11. reactance of a capacitor(circuit)XC=12⋅π⋅ν⋅C
  12. XC=12⋅3.14159265⋅21.965469⋅3.5×10−5
  13. XC=207.019668Ω
  14. impedance of a series circuit(circuit)Z2=R2+(XL-XC)2
  15. Z2=202+(207.019668-207.019668)2
  16. Z=20Ω
  17. current through the impedance(circuit)V=I⋅Z
  18. 200=I⋅20
  19. I=10A
  20. power spent in the resistance(circuit)P=I2⋅R
  21. P=102⋅20
  22. P=2000W
  23. P≈2.0×103Wto 2 significant figures
Conclusion

At its natural frequency the inductor and the capacitor cancel each other, so the resistance alone limits the current and takes all the power.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb
1require_relative "../physics"23question :an_lcr_circuit_at_its_natural_frequency do4  kind :numerical5  describes "A series LCR circuit with R = %{resistance}, L = %{inductance} and C = %{capacitance} is connected " \6            "to a variable-frequency %{supply} ac supply. When the frequency of the supply equals the natural " \7            "frequency of the circuit, %{power}, in watts?"8  given resistance: "20 Ω"9  given inductance: "1.5 H"10  given capacitance: "35 μF"11  given supply: "200 V"12  asks power: "what is the average power transferred to the circuit in one complete cycle"13end
solution.rb
1require_relative "../questions/an_lcr_circuit_at_its_natural_frequency"2require_relative "../science/alternating_current/series_lcr"34solve :an_lcr_circuit_at_its_natural_frequency, using: SeriesLcr do5  objects :circuit67  take resistance: circuit.resistance8  take inductance: circuit.inductance9  take capacitance: circuit.capacitance10  take supply: circuit.rms_voltage1112  deduce_from "equals the natural frequency", circuit.frequency => circuit.resonant_frequency1314  power = circuit.power_drawn15  answer power:, in: "W", figures: 21617  conclusion "At its natural frequency the inductor and the capacitor cancel each other, so the resistance " \18             "alone limits the current and takes all the power."19end
playground.rb
1require_relative "../solutions/an_lcr_circuit_at_its_natural_frequency"23play_with :an_lcr_circuit_at_its_natural_frequency do4  title "An LCR Circuit at Its Natural Frequency"5end
The laws it drew on
quantitywrittenunitswithinfrom
piπ—0 … ∞Reactance
rms voltageVV0 … ∞PeakAndRms
peak voltagevmV—PeakAndRms
rms currentIA0 … ∞PeakAndRms
peak currentimA—PeakAndRms
frequencyνHz0 … ∞Reactance
angular frequencyωrad/s—Reactance
inductanceLH—Reactance
capacitanceCF0 … ∞Reactance
inductive reactanceXLΩ0 … ∞Reactance
capacitive reactanceXCΩ0 … ∞Reactance
voltage across the inductorVLV—Reactance
voltage across the capacitorVCV—Reactance
resistanceRΩ0 … ∞SeriesLcr
impedanceZΩ0 … 1e+09SeriesLcr
power drawnPW—SeriesLcr
resonant frequencyνrHz—SeriesLcr
phase angleφ—-3.142 … 3.142SeriesLcr
resonant angular frequencyωrrad/s—SeriesLcr
peak voltage of groundvm,gV—
rms voltage of groundVgV0 … ∞
peak voltage of circuitvmV—
rms voltage of circuitVV0 … ∞
peak current of groundim,gA—
rms current of groundIgA0 … ∞
peak current of circuitimA—
rms current of circuitIA0 … ∞
angular frequency of groundωgrad/s—
frequency of groundνgHz0 … ∞
angular frequency of circuitωrad/s—
frequency of circuitνHz0 … ∞
inductive reactance of groundXL,gΩ0 … ∞
inductance of groundLgH—
inductive reactance of circuitXLΩ0 … ∞
inductance of circuitLH—
capacitive reactance of groundXC,gΩ0 … ∞
capacitance of groundCgF0 … ∞
capacitive reactance of circuitXCΩ0 … ∞
capacitance of circuitCF0 … ∞
voltage across the inductor of groundVL,gV—
voltage across the inductor of circuitVLV—
voltage across the capacitor of groundVC,gV—
voltage across the capacitor of circuitVCV—
impedance of groundZgΩ0 … 1e+09
resistance of groundRgΩ0 … ∞
impedance of circuitZΩ0 … 1e+09
resistance of circuitRΩ0 … ∞
power drawn of groundPgW—
power drawn of circuitPW—
phase angle of groundφg—-3.142 … 3.142
phase angle of circuitφ—-3.142 … 3.142
resonant frequency of groundνr,gHz—
resonant frequency of circuitνrHz—
resonant angular frequency of groundωr,grad/s—
resonant angular frequency of circuitωrrad/s—
vm,c=2⋅Vc
peak is root two times rms
im,c=2⋅Ic
peak current is root two times rms
ωc=2⋅π⋅νc
angular frequency and frequency
XL,c=2⋅π⋅νc⋅Lc
reactance of an inductor
XC,c=12⋅π⋅νc⋅Cc
reactance of a capacitor
VL,c=Ic⋅XL,c
current through the inductor
VC,c=Ic⋅XC,c
current through the capacitor
Zc2=Rc2+(XL,c-XC,c)2
impedance of a series circuit
Vc=Ic⋅Zc
current through the impedance
Pc=Ic2⋅Rc
power spent in the resistance
φc=atan⁡(XC,c-XL,cRc)
phase of the current
νr,c=12⋅π⋅Lc⋅Cc
resonance of a series circuit
ωr,c=2⋅π⋅νr,c
resonance in radians a second
vm,g=2⋅Vg
peak is root two times rms(ground)
vm=2⋅V
peak is root two times rms(circuit)
im,g=2⋅Ig
peak current is root two times rms(ground)
im=2⋅I
peak current is root two times rms(circuit)
ωg=2⋅π⋅νg
angular frequency and frequency(ground)
ω=2⋅π⋅ν
angular frequency and frequency(circuit)
XL,g=2⋅π⋅νg⋅Lg
reactance of an inductor(ground)
XL=2⋅π⋅ν⋅L
reactance of an inductor(circuit)
XC,g=12⋅π⋅νg⋅Cg
reactance of a capacitor(ground)
XC=12⋅π⋅ν⋅C
reactance of a capacitor(circuit)
VL,g=Ig⋅XL,g
current through the inductor(ground)
VL=I⋅XL
current through the inductor(circuit)
VC,g=Ig⋅XC,g
current through the capacitor(ground)
VC=I⋅XC
current through the capacitor(circuit)
Zg2=Rg2+(XL,g-XC,g)2
impedance of a series circuit(ground)
Z2=R2+(XL-XC)2
impedance of a series circuit(circuit)
Vg=Ig⋅Zg
current through the impedance(ground)
V=I⋅Z
current through the impedance(circuit)
Pg=Ig2⋅Rg
power spent in the resistance(ground)
P=I2⋅R
power spent in the resistance(circuit)
φg=atan⁡(XC,g-XL,gRg)
phase of the current(ground)
φ=atan⁡(XC-XLR)
phase of the current(circuit)
νr,g=12⋅π⋅Lg⋅Cg
resonance of a series circuit(ground)
νr=12⋅π⋅L⋅C
resonance of a series circuit(circuit)
ωr,g=2⋅π⋅νr,g
resonance in radians a second(ground)
ωr=2⋅π⋅νr
resonance in radians a second(circuit)
ν=νr
equals the natural frequency