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A Resistor and Capacitor in Series

solved SeriesLcr

Q. A resistor ofXL=0Ω 200 ΩR=200Ω and a capacitor of 15.0 μFC=15.0μF are connected in series to a 220 VV=220V, 50 Hzν=50Hz ac source. What is the current in the circuit, in amperes?

R = 200 ΩC = 1.5×10⁻⁵ FV = 220 Vν = 50 HzI = ?Z = ?P = ?φ = ?
Inputs
R
200 Ω
C
15 μF
V
220 V
ν
50 Hz
Outputs
currentI?

How it's solved

Solved with SeriesLcr

Given

  • R=200Ω
  • C=15.0μF
  • V=220V
  • ν=50Hz
  • XL=0Ω from “A resistor of”

Find I.

Working
  1. C=15.0μF=1.5×10−5F
  2. reactance of a capacitor(circuit)XC=12⋅π⋅ν⋅C
  3. XC=12⋅3.14159265⋅50⋅1.5×10−5
  4. XC=212.206591Ω
  5. impedance of a series circuit(circuit)Z2=R2+(XL-XC)2
  6. Z2=2002+(0-212.206591)2
  7. Z=291.601847Ω
  8. current through the impedance(circuit)V=I⋅Z
  9. 220=I⋅291.601847
  10. I=0.7545A
  11. I≈0.754Ato 3 significant figures
Conclusion

The resistor's and the capacitor's oppositions are a quarter cycle apart, so they add like the sides of a right triangle into the impedance.

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 :a_resistor_and_capacitor_in_series do4  kind :numerical5  describes "A resistor of %{resistance} and a capacitor of %{capacitance} are connected in series to a %{supply}, " \6            "%{frequency} ac source. %{current}, in amperes?"7  given resistance: "200 Ω"8  given capacitance: "15.0 μF"9  given supply: "220 V"10  given frequency: "50 Hz"11  asks current: "What is the current in the circuit"12end
solution.rb
1require_relative "../questions/a_resistor_and_capacitor_in_series"2require_relative "../science/alternating_current/series_lcr"34solve :a_resistor_and_capacitor_in_series, using: SeriesLcr do5  objects :circuit67  take resistance: circuit.resistance8  take capacitance: circuit.capacitance9  take supply: circuit.rms_voltage10  take frequency: circuit.frequency1112  deduce_from "A resistor of", circuit.inductive_reactance => "0 Ω"1314  current = circuit.rms_current15  answer current:, in: "A", figures: 31617  conclusion "The resistor's and the capacitor's oppositions are a quarter cycle apart, so they add like the " \18             "sides of a right triangle into the impedance."19end
playground.rb
1require_relative "../solutions/a_resistor_and_capacitor_in_series"23play_with :a_resistor_and_capacitor_in_series do4  title "A Resistor and Capacitor in Series"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)