play phryzby with physics

A Capacitor on an AC Supply

solved Reactance

Q. A 60 μFC=60μF capacitor is connected to a 110 VVC=110V, 60 Hzν=60Hz ac supply. What is the rms value of the current in the circuit, in amperes?

C = 6×10⁻⁵ Fν = 60 HzI = ?
Inputs
C
60 μF
VC
110 V
ν
60 Hz
Outputs
currentI?

How it's solved

Solved with Reactance

Given

  • C=60μF
  • VC=110V
  • ν=60Hz

Find I.

Working
  1. C=60μF=6×10−5F
  2. reactance of a capacitor(circuit)XC=12⋅π⋅ν⋅C
  3. XC=12⋅3.14159265⋅60⋅6×10−5
  4. XC=44.2097064Ω
  5. current through the capacitor(circuit)VC=I⋅XC
  6. 110=I⋅44.2097064
  7. I=2.488A
  8. I≈2.49Ato 3 significant figures
Conclusion

The capacitor alone opposes the current, by a reactance that shrinks as the frequency and the capacitance grow.

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_capacitor_on_an_ac_supply do4  kind :numerical5  describes "A %{capacitance} capacitor is connected to a %{supply}, %{frequency} ac supply. %{current}, in " \6            "amperes?"7  given capacitance: "60 μF"8  given supply: "110 V"9  given frequency: "60 Hz"10  asks current: "What is the rms value of the current in the circuit"11end
solution.rb
1require_relative "../questions/a_capacitor_on_an_ac_supply"2require_relative "../science/alternating_current/reactance"34solve :a_capacitor_on_an_ac_supply, using: Reactance do5  objects :circuit67  take capacitance: circuit.capacitance8  take supply: circuit.voltage_across_the_capacitor9  take frequency: circuit.frequency1011  current = circuit.rms_current12  answer current:, in: "A", figures: 31314  conclusion "The capacitor alone opposes the current, by a reactance that shrinks as the frequency and the " \15             "capacitance grow."16end
playground.rb
1require_relative "../solutions/a_capacitor_on_an_ac_supply"23play_with :a_capacitor_on_an_ac_supply do4  title "A Capacitor on an AC Supply"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
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—
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
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)