The Inductor's Drop at Resonance

solved SeriesLcr

Q. A series LCR circuit is connected to a variable frequency 230 V $V = 230$ V source, with L = 5.0 H $L = 5.0$ H, C = 80 μF $C = 80$ μF and R = 40 Ω $R = 40$ Ω. The source is tuned to the frequency which drives the circuit in resonance $ν = ν_{r}$ . What is the rms potential drop across the inductor, in volts?

Inputs

supply

V

230 V

inductance

L

5 H

capacitance

C

80 μF

resistance

R

40 Ω

Outputs

drop

V_{L}

How it's solved

Solved with SeriesLcr

Given

$V = 230$ V
$L = 5.0$ H
$C = 80$ μF
$R = 40$ Ω
$ν = ν_{r}$ from “drives the circuit in resonance”

Find $V_{L}$ .

Working

$C = 80$ μF $= 8 \times 10^{−5}$ F
resonance of a series circuit(circuit) $ν_{r} = \frac{1}{2 \cdot π \cdot \sqrt{L \cdot C}}$
$ν_{r} = \frac{1}{2 \cdot 3.14159265 \cdot \sqrt{5 \cdot 8 \times 10^{−5}}}$
$ν_{r} = 7.95774715$ Hz
drives the circuit in resonance $ν = ν_{r}$
$ν = 7.95774715$
$ν = 7.95774715$ Hz
reactance of an inductor(circuit) $X_{L} = 2 \cdot π \cdot ν \cdot L$
$X_{L} = 2 \cdot 3.14159265 \cdot 7.95774715 \cdot 5$
$X_{L} = 250$ Ω
reactance of a capacitor(circuit) $X_{C} = \frac{1}{2 \cdot π \cdot ν \cdot C}$
$X_{C} = \frac{1}{2 \cdot 3.14159265 \cdot 7.95774715 \cdot 8 \times 10^{−5}}$
$X_{C} = 250$ Ω
impedance of a series circuit(circuit) $Z^{2} = R^{2} + {(X_{L} - X_{C})}^{2}$
$Z^{2} = 40^{2} + {(250 - 250)}^{2}$
$Z = 40$ Ω
current through the impedance(circuit) $V = I \cdot Z$
$230 = I \cdot 40$
$I = 5.75$ A
current through the inductor(circuit) $V_{L} = I \cdot X_{L}$
$V_{L} = 5.75 \cdot 250$
$V_{L} = 1438$ V
$V_{L} \approx 1437.5$ Vto 5 significant figures

Conclusion

At resonance the inductor and the capacitor each drop far more than the source supplies, but half a cycle apart, so across the two together the drop is nothing.

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 :the_inductor_s_drop_at_resonance do4  kind :numerical5  describes "A series LCR circuit is connected to a variable frequency %{supply} source, with L = %{inductance}, " \6            "C = %{capacitance} and R = %{resistance}. The source is tuned to the frequency which drives the " \7            "circuit in resonance. %{drop}, in volts?"8  given supply: "230 V"9  given inductance: "5.0 H"10  given capacitance: "80 μF"11  given resistance: "40 Ω"12  asks drop: "What is the rms potential drop across the inductor"13end

solution.rb

1require_relative "../questions/the_inductor_s_drop_at_resonance"2require_relative "../science/alternating_current/series_lcr"34solve :the_inductor_s_drop_at_resonance, using: SeriesLcr do5  objects :circuit67  take supply: circuit.rms_voltage8  take inductance: circuit.inductance9  take capacitance: circuit.capacitance10  take resistance: circuit.resistance1112  deduce_from "drives the circuit in resonance", circuit.frequency => circuit.resonant_frequency1314  drop = circuit.voltage_across_the_inductor15  answer drop:, in: "V", figures: 51617  conclusion "At resonance the inductor and the capacitor each drop far more than the source supplies, but " \18             "half a cycle apart, so across the two together the drop is nothing."19end

playground.rb

1require_relative "../solutions/the_inductor_s_drop_at_resonance"23play_with :the_inductor_s_drop_at_resonance do4  title "The Inductor's Drop at Resonance"5end

The laws it drew on

quantity	written	units	within	from
pi	$π$	—	0 … ∞	Reactance
rms voltage	$V$	V	0 … ∞	PeakAndRms
peak voltage	$v_{m}$	V	—	PeakAndRms
rms current	$I$	A	0 … ∞	PeakAndRms
peak current	$i_{m}$	A	—	PeakAndRms
frequency	$ν$	Hz	0 … ∞	Reactance
angular frequency	$ω$	rad/s	—	Reactance
inductance	$L$	H	—	Reactance
capacitance	$C$	F	0 … ∞	Reactance
inductive reactance	$X_{L}$	Ω	0 … ∞	Reactance
capacitive reactance	$X_{C}$	Ω	0 … ∞	Reactance
voltage across the inductor	$V_{L}$	V	—	Reactance
voltage across the capacitor	$V_{C}$	V	—	Reactance
resistance	$R$	Ω	0 … ∞	SeriesLcr
impedance	$Z$	Ω	0 … 1e+09	SeriesLcr
power drawn	$P$	W	—	SeriesLcr
resonant frequency	$ν_{r}$	Hz	—	SeriesLcr
phase angle	$φ$	—	-3.142 … 3.142	SeriesLcr
resonant angular frequency	$ω_{r}$	rad/s	—	SeriesLcr
peak voltage of ground	$v_{m,g}$	V	—
rms voltage of ground	$V_{g}$	V	0 … ∞
peak voltage of circuit	$v_{m}$	V	—
rms voltage of circuit	$V$	V	0 … ∞
peak current of ground	$i_{m,g}$	A	—
rms current of ground	$I_{g}$	A	0 … ∞
peak current of circuit	$i_{m}$	A	—
rms current of circuit	$I$	A	0 … ∞
angular frequency of ground	$ω_{g}$	rad/s	—
frequency of ground	$ν_{g}$	Hz	0 … ∞
angular frequency of circuit	$ω$	rad/s	—
frequency of circuit	$ν$	Hz	0 … ∞
inductive reactance of ground	$X_{L,g}$	Ω	0 … ∞
inductance of ground	$L_{g}$	H	—
inductive reactance of circuit	$X_{L}$	Ω	0 … ∞
inductance of circuit	$L$	H	—
capacitive reactance of ground	$X_{C,g}$	Ω	0 … ∞
capacitance of ground	$C_{g}$	F	0 … ∞
capacitive reactance of circuit	$X_{C}$	Ω	0 … ∞
capacitance of circuit	$C$	F	0 … ∞
voltage across the inductor of ground	$V_{L,g}$	V	—
voltage across the inductor of circuit	$V_{L}$	V	—
voltage across the capacitor of ground	$V_{C,g}$	V	—
voltage across the capacitor of circuit	$V_{C}$	V	—
impedance of ground	$Z_{g}$	Ω	0 … 1e+09
resistance of ground	$R_{g}$	Ω	0 … ∞
impedance of circuit	$Z$	Ω	0 … 1e+09
resistance of circuit	$R$	Ω	0 … ∞
power drawn of ground	$P_{g}$	W	—
power drawn of circuit	$P$	W	—
phase angle of ground	$φ_{g}$	—	-3.142 … 3.142
phase angle of circuit	$φ$	—	-3.142 … 3.142
resonant frequency of ground	$ν_{r,g}$	Hz	—
resonant frequency of circuit	$ν_{r}$	Hz	—
resonant angular frequency of ground	$ω_{r,g}$	rad/s	—
resonant angular frequency of circuit	$ω_{r}$	rad/s	—

v_{m,c} = \sqrt{2} \cdot V_{c}

peak is root two times rms

i_{m,c} = \sqrt{2} \cdot I_{c}

peak current is root two times rms

ω_{c} = 2 \cdot π \cdot ν_{c}

angular frequency and frequency

X_{L,c} = 2 \cdot π \cdot ν_{c} \cdot L_{c}

reactance of an inductor

X_{C,c} = \frac{1}{2 \cdot π \cdot ν_{c} \cdot C_{c}}

reactance of a capacitor

V_{L,c} = I_{c} \cdot X_{L,c}

current through the inductor

V_{C,c} = I_{c} \cdot X_{C,c}

current through the capacitor

{Z_{c}}^{2} = {R_{c}}^{2} + {(X_{L,c} - X_{C,c})}^{2}

impedance of a series circuit

V_{c} = I_{c} \cdot Z_{c}

current through the impedance

P_{c} = {I_{c}}^{2} \cdot R_{c}

power spent in the resistance

φ_{c} = atan (\frac{X_{C,c} - X_{L,c}}{R_{c}})

phase of the current

ν_{r,c} = \frac{1}{2 \cdot π \cdot \sqrt{L_{c} \cdot C_{c}}}

resonance of a series circuit

ω_{r,c} = 2 \cdot π \cdot ν_{r,c}

resonance in radians a second

v_{m,g} = \sqrt{2} \cdot V_{g}

peak is root two times rms(ground)

v_{m} = \sqrt{2} \cdot V

peak is root two times rms(circuit)

i_{m,g} = \sqrt{2} \cdot I_{g}

peak current is root two times rms(ground)

i_{m} = \sqrt{2} \cdot I

peak current is root two times rms(circuit)

ω_{g} = 2 \cdot π \cdot ν_{g}

angular frequency and frequency(ground)

ω = 2 \cdot π \cdot ν

angular frequency and frequency(circuit)

X_{L,g} = 2 \cdot π \cdot ν_{g} \cdot L_{g}

reactance of an inductor(ground)

X_{L} = 2 \cdot π \cdot ν \cdot L

reactance of an inductor(circuit)

X_{C,g} = \frac{1}{2 \cdot π \cdot ν_{g} \cdot C_{g}}

reactance of a capacitor(ground)

X_{C} = \frac{1}{2 \cdot π \cdot ν \cdot C}

reactance of a capacitor(circuit)

V_{L,g} = I_{g} \cdot X_{L,g}

current through the inductor(ground)

V_{L} = I \cdot X_{L}

current through the inductor(circuit)

V_{C,g} = I_{g} \cdot X_{C,g}

current through the capacitor(ground)

V_{C} = I \cdot X_{C}

current through the capacitor(circuit)

{Z_{g}}^{2} = {R_{g}}^{2} + {(X_{L,g} - X_{C,g})}^{2}

impedance of a series circuit(ground)

Z^{2} = R^{2} + {(X_{L} - X_{C})}^{2}

impedance of a series circuit(circuit)

V_{g} = I_{g} \cdot Z_{g}

current through the impedance(ground)

V = I \cdot Z

current through the impedance(circuit)

P_{g} = {I_{g}}^{2} \cdot R_{g}

power spent in the resistance(ground)

P = I^{2} \cdot R

power spent in the resistance(circuit)

φ_{g} = atan (\frac{X_{C,g} - X_{L,g}}{R_{g}})

phase of the current(ground)

φ = atan (\frac{X_{C} - X_{L}}{R})

phase of the current(circuit)

ν_{r,g} = \frac{1}{2 \cdot π \cdot \sqrt{L_{g} \cdot C_{g}}}

resonance of a series circuit(ground)

ν_{r} = \frac{1}{2 \cdot π \cdot \sqrt{L \cdot C}}

resonance of a series circuit(circuit)

ω_{r,g} = 2 \cdot π \cdot ν_{r,g}

resonance in radians a second(ground)

ω_{r} = 2 \cdot π \cdot ν_{r}

resonance in radians a second(circuit)

ν = ν_{r}

drives the circuit in resonance