An LCR Circuit's Phase

solved SeriesLcr

Q. A sinusoidal voltage of peak value 283 V $v_{m} = 283$ V and frequency 50 Hz $ν = 50$ Hz is applied to a series LCR circuit in which R = 3 Ω $R = 3$ Ω, L = 25.48 mH $L = 25.48$ mH and C = 796 μF $C = 796$ μF. What is the phase difference between the voltage across the source and the current, in degrees?

Inputs

peak

v_{m}

283 V

frequency

ν

50 Hz

resistance

R

3 Ω

inductance

L

25.48 mH

capacitance

C

796 μF

Outputs

phase

φ

How it's solved

Solved with SeriesLcr

Given

$v_{m} = 283$ V
$ν = 50$ Hz
$R = 3$ Ω
$L = 25.48$ mH
$C = 796$ μF

Find $φ$ .

Working

$L = 25.48$ mH $= 0.02548$ H
$C = 796$ μF $= 7.96 \times 10^{−4}$ F
reactance of a capacitor(circuit) $X_{C} = \frac{1}{2 \cdot π \cdot ν \cdot C}$
$X_{C} = \frac{1}{2 \cdot 3.14159265 \cdot 50 \cdot 7.96 \times 10^{−4}}$
$X_{C} = 3.99886792$ Ω
reactance of an inductor(circuit) $X_{L} = 2 \cdot π \cdot ν \cdot L$
$X_{L} = 2 \cdot 3.14159265 \cdot 50 \cdot 0.02548$
$X_{L} = 8.00477808$ Ω
phase of the current(circuit) $φ = atan (\frac{X_{C} - X_{L}}{R})$
$φ = atan (\frac{3.99886792 - 8.00477808}{3})$
$φ = -53.17°$
$φ \approx −53.2$ °to 3 significant figures

Conclusion

The inductor's reactance outweighs the capacitor's, so the current lags the voltage, by the angle whose tangent is the net reactance over the resistance.

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_s_phase do4  kind :numerical5  describes "A sinusoidal voltage of peak value %{peak} and frequency %{frequency} is applied to a series LCR " \6            "circuit in which R = %{resistance}, L = %{inductance} and C = %{capacitance}. %{phase}, in degrees?"7  given peak: "283 V"8  given frequency: "50 Hz"9  given resistance: "3 Ω"10  given inductance: "25.48 mH"11  given capacitance: "796 μF"12  asks phase: "What is the phase difference between the voltage across the source and the current"13end

solution.rb

1require_relative "../questions/an_lcr_circuit_s_phase"2require_relative "../science/alternating_current/series_lcr"34solve :an_lcr_circuit_s_phase, using: SeriesLcr do5  objects :circuit67  take peak: circuit.peak_voltage8  take frequency: circuit.frequency9  take resistance: circuit.resistance10  take inductance: circuit.inductance11  take capacitance: circuit.capacitance1213  phase = circuit.phase_angle14  answer phase:, in: "deg", figures: 31516  conclusion "The inductor's reactance outweighs the capacitor's, so the current lags the voltage, by the angle whose tangent is the net reactance over the resistance."17end

playground.rb

1require_relative "../solutions/an_lcr_circuit_s_phase"23play_with :an_lcr_circuit_s_phase do4  title "An LCR Circuit's Phase"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)