A Resistor and Capacitor in Series

Q. A resistor of 200 Ω and a capacitor of 15.0 μF are connected in series to a 220 V, 50 Hz ac source. What is the current in the circuit, in amperes? expects 0.754 A

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The equations it used

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Q. A resistor of $X_{L} = 0$ Ω 200 Ω $R = 200$ Ω and a capacitor of 15.0 μF $C = 15.0$ μF are connected in series to a 220 V $V = 220$ V, 50 Hz $ν = 50$ Hz ac source. What is the current in the circuit, in amperes?

The playground

Inputs

resistance

R

200 Ω

capacitance

C

15 μF

supply

V

220 V

frequency

ν

50 Hz

Outputs

current

I

The solution

Solved with SeriesLcr

Given

$R = 200$ Ω
$C = 15.0$ μF
$V = 220$ V
$ν = 50$ Hz
$X_{L} = 0$ Ω from “A resistor of”

Find $I$ .

Working

$C = 15.0$ μF $= 1.5 \times 10^{−5}$ F
reactance of a capacitor(circuit) $X_{C} = \frac{1}{2 \cdot π \cdot ν \cdot C}$
$X_{C} = \frac{1}{2 \cdot 3.14159265 \cdot 50 \cdot 1.5 \times 10^{−5}}$
$X_{C} = 212.206591$ Ω
impedance of a series circuit(circuit) $Z^{2} = R^{2} + {(X_{L} - X_{C})}^{2}$
$Z^{2} = 200^{2} + {(0 - 212.206591)}^{2}$
$Z = 291.601847$ Ω
current through the impedance(circuit) $V = I \cdot Z$
$220 = I \cdot 291.601847$
$I = 0.7545$ A
$I \approx 0.754$ Ato 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 Ruby

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

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

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

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)