The Conduction Current into a Capacitor

solved Reactance

Q. A parallel plate capacitor made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF $C = 100$ pF. The capacitor is connected to a 230 V $V_{C} = 230$ V ac supply with a (angular) frequency of 300 rad/s $ω = 300$ rad/s. What is the rms value of the conduction current, in μA?

Inputs

capacitance

C

100 pF

supply

V_{C}

230 V

angular

ω

300 rad/s

Outputs

current

I

How it's solved

Solved with Reactance

Given

$C = 100$ pF
$V_{C} = 230$ V
$ω = 300$ rad/s

Find $I$ .

Working

$C = 100$ pF $= 1 \times 10^{−10}$ F
angular frequency and frequency(capacitor) $ω = 2 \cdot π \cdot ν$
$300 = 2 \cdot 3.14159265 \cdot ν$
$ν = 47.7464829$ Hz
reactance of a capacitor(capacitor) $X_{C} = \frac{1}{2 \cdot π \cdot ν \cdot C}$
$X_{C} = \frac{1}{2 \cdot 3.14159265 \cdot 47.7464829 \cdot 1 \times 10^{−10}}$
$X_{C} = 3.33333333 \times 10^{7}$ Ω
current through the capacitor(capacitor) $V_{C} = I \cdot X_{C}$
$230 = I \cdot 3.33333333 \times 10^{7}$
$I = 6.9$ μA
$I = 6.9 \times 10^{−6}$ A $= 6.9$ μA
$I \approx 6.9$ μAto 2 significant figures

Conclusion

A small capacitor on a slowly alternating supply has a huge reactance, so only a tiny current flows in and out of its plates.

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_conduction_current_into_a_capacitor do4  kind :numerical5  describes "A parallel plate capacitor made of circular plates each of radius R = 6.0 cm has a capacitance C = " \6            "%{capacitance}. The capacitor is connected to a %{supply} ac supply with a (angular) frequency of " \7            "%{angular}. %{current}, in μA?"8  given capacitance: "100 pF"9  given supply: "230 V"10  given angular: "300 rad/s"11  asks current: "What is the rms value of the conduction current"12end

solution.rb

1require_relative "../questions/the_conduction_current_into_a_capacitor"2require_relative "../science/alternating_current/reactance"34solve :the_conduction_current_into_a_capacitor, using: Reactance do5  objects :capacitor67  take capacitance: capacitor.capacitance8  take supply: capacitor.voltage_across_the_capacitor9  take angular: capacitor.angular_frequency1011  current = capacitor.rms_current12  answer current:, in: "μA", figures: 21314  conclusion "A small capacitor on a slowly alternating supply has a huge reactance, so only a tiny current " \15             "flows in and out of its plates."16end

playground.rb

1require_relative "../solutions/the_conduction_current_into_a_capacitor"23play_with :the_conduction_current_into_a_capacitor do4  title "The Conduction Current into a Capacitor"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
peak voltage of ground	$v_{m,g}$	V	—
rms voltage of ground	$V_{g}$	V	0 … ∞
peak voltage of capacitor	$v_{m}$	V	—
rms voltage of capacitor	$V$	V	0 … ∞
peak current of ground	$i_{m,g}$	A	—
rms current of ground	$I_{g}$	A	0 … ∞
peak current of capacitor	$i_{m}$	A	—
rms current of capacitor	$I$	A	0 … ∞
angular frequency of ground	$ω_{g}$	rad/s	—
frequency of ground	$ν_{g}$	Hz	0 … ∞
angular frequency of capacitor	$ω$	rad/s	—
frequency of capacitor	$ν$	Hz	0 … ∞
inductive reactance of ground	$X_{L,g}$	Ω	0 … ∞
inductance of ground	$L_{g}$	H	—
inductive reactance of capacitor	$X_{L}$	Ω	0 … ∞
inductance of capacitor	$L$	H	—
capacitive reactance of ground	$X_{C,g}$	Ω	0 … ∞
capacitance of ground	$C_{g}$	F	0 … ∞
capacitive reactance of capacitor	$X_{C}$	Ω	0 … ∞
capacitance of capacitor	$C$	F	0 … ∞
voltage across the inductor of ground	$V_{L,g}$	V	—
voltage across the inductor of capacitor	$V_{L}$	V	—
voltage across the capacitor of ground	$V_{C,g}$	V	—
voltage across the capacitor of capacitor	$V_{C}$	V	—

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

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(capacitor)

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(capacitor)

ω_{g} = 2 \cdot π \cdot ν_{g}

angular frequency and frequency(ground)

ω = 2 \cdot π \cdot ν

angular frequency and frequency(capacitor)

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(capacitor)

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(capacitor)

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

current through the inductor(ground)

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

current through the inductor(capacitor)

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

current through the capacitor(ground)

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

current through the capacitor(capacitor)