The Conduction Current into a Capacitor

Q. A parallel plate capacitor made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad/s. What is the rms value of the conduction current, in μA? expects 6.9 μA

What the draft establishes

Fills in once a draft is checked.

The web of concepts

given read from the words equation asked

The equations it used

None yet.

Check an answer

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?

The playground

Inputs

capacitance

C

100 pF

supply

V_{C}

230 V

angular

ω

300 rad/s

Outputs

current

I

The solution

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 Ruby

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

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

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

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)