Three Capacitors in Series on 120 Volts

solved Capacitance

Q. Three capacitors each of capacitance 9 pF $C_{o} = 9$ pF are connected in series $Q_{o} = Q_{r}$ $C_{r} = \frac{1}{\frac{3}{C_{o}}}$ . What is the potential difference across each capacitor if the combination is connected to a 120 V $V_{r} = 120$ V supply, in volts?

Inputs

each

C_{o}

9 pF

supply

V_{r}

120 V

Outputs

voltage

V_{o}

How it's solved

Solved with Capacitance

Given

$C_{o} = 9$ pF
$V_{r} = 120$ V
$C_{r} = \frac{1}{\frac{3}{C_{o}}}$ from “connected in series”
$Q_{o} = Q_{r}$ from “in series”

Find $V_{o}$ .

Working

$C_{o} = 9$ pF $= 9 \times 10^{−12}$ F
connected in series $C_{r} = \frac{1}{\frac{3}{C_{o}}}$
$C_{r} = \frac{1}{\frac{3}{9 \times 10^{−12}}}$
$C_{r} = 3 \times 10^{−12}$ F
charge on a capacitor(row) $Q_{r} = C_{r} \cdot V_{r}$
$Q_{r} = 3 \times 10^{−12} \cdot 120$
$Q_{r} = 3.6 \times 10^{−10}$ C
in series $Q_{o} = Q_{r}$
$Q_{o} = 3.6 \times 10^{−10}$
$Q_{o} = 3.6 \times 10^{−10}$ C
charge on a capacitor(one) $Q_{o} = C_{o} \cdot V_{o}$
$3.6 \times 10^{−10} = 9 \times 10^{−12} \cdot V_{o}$
$V_{o} = 40$ V
$V_{o} \approx 40$ Vto 2 significant figures

Conclusion

Capacitors in series all carry the same charge, so equal ones share the supply's voltage equally.

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 :three_capacitors_in_series_on_120_volts do4  kind :numerical5  describes "Three capacitors each of capacitance %{each} are connected in series. %{voltage} if the combination " \6            "is connected to a %{supply} supply, in volts?"7  given each: "9 pF"8  given supply: "120 V"9  asks voltage: "What is the potential difference across each capacitor"10end

solution.rb

1require_relative "../questions/three_capacitors_in_series_on_120_volts"2require_relative "../science/electrostatic_potential_and_capacitance/capacitance"34solve :three_capacitors_in_series_on_120_volts, using: Capacitance do5  objects :one, :row67  take each: one.capacitance8  take supply: row.voltage910  deduce_from "connected in series", row.capacitance => 1 / (3 / one.capacitance)11  deduce_from "in series", one.charge => row.charge1213  voltage = one.voltage14  answer voltage:, in: "V", figures: 21516  conclusion "Capacitors in series all carry the same charge, so equal ones share the supply's voltage equally."17end

playground.rb

1require_relative "../solutions/three_capacitors_in_series_on_120_volts"23play_with :three_capacitors_in_series_on_120_volts do4  title "Three Capacitors in Series on 120 Volts"5end

The laws it drew on

quantity	written	units	within	from
permittivity of free space	$ε_{0}$	F/m	—	Capacitance
capacitance	$C$	F	0 … ∞	Capacitance
charge	$Q$	C	—	Capacitance
voltage	$V$	V	—	Capacitance
energy stored	$U$	J	—	Capacitance
area	$A$	m²	—	Capacitance
plate separation	$d$	m	—	Capacitance
dielectric constant	$K$	—	—	Capacitance
charge of ground	$Q_{g}$	C	—
capacitance of ground	$C_{g}$	F	0 … ∞
voltage of ground	$V_{g}$	V	—
charge of one	$Q_{o}$	C	—
capacitance of one	$C_{o}$	F	0 … ∞
voltage of one	$V_{o}$	V	—
charge of row	$Q_{r}$	C	—
capacitance of row	$C_{r}$	F	0 … ∞
voltage of row	$V_{r}$	V	—
dielectric constant of ground	$K_{g}$	—	—
area of ground	$A_{g}$	m²	—
plate separation of ground	$d_{g}$	m	—
dielectric constant of one	$K_{o}$	—	—
area of one	$A_{o}$	m²	—
plate separation of one	$d_{o}$	m	—
dielectric constant of row	$K_{r}$	—	—
area of row	$A_{r}$	m²	—
plate separation of row	$d_{r}$	m	—
energy stored of ground	$U_{g}$	J	—
energy stored of one	$U_{o}$	J	—
energy stored of row	$U_{r}$	J	—

Q_{c} = C_{c} \cdot V_{c}

charge on a capacitor

C_{c} = \frac{K_{c} \cdot ε_{0} \cdot A_{c}}{d_{c}}

a parallel plate capacitor

U_{c} = \frac{C_{c} \cdot {V_{c}}^{2}}{2}

energy in a capacitor

Q_{g} = C_{g} \cdot V_{g}

charge on a capacitor(ground)

Q_{o} = C_{o} \cdot V_{o}

charge on a capacitor(one)

Q_{r} = C_{r} \cdot V_{r}

charge on a capacitor(row)

C_{g} = \frac{K_{g} \cdot ε_{0} \cdot A_{g}}{d_{g}}

a parallel plate capacitor(ground)

C_{o} = \frac{K_{o} \cdot ε_{0} \cdot A_{o}}{d_{o}}

a parallel plate capacitor(one)

C_{r} = \frac{K_{r} \cdot ε_{0} \cdot A_{r}}{d_{r}}

a parallel plate capacitor(row)

U_{g} = \frac{C_{g} \cdot {V_{g}}^{2}}{2}

energy in a capacitor(ground)

U_{o} = \frac{C_{o} \cdot {V_{o}}^{2}}{2}

energy in a capacitor(one)

U_{r} = \frac{C_{r} \cdot {V_{r}}^{2}}{2}

energy in a capacitor(row)

C_{r} = \frac{1}{\frac{3}{C_{o}}}

connected in series

Q_{o} = Q_{r}

in series