Three Capacitors in Series

solved Capacitance

Q. Three capacitors each of capacitance 9 pF $C_{o} = 9$ pF are connected in series $C_{r} = \frac{1}{\frac{3}{C_{o}}}$ . What is the total capacitance of the combination, in pF?

Inputs

each

C_{o}

9 pF

Outputs

total

C_{r}

How it's solved

Solved with Capacitance

Given

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

Find $C_{r}$ .

Working

$C_{o} = 9$ pF $= 9 \times 10^{−12}$ F
connected in series $C_{r} = \frac{1}{\frac{3}{C_{o}}}$
$C_{r} = \frac{1}{\frac{3}{9 \times 10^{−12}}}$
$C_{r} = 3$ pF
$C_{r} = 3 \times 10^{−12}$ F $= 3$ pF
$C_{r} \approx 3$ pFto 1 significant figure

Conclusion

In series the reciprocals of the capacitances add, so three equal capacitors hold a third as much as one.

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 do4  kind :numerical5  describes "Three capacitors each of capacitance %{each} are connected in series. %{total}, in pF?"6  given each: "9 pF"7  asks total: "What is the total capacitance of the combination"8end

solution.rb

1require_relative "../questions/three_capacitors_in_series"2require_relative "../science/electrostatic_potential_and_capacitance/capacitance"34solve :three_capacitors_in_series, using: Capacitance do5  objects :one, :row67  take each: one.capacitance89  deduce_from "connected in series", row.capacitance => 1 / (3 / one.capacitance)1011  total = row.capacitance12  answer total:, in: "pF", figures: 11314  conclusion "In series the reciprocals of the capacitances add, so three equal capacitors hold a third as much " \15             "as one."16end

playground.rb

1require_relative "../solutions/three_capacitors_in_series"23play_with :three_capacitors_in_series do4  title "Three Capacitors in Series"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