Three Capacitors in Series on 120 Volts

Q. Three capacitors each of capacitance 9 pF are connected in series. What is the potential difference across each capacitor if the combination is connected to a 120 V supply, in volts? expects 40 V

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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?

The playground

Inputs

each

C_{o}

9 pF

supply

V_{r}

120 V

Outputs

voltage

V_{o}

The solution

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 Ruby

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

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

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

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