Four Capacitors on 500 Volts

Q. A network of four 10 μF capacitors is connected to a 500 V supply: three of them in series, and the fourth across that row. What is the equivalent capacitance of the network, in μF? expects 13.3 μF

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The web of concepts

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Q. A network of four 10 μF $C_{o} = 10$ μF capacitors is connected to a 500 V $V_{n} = 500$ V supply: three of them in series $C_{r} = \frac{1}{\frac{3}{C_{o}}}$ , and the fourth across that row $C_{n} = C_{r} + C_{o}$ . What is the equivalent capacitance of the network, in μF?

The playground

Inputs

each

C_{o}

10 μF

supply

V_{n}

500 V

Outputs

equivalent

C_{n}

The solution

Solved with Capacitance

Given

$C_{o} = 10$ μF
$V_{n} = 500$ V
$C_{r} = \frac{1}{\frac{3}{C_{o}}}$ from “three of them in series”
$C_{n} = C_{r} + C_{o}$ from “the fourth across that row”

Find $C_{n}$ .

Working

$C_{o} = 10$ μF $= 1 \times 10^{−5}$ F
three of them in series $C_{r} = \frac{1}{\frac{3}{C_{o}}}$
$C_{r} = \frac{1}{\frac{3}{1 \times 10^{−5}}}$
$C_{r} = 3.33333333 \times 10^{−6}$ F
the fourth across that row $C_{n} = C_{r} + C_{o}$
$C_{n} = 3.33333333 \times 10^{−6} + 1 \times 10^{−5}$
$C_{n} = 13.33$ μF
$C_{n} = 1.33333333 \times 10^{−5}$ F $= 13.3333333$ μF
$C_{n} \approx 13.3$ μFto 3 significant figures

Conclusion

In series the capacitances add as reciprocals, so the row of three holds a third as much; across it, the fourth simply adds.

The Ruby

1require_relative "../physics"23question :four_capacitors_on_500_volts do4  kind :numerical5  describes "A network of four %{each} capacitors is connected to a %{supply} supply: three of them in series, and " \6            "the fourth across that row. %{equivalent}, in μF?"7  given each: "10 μF"8  given supply: "500 V"9  asks equivalent: "What is the equivalent capacitance of the network"10end

1require_relative "../questions/four_capacitors_on_500_volts"2require_relative "../science/electrostatic_potential_and_capacitance/capacitance"34solve :four_capacitors_on_500_volts, using: Capacitance do5  objects :one, :row, :network67  take each: one.capacitance8  take supply: network.voltage910  deduce_from "three of them in series", row.capacitance => 1 / (3 / one.capacitance)11  deduce_from "the fourth across that row", network.capacitance => row.capacitance + one.capacitance1213  equivalent = network.capacitance14  answer equivalent:, in: "μF", figures: 31516  conclusion "In series the capacitances add as reciprocals, so the row of three holds a third as much; " \17             "across it, the fourth simply adds."18end

1require_relative "../solutions/four_capacitors_on_500_volts"23play_with :four_capacitors_on_500_volts do4  title "Four Capacitors on 500 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	—
charge of network	$Q_{n}$	C	—
capacitance of network	$C_{n}$	F	0 … ∞
voltage of network	$V_{n}$	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	—
dielectric constant of network	$K_{n}$	—	—
area of network	$A_{n}$	m²	—
plate separation of network	$d_{n}$	m	—
energy stored of ground	$U_{g}$	J	—
energy stored of one	$U_{o}$	J	—
energy stored of row	$U_{r}$	J	—
energy stored of network	$U_{n}$	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)

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

charge on a capacitor(network)

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)

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

a parallel plate capacitor(network)

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)

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

energy in a capacitor(network)

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

three of them in series

C_{n} = C_{r} + C_{o}

the fourth across that row