Introduction: Charge, Voltage, and Current
Hydraulic systems are like electric circuits:
- volume (liters) is like charge (coulombs)
- flow rate
(liters per second, or LPS) is like current (coulombs per second, or amps)
- pressure (kPa) is like voltage (volts)
- energy is pressure x volume (kPa x liters = joules)
or voltage x charge (volts x coulombs = joules)
- power is pressure x flow rate (kPa x LPS = watts)
or voltage x current (volts x amps = watts)
The hydraulic analogy is useful for
visualizing what happens in electric circuits, especially the behavior
of wires, batteries, resistors, capacitors, inductors, and diodes.
There are hydraulic models for each of these circuit components.
In a hydraulic circuit, a pump brings up water from a reservoir at zero
pressure and pressurizes it; this requires energy. The water flows under
pressure. When the water under pressure is released, the pressure drops
back to zero and the stored energy is released.
In an electric circuit, a battery brings up charge from ground at zero
voltage and raises the charge to a higher voltage; this requires energy.
The charge flows under the increased voltage. When the charge is
released, the voltage drops back to zero and the stored energy is
Energy and Power
In a hydraulic circuit, water under pressure flows through a pipe, with:
Energy = Pressure x Volume
Power = Pressure x Flow Rate
In an electric
circuit, charge at a specific voltage flows through a wire, with:
Voltage x Charge
Power = Voltage x Current
If you use standard units
(kPa, liters, LPS for hydraulic systems and volts, coulombs, and amps
for electric circuits), the energy units are Joules and the power units
Hydraulic energy is calculated as work (force x distance), for example, a
piston pushing against water under pressure.
In this case, the force is the pressure multiplied by the area of the top of the piston. Thus:
Energy = Force x Distance
= (Pressure x Area) x Distance
= Pressure x (Area x Distance)
= Pressure x Volume
= 300 kPa x 2 liters = 600 joules
where the pressure is 300 kPa and the the piston displaces 2 liters of water.
Similarly, electric energy
is calculated as a force that moves an amount of charge in an electric
field, changing the voltage where the charge rests. For example, if you rub a piece of amber with fur, the amber
acquires a negative charge. If you put the amber between two metal
plates with 100 volts of potential difference, the amber is attracted to
the positive plate and repelled by the negative plate.
In this case, if the charge on the amber is 1 coulomb, the amount of
energy required to move the amber from the bottom plate to the top plate
Energy = Voltage x Charge
= 100 volts x 1 coulomb
= 100 joules
In a hydraulic circuit, a pump operates continuously with a power of pressure
to pressurize and move water. When the
pressurized water is released and allowed to fall to a lower pressure, the stored energy is also released.
Similarly, in an electric circuit, a battery operates continuously with a
power of voltage
to "pressurize" and move charge. When the
"pressurized" charge is released and allowed to fall to a lower voltage, the stored energy is also released.
A hydraulic resistor is a narrow section of a pipe that causes the
pressure top drop in the direction of flow. Similarly, and electronic
resistor is a partial blockage in a wire that causes a drop in voltage
in the direction of flow.
According to Ohm's law, the drop in pressure (voltage) across a resistor
is proportional to the flow rate (current) through the resistor.
electronic resistor, the constant of proportionality is called the
resistance, and the units are volts per amp, or Ohms. For example, a
1-ohm resistor has a voltage drop of 1 volt for 1 amp of current, 2
volts for 2 amps of current, and so on.
You can imagine a similar
behavior for a hydraulic resistor. The more water you push through the
narrow point in the pipe, the greater the pressure drop across the
Flow through a resistor converts potential energy into heat. The rate at which heat is
produced is the pressure drop in kPa multiplied by the flow rate in
liters per second; or the voltage drop in volts multiplied by the
current in amps. In both cases, the power units are Watts.
Here I compare the behavior of hydraulic and electronic capacitors.
A hydraulic capacitor is a cylinder divided by a flexible rubber sheet.
Here's an oblique view and a cross-section side view (the latter is
easier to draw).
When you apply pressure to one side of the capacitor, the rubber
membrane is displaced in the direction of higher to lower pressure.
volume of water stored is proportional to the pressure applied:
Volume = Capacitance x Pressure
2 liters = 0.01 liters/kPa x 200 kPa
The constant of proportionality is called the capacitance, which has units of liters per kPa.
Consider the change over time:
over time) = Capacitance x (change in
Pressure over time)
Then you get:
Flow Rate = Capacitance x (dP/dt)
where dP/dt means the change in pressure per unit of time.
For example, if you increase the pressure at a rate of 100 kPa per
second, the flow rate into the capacitor is 1 liter per second.
1 LPS = 0.01 liter/kPa x 100 kPa/sec
An electronic capacitor is a set of two parallel metal plates separated
by an insulator.
When you apply a voltage across of the capacitor, charge is stored in
The amount of charged stored is proportional to the voltage applied:
Charge = Capacitance x Voltage
2 coulombs = 0.01 farad x 200 volts
The constant of proportionality is called the capacitance, and it has
units of coulombs per volt, also known as Farads.
Consider the change over time:
over time) = Capacitance x (change in
Voltage over time)
Then you get:
Current = Capacitance x (dv/dt)
where dv/dt means the change in voltage per unit of time.
For example, if you increase the voltage at a rate of 100 volts per
second, the flow rate into the capacitor is 1 coulomb per second (1 amp).
1 amp = 0.01 F x 100 volts/sec
Energy Stored in a Capacitor
The energy stored in a charged capacitor is:
Energy = 1/2 x Pressure x Volume (hydraulic)
Energy = 1/2 x Voltage x Charge (electronic)
For the hydraulic capacitor, the stored energy comes from the pump that pushed the water into the capacitor. The average
that pushed in the water is one-half the current pressure (the
pressure ramped up from zero to the current pressure as the capacitor
Similarly, for the electronic
capacitor, the stored energy comes from the voltage source that
pushed the charge into the capacitor. The average
voltage that pushed in the charge
is one-half the current voltage (the voltage ramped up from zero to
the current voltage).
You can extract the energy by connecting a resistor between the two ends
of the capacitor. This causes a current to flow through the resistor.
The voltage and
current both decay exponentially until the voltage reaches zero and all the
energy is dissipated as heat in the resistor.
Conservation of Charge and the Electric Field
Charge is conserved, even as a capacitor is charged and
discharged. When you connect a battery to a capacitor, the battery draws
positive charge from one side of the capacitor, leaving a deficit or
negative charge on that side, and puts the positive charge on the other
For example, connecting a 100 V battery to a 1 Farad capacitor
draws 50 coulombs of charge into the positive side of the capacitor,
leaving -50 coulombs on the negative side.
Then the capacitor has a net
charge of zero (50 coulombs + -50 coulombs) but a difference of 100 coulombs on the two
When the battery is disconnected, the voltage difference remains
because the charge is trapped in the capacitor.
The physical attraction
of opposite charges holds the positive and negative charge in place.
That's why a capacitor is made with closely spaces parallel plates -- to
get the opposite charges as close together as possible, over as large an area as possible.
An electric field exists in the space between the two plates, which
causes any charged object in that space to feel a force. For example, a
free electron is attracted to the positively charged plate and repelled
by the negatively charged plate.
If you short-circuit the two terminals of the
capacitor, the charge rushes back from the positively charged side to
the negatively charged side, bringing the charge back to zero on both sides.
Here I compare the behavior of hydraulic and electronic inductors.
A hydraulic inductor is a water wheel connected through a rigid axle to a
heavy stone flywheel, with a housing that forces the water in a
clockwise direction when the water flows from left to right.
Here's a conceptual view and a cross-section side view:
Applying pressure to the inductor presses water
against the paddles of the water wheel. This starts the turning of the
water wheel and the connected flywheel. As pressure is applied over
time, the flywheel gradually turns faster and faster. Once the flywheel
is turning, it tends to keep turning even when the pressure is removed,
keeping the water flowing at a constant speed in the absence of outside
The momentum of the flywheel is proportional to the flow rate.
Momentum = Inductance x Flow Rate
60 kPa-sec = 20 kPa-sec/LPS x 3 LPS
The constant of proportionality is called the inductance.
Consider change over time:
Momentum over time) = Inductance x (change in
Flow Rate over time)
The change in momentum over time is the same as the pressure applied:
Pressure = Inductance x (dF/dt)
The notation dF/dt means the change in flow rate over time (units: liters per second per second).
For example, if you apply a 20 kPa the inductor, the flow rate increases by one LPS per second.
20 kPa= 20 kPa-sec/LPS x 1 LPS/second
- Applying a constant pressure of 20 kPa to the inductor
causes the flow rate to increase by 1 liter per second for each second
that the pressure is applied.
- Applying zero pressure causes the flow to
continue at a constant rate.
- Applying a negative pressure of 20
kPa causes the flow to decrease by 1 liter per second for each second
that the negative pressure is applied.
This is very much like people going through a revolving door.
If the people are in a big hurry, they push on the revolving door (apply
positive pressure), which causes the door to speed up and gain
If the people are relaxed and walk through at the
same speed that the door is moving, the door keeps turning at the same speed and has a constant momentum.
If some very slow people come
through, the door catches up to them and pushes them through (negative
pressure), which causes the door to slow down and lose momentum.
An electronic inductor is a coil of wire wrapped around a cylindrical
Applying a voltage to the inductor causes magnetic flux to build
up in the coil. Magnetic flux is just like the momentum of the flywheel
in the hydraulic inductor, and applying a positive voltage is like people in a hurry pushing on the revolving door.
As the voltage is
applied over time, the
magnetic flux gradually gets larger and larger. Once the magnetic flux
is built up, it tends to keep the current flowing even when the voltage
source is removed, keeping the current constant in the absence of
The magnetic flux is proportional to the current through the coil:
Flux = Inductance x Current
60 webers = 20 henrys x 3 amps
The constant of proportionality is called the inductance (units: webers per amp, or henrys)
Consider change over time:
Flux over time) = Inductance x (change in
Current over time)
The change in flux over time is equal the voltage applied (per Maxwell's equations):
Voltage = Inductance x (di/dt)
The notation di/dt means the change in current over time (units: coulombs per second per second, or amps per second).
For example, if you apply a 20 volts the inductor, the flow rate increases by one amp per second.
20 volts = 20 henrys x 1 amp/second
- Applying a 20 volts to the inductor
causes the current to increase by 1 amp for each second
that the voltage is applied.
- Applying zero volts causes the current to
continue at a constant rate.
- Applying -20 volts causes the current to decrease by 1 amp for each second
that the negative voltage is applied.
Energy Stored in an Inductor
You might recall from your Physics 1 class that kinetic energy if a moving object is
proportional to the square of its velocity. The same principle applies
to an inductor. The energy stored in the flywheel of the hydraulic
inductor, or the magnetic flux of the electronic inductor, is
proportional to the square of the flow rate:
Energy = 1/2 x Inductance x Flow-Rate2 (hydraulic)
Energy = 1/2 x Inductance x Current2 (electronic)
For the hydraulic inductor, the stored energy comes from the pump that
pushed the water through the inductor and accelerated the
flywheel. For the electronic inductor, the stored energy comes from the
pushed the charge through the inductor and built up the magnetic flux
through the coil.
You can extract the energy as heat by connecting a resistor between the
two ends of the inductor. This causes a voltage to develop across the
resistor and current to flow through it. The voltage and current decay exponentially until the flux
reaches zero and all the energy is dissipated as heat in the resistor.
Inductor-Capacitor (L-C) Oscillator
Connecting a capacitor to an inductor creates an oscillator. Once
started, charge flows back and forth between
the two sides of the capacitor at a fixed frequency, even after maximum
charge, voltage, and current decrease due to energy loss.
For the hydraulic inductor-capacitor circuit, you start by pumping water
into the left side of the capacitor, pressurizing the water. The
potential energy is stored in the stretched rubber sheet separating the
two sides of the capacitor. You remove the pump and allow the
circuit to run freely.
The rubber sheet forces the water through the inductor, causing the
inductor's flywheel to spin faster and faster. When the rubber sheet
reaches the neutral position, the pressure is zero but the flywheel momentum and current are at
As the flywheel continues to turn, it pressurizes the
water and forces it into the opposite side of the capacitor, stretching
the rubber sheet in the opposite direction. When the flywheel runs out
of energy and stops, the capacitor is fully pressurized in the opposite
The same process runs in reverse, returning the circuit to
the original starting position if there is no friction. With loss of
energy to friction, the rubber membrane comes back into a less-stretched
maximum position, but it still oscillates with the same period.
For the electronic inductor-capacitor (L-C) circuit, you start by charging the capacitor,
raising the charge to the same voltage as a battery. The potential
energy is stored in the electric field that holds the charge on the two
sides of the capacitor. You disconnect the battery, connect the
capacitor to the inductor, and allow the circuit to run freely.
The voltage of the capacitor forces the accumulated charge through the
inductor, causing the inductor's magnetic flux to build up. The flux is
like the momentum of the flywheel in the hydraulic circuit. When the
capacitor runs out of charge, the voltage is zero but the flux and current are at their maximum.
The magnetic flux forces the current to continue flowing
into the opposite side of the capacitor, raising the voltage and
building up the charge across the capacitor in the opposite direction.
When the magnetic flux runs out of energy and the current stops flowing,
the capacitor is fully charged in the opposite direction.
process runs in reverse, returning the circuit to the original starting
point if there is no loss to wire resistance. With loss of energy
resistance, the capacitor carries less charge at the maximum point, but
it still oscillates with the same
If you take a charged capacitor as the
starting point (time=0), the voltage across the two devices is the
cosine function and the current through the circuit is the sine
Period = 2π SQRT(Inductance x Capacitance)
Frequency = 1/(Period)
The frequency of the oscillator does not depend on the amount of
charge. Charging the capacitor to a higher voltage increases the
amplitude of the voltage and current but does not affect the frequency.
This is just like a guitar string, which vibrates at the same frequency
whether plucked gently or strongly, producing the same note in either
case. Only the volume is affected. Like a guitar string, the amplitude dies away due to friction/resistance.
L-C Oscillator in a Crystal Radio
The oscillator can be used as a tuner for a crystal radio. The radio
consists of an L-C oscillator, an antenna, a germanium diode, and an
earphone. There is no battery or other internal power source. The inductor has a ferrite core that can slide in and out of
the coil, which modifies the inductance and therefore the oscillation frequency.
Here is a schematic diagram of the crystal radio circuit:
If the inductor-capacitor combination is tuned
to oscillate at 810 kHz, the 810 kHz radio signal is maintained in the
oscillating circuit, whereas signals from other stations fade out due to wire resistance. The
diode converts the amplitude-modulated AM radio signal into an audio
signal that can be heard in the earphone.
The energy for the radio comes
entirely from the transmitted radio signal; no battery is needed. Only a
tiny amount of power from the radio signal is captured by the antenna,
so only one person can
listen to the faint audio signal through the earphone.
Future topics to be covered:
- How good is the hydraulic analogy?
Comparison of capacitor and inductor behavior
- Dielectric materials in capacitors
- Ferromagnetic materials in inductors
Exponential decay in RC and RL circuits
Other hydraulic analogy websites:
Water circuit analogy to electric circuit
from HyperPhysics by C. Rod Nave, Georgia State University
Excellent resource for physics students
Hydraulic analogy, Wikipedia
Brief Wikipedia article, good overview
Understanding Electricity with Hydraulics
Describes hydraulic models for diodes, transistors, and op amps
Circuit Analysis, Khan Academy
Math analysis of electric circuits, including LC oscillator
Crystal radio, Wikipedia
Good info on crystal radio operation and history