**For Physics & Engineering Students **

Hydraulic Resistor, Capacitor, & Inductor Models

V=IR, I=C(dv/dt), V=L(di/dt)

IntroductionRead |
Energy, PowerRead |
Resistor (easy)Read |
Capacitor (med)Read |
Inductor (hard)Read |
Oscillator (fun)Read |

**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)

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 released.

- Energy = Pressure x Volume
- Power = Pressure x Flow Rate

- Energy = Voltage x Charge
- Power = Voltage x Current

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.

Energy = Voltage x Charge

= 100 volts x 1 coulomb

= 100 joules

In a hydraulic circuit, a pump operates continuously with a power of

Similarly, in an electric circuit, a battery operates continuously with a power of

According to Ohm's law, the drop in pressure (voltage) across a resistor is proportional to the flow rate (current) through the resistor.

For an 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 resistive point.

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.

Watch video

The
volume of water stored is proportional to the pressure applied:

The constant of proportionality is called the capacitance, which has units of liters per kPa.Volume = Capacitance x Pressure

2 liters = 0.01 liters/kPa x 200 kPa

Consider the change 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

When you apply a voltage across of the capacitor, charge is stored in the capacitor.

The amount of charged stored is proportional to the voltage applied:

The constant of proportionality is called the capacitance, and it has units of coulombs per volt, also known as Farads.Charge = Capacitance x Voltage

2 coulombs = 0.01 farad x 200 volts

Consider the change 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 = 1/2 x Pressure x Volume (hydraulic)For the hydraulic capacitor, the stored energy comes from the pump that pushed the water into the capacitor. The

Energy = 1/2 x Voltage x Charge (electronic)

Similarly, for the electronic capacitor, the stored energy comes from the voltage source that pushed the charge into the capacitor. The

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.

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 sides. 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.

Watch Video

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 forces.

The momentum of the flywheel is proportional to the flow rate.

Momentum = Inductance x Flow RateThe constant of proportionality is called the inductance.

60 kPa-sec = 20 kPa-sec/LPS x 3 LPS

Consider change 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/secondTo summarize:

- 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.

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 momentum.

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.

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 outside forces.

The magnetic flux is proportional to the current through the coil:

Flux = Inductance x CurrentThe constant of proportionality is called the inductance (units: webers per amp, or henrys)

60 webers = 20 henrys x 3 amps

Consider change 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/secondTo summarize:

- 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 = 1/2 x Inductance x Flow-RateFor 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 battery that pushed the charge through the inductor and built up the magnetic flux through the coil.^{2}(hydraulic)

Energy = 1/2 x Inductance x Current^{2}(electronic)

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.

Watch Video

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 their maximum.

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 direction.

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.

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.

The same process runs in reverse, returning the circuit to the original starting point if there is no loss to wire resistance. With loss of energy to resistance, the capacitor carries less charge at the maximum point, but it still oscillates with the same period.

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 function.

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.

- Period = 2π SQRT(Inductance x Capacitance)

- Frequency = 1/(Period)

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?
- Kirchhoff's Laws
- Comparison of capacitor and inductor behavior
- Dielectric materials in capacitors

- Ferromagnetic materials in inductors

- Exponential decay in RC and RL circuits
- Diode behavior
- Alternating current
- Voltage transformer
- Power transmission

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