Electromechanics

Question 1

Describe the first and second Kirchhoff law.

Answer 1

The first law states that the sum of the currents into a
given node equals the sum of the currents out of that node.

The second law states that the sum of electromotive forces in a loop in the network equals the sum of potential drops, or voltages across each of the resistances, in the loop.

Question 2

Give an example for the application of the Kirchhoff laws.

Answer 2

Calculate suitable values for limiting resistors

Explain how resistors can be used to form voltage dividers

Select suitable resistors for given applications

Question 3

Describe Ohms law.

Answer 3

Ohms law with
Resistance 𝑹:
𝑼 = 𝑹 ∙ I

Ohm believed that the communication of electricity
occurred between “contiguous particles” which is the
term Ohm himself uses.
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Ohm’s law states that the current through a conductor between two points is directly proportional to the voltage across the two points.

Question 4

Define capacitance.

Answer 4

Capacitors are measured in the
amount of charge it can store, the SI
unit is the farad.
A capacitor has a value of 1 farad if 1
coulomb of charge change the
voltage across it by 1V.

Question 5

Calculate the electrostatic force between the plates of charged capacitor.

Answer 5

F=W/x=1/2 ϵ_r ϵ_0 A∙1/x^2 U_0^2 = 1/2 C/x U^2

Question 6

Explain how an electrostatic motor works.

Answer 6

If high voltage (5000V) is applied between the two
stator bottle foils, one stator bottle acquires a
negative charge imbalance, while the other one
becomes positive. Also, a tiny spark jumps from the
tip of each commutator brush to one of the foil
sectors on the rotor bottle.
The sector under the positive brush becomes positive,
the one under the negative brush becomes negative.
The rotor’s foil sectors are then repelled from the
alike-charged stator bottle and attracted to the unlike
charged stator bottle.
This sideway electrostatic force causes the center
bottle to rotate, which brings new foil sectors under
the brushes. The force is continuous…

–
An electrostatic motor or capacitor motor is a type of electric motor based on the attraction and repulsion of electric charge.

Question 7

In which circumstances is an electrostatic motor more efficient than an electromagnetic motor?

Answer 7

Electrostatic motors at small sizes are
advantageous because the electric charge
and with it the torque is produced at the
surface (𝑆 = 4𝜋𝑟2).

Question 8

Explain why there is current flow when the poles of a capacitor are connected to a battery.

Answer 8

Conductor:
Again a battery supplies two plates with current I such that in the conductor (conductivity𝛾=1/𝜌) a current field with current density 𝑆 = 𝛾 ∙ 𝐸 Is build up. The electric field 𝐸 moves positive charges from right plate to left plate.

Insulator:
The current is a consequence of the polarization of particles in the insulator. As soon as the polarization
has come to a saturation the current flow through the wires comes to an end.
Current density is zero but the electric field 𝐸 is non zero.

Question 9

What is the amount of energy stored in a capacitor?

Answer 9

Work of electric field when moving a charge 𝑄 to
a distance 𝑧 is:
𝑊𝑒𝑙 = −𝐹𝑒𝑙 ∙ 𝑧
This work is stored as potential energy Π𝑒𝑙.
Dividing through 𝑄 gives the electric potential:
𝑉𝑒𝑙 =𝐹𝑒𝑙/𝑄 ∙ 𝑧 = 𝐸 ∙ z

–
Energy = 1/2 CU^2 = 1/2 QU = 1/2 (eps_0eps_rA*U)/d

Question 10

Show the magnetic field of a simple permanent magnet

Answer 10

Lines N –> S

Question 11

Sketch the magnetic field of a solenoid.

Answer 11

Right hand rule

Question 12

. Given the direction of current in a solenoid, show the direction of the flux lines.

Answer 12

Right hand rule.

Question 13

What is the B-Field?

Answer 13

note that the physical name for the
magnetic field is “B-field”.

–
Magnetic flux density

Question 14

What is the magnetic field strength?

Answer 14

Again in analogy to the electrostatic field (electric field strength E) we define the magnetic field strength H:
𝑉12 = Integral1-2(𝐻 ∙ 𝑑𝑠)

–
For a closed loop:
Theta = line integral (H_vec) ds

Question 15

Write down Amperes law.

Answer 15

Again in analogy to the electrostatic field (electric field strength E) we define the magnetic field strength H:
𝑉12 = Integral1-2(𝐻 ∙ 𝑑𝑠)

For a closed path we find, that the integral (amperes law) does not vanish if there is magnetism:
Θ = Integral round (𝐻 ∙ 𝑑𝑠) = 𝑁 ∙ 𝐼
The physical quantity Θ is called magneto-motive force and has the unit [ampere turns].

Question 16

Sketch the magnetic field strength in and outside a conductor with current i.

Answer 16

Outside:
hyperbolic

Inside:
linear

Question 17

Explain the hysteresis effect in an electromagnetic circuit.

Answer 17

Hysteresis and magnetization are material properties which have to be found with an appliance like the toroid. Magnetic field strength 𝐻 and flux density 𝐵 often are proportional:
𝐵 = 𝜇𝐻 = 𝜇𝑟𝜇0𝐻
𝐵 = 𝜇𝑟𝜇0𝐻
Relative permeability, sometimes denoted by the symbol 𝜇𝑟, is the ratio of the permeability of a specific medium to the permeability of free space 𝜇0.
𝜇0 = 4𝜋10^−7(Ns/Am)
If 𝜇𝑟&raquo_space; 1 the material is called ferromagnetic. If 𝜇𝑟 > 1 the material is paramagnetic and if 𝜇𝑟 < 1 the material denotes as diamagnetic

–
Magnetic hysteresis occurs when an external magnetic field is applied to a ferromagnet such as iron and the atomic dipoles align themselves with it.
Even when the field is removed, part of the alignment will be retained: the material has become magnetized. Once magnetized,
the magnet will stay magnetized indefinitely. To demagnetize it requires heat or a magnetic field in the opposite direction.

Question 18

What does the Lenz pendulum show us?

Answer 18

With the Lenz’ Pendulum we can show, that changing flux also changes current. As long as the ring of aluminum is open, the pendulum can move freely. After closing the ring current can flow and because of the resistance of the coil dissipates energy. This results in a damped vibration of the pendulum. Because there is current there must be an electric field E.

Question 19

Explain the function of a transformer.

Answer 19

u1/u2 = N1/N2

–
A transformer is a passive electrical device that transfers electrical energy from one electrical circuit to another, or multiple circuits. A varying current in any one coil of the transformer produces a varying magnetic flux in the transformer’s core, which induces a varying electromotive force across any other coils wound around the same core.

Question 20

Calculate the magnetic energy stored in an electromagnet.

Answer 20

Energy stored in the magnetic field:
𝑊𝑚𝑎𝑔 =1/2𝐿𝑖^2
= 1/2 1/L Phi^2

Question 21

Calculate the electromagnetic static force applied to the movable part of a variable inductance motor.

Answer 21

Example from lecture slides:
F_y = - mü_0AN^2I^2/(2(s+y)^2)

Question 22

Explain the principle of magnetic bearings.

Answer 22

support moving machinery without physical contact,
• can levitate a rotating shaft and permit relative motion
without friction or wear.
• Long considered a promising advancement, they are now moving into actual service
• Applications are electric power generation, petroleum refining, machine tool operation and natural gas pipelines

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The magnetic bearing consists of a electromagnet, that is excited by a current.
The magnetic field induces a force on the rotor and the position is measured by a sensor.
Then, the position of the rotor is controlled by a controller that controls the current and thus the force on the rotor.
By using multiple electromagnets, a bearing in all radial directions is given.

Question 23

What are the elements a magnetic bearing consists of?

Answer 23

Electromagnet, Amplifier, Rotor, Sensor, Controller

Question 24

How can we linearize the characteristics of a magnetic bearing?

Answer 24

Usually two radial magnets are connected to produce the radial force, such that the nonlinearities in the characteristics eliminate each other.

Question 25

What sensors can we use in a magnetic bearing?

Answer 25

Current Measurements with Hall Sensor, Magnetic Displacement and Velocity Sensor, Capacitive Displacement Sensors, Eddy Current Displacement Sensors, Inductive Displacement Sensors

Question 26

Why are switching amplifiers advantageous for magnetic bearings?

Answer 26

Because the voltage drop 𝑈𝑠 at the conducting
switching transistor is very small, the losses
of digital amplifiers are considerably
smaller than the losses of analogue
Amplifiers.

Question 27

Why is the use of digital controllers preferable in magnetic bearing applications?

Answer 27

All control strategies can be tested with the same hardware

• In the final implementation even very complex control strategies are realizable.

• Besides the stabilization task for the rotor additional features can be realized (adaptive control, target positioning, force free run of the unbalanced rotor,
monitoring of load, start-stop ramp, etc.)

• Calibration of the system is far easier. Even auto calibration algorithm are possible
• Automatic monitoring and diagnostics of the system’s condition
• Late changes in the control algorithm are realizable far easier than with analogue control