Tuesday 13 June 2017

Basics of Electrical Engineering

Basics of Electrical Engineering

Electric Charge

What is electric charge?

Electric charge, basic property of matter carried by some elementary particles. Electric charge, which can be positive or negative, occurs in discrete natural units and is neither created nor destroyed.

Electric charges are of two general types: positive and negative. Two objects that have an excess of one type of charge exert a force of repulsion on each other when relatively close together. Two objects that have excess opposite charges, one positively charged and the other negatively charged, attract each other when relatively near. 
There are 2 types of electric charge:

Positive charge (+)

Positive charge has more protons than electrons (Np>Ne).
Positive charge is denoted with plus (+) sign.
The positive charge attracts other negative charges and repels other positive charges.
The positive charge is attracted by other negative charges and repelled by other positive charges.

Negative charge (-)

Negative charge has more electrons than protons (Ne>Np).
Negative charge is denoted with minus (-) sign.
Negative charge attracts other positive charges and repels other negative charges.
The negative charge is attracted by other positive charges and repelled by other negative charges.

Electric force (F) direction according to charge type

q1/q2 chargesForce on q1 chargeForce on q2 charge
- / -←⊝⊝→repletion
+ / +←⊕⊕→repletion
- / +⊝→←⊕attraction
+ / -⊕→←⊝attraction

Charge of elementary particles

ParticleCharge (C)Charge (e)
Electron1.602×10-19 C
-e
Proton1.602×10-19 C
+e
Neutron0 C0

Coulomb unit

The electric charge is measured with the unit of Coulomb [C].
One coulomb has the charge of 6.242×1018 electrons:
1C = 6.242×1018 e

Electric charge calculation

When electric current flows for a specified time, we can calculate the charge:

Constant current

Q =  t
is the electric charge, measured in coulombs [C].
I is the current, measured in amperes [A].
t is the time period, measured in seconds [s].

Momentary current

Q(t)=\int_{0}^{t}i(\tau )d\tau
is the electric charge, measured in coulombs [C].
i(t) is the momentary current, measured in amperes [A].
t is the time period, measured in seconds [s].

Electrical Voltage

Electrical voltage is defined as electric potential difference between two points of an electric field.
Using water pipe analogy, we can visualize the voltage as height difference that makes the water flow down.
V = φ2 - φ1
V is the voltage between point 2 and 1 in volts (V).
φ2 is the electric potential at point #2 in volts (V).
φ1 is the electric potential at point #1 in volts (V).

In an electrical circuit, the electrical voltage V in volts (V) is equal to the energy consumption E in joules (J)
divided by the electric charge Q in coulombs (C).
V=\frac{E}{Q}
V is the voltage measured in volts (V)
E is the energy measured in joules (J)
Q is the electric charge measured in coulombs (C)

Voltage in series

The total voltage of several voltage sources or voltage drops in series is their sum.
VT = V+ V+ V+...
VT - the equivalent voltage source or voltage drop in volts (V).
V1 - voltage source or voltage drop in volts (V).
V2 - voltage source or voltage drop in volts (V).
V3 - voltage source or voltage drop in volts (V).

Voltage in parallel

Voltage sources or voltage drops in parallel have equal voltage.
VT = V= V= V=...
VT - the equivalent voltage source or voltage drop in volts (V).
V1 - voltage source or voltage drop in volts (V).
V2 - voltage source or voltage drop in volts (V).
V3 - voltage source or voltage drop in volts (V).

Voltage divider

For electrical circuit with resistors (or other impedance) in series, the voltage drop Vi on resistor Riis:
V_i=V_T\: \frac{R_i}{R_1+R_2+R_3+...}

Kirchhoff's voltage law (KVL)

The sum of voltage drops at a current loop is zero.
∑ Vk = 0

DC circuit

Direct current (DC) is generated by a constant voltage source like a battery or DC voltage source.
The voltage drop on a resistor can be calculated from the resistor's resistance and the resistor's current, using Ohm's law:
Voltage calculation with Ohm's law
VR = IR × R
VR - voltage drop on the resistor measured in volts (V)
IR - current flow through the resistor measured in amperes (A)
R - resistance of the resistor measured in ohms (Ω)

AC circuit

Alternating current is generated by a sinusoidal voltage source.
Ohm's law
VZ = IZ × Z
VZ - voltage drop on the load measured in volts (V)
IZ - current flow through the load measured in amperes (A)
Z - impedance of the load measured in ohms (Ω)
Momentary voltage
v(t) = Vmax × sin(ωt)
v(t) - voltage at time t, measured in volts (V).
Vmax - maximal voltage (=amplitude of sine), measured in volts (V).
ω      - angular frequency measured in radians per second (rad/s).
t        - time, measured in seconds (s).
θ       - phase of sine wave in radians (rad).
RMS (effective) voltage
Vrms = Veff  =  Vmax / √2 ≈ 0.707 Vmax
Vrms -  RMS voltage, measured in volts (V).
Vmax - maximal voltage (=amplitude of sine), measured in volts (V).
Peak-to-peak voltage
Vp-p = 2Vmax

Voltage drop

Voltage drop is the drop of electrical potential or potential difference on the load in an electrical circuit.

Voltage Measurement

Electrical voltage is measured with Voltmeter. The Voltmeter is connected in parallel to the measured component or circuit.
The voltmeter has very high resistance, so it almost does not affect the measured circuit.

Voltage by Country

AC voltage supply may vary for each country.
European countries use 230V while north America countries use 120V.

CountryVoltage
[Volts]
Frequency
[Hertz]
Australia230V50Hz
Brazil110V60Hz
Canada120V60Hz
China220V50Hz
France230V50Hz
Germany230V50Hz
India230V50Hz
Ireland230V50Hz
Israel230V50Hz
Italy230V50Hz
Japan100V50/60Hz
New Zealand230V50Hz
Philippines220V60Hz
Russia220V50Hz
South Africa220V50Hz
Thailand220V50Hz
UK230V50Hz
USA120V60Hz

Electric Current

Electric current definition and calculations.

Electric current definition

Electrical current is the flow rate of electric charge in electric field, usually in electrical circuit.
Using water pipe analogy, we can visualize the electrical current as water current that flows in a pipe.
The electrical current is measured in ampere (amp) unit.

Electric current calculation

Electrical current is measured by the rate of electric charge flow in an electrical circuit:
i(t) = dQ(t) / dt
The momentary current is given by the derivative of the electric charge by time.
i(t) is the momentary current I at time t in amps (A).
Q(t) is the momentary electric charge in coulombs (C).
t is the time in seconds (s).

When the current is constant:
I = ΔQ / Δt
I is the current in amps (A).
ΔQ is the electric charge in coulombs (C), that flows at time duration of Δt.
Δt is the time duration in seconds (s).

Example
When 5 coulombs flow through a resistor for duration of 10 seconds,
the current will be calculated by:
I = ΔQ / Δ = 5C / 10s = 0.5A

Current calculation with Ohm's law

The current Iin anps (A) is equal to the resistor's voltage Vin volts (V) divided by the resistanceR in ohms (Ω).
IR = VR / R
Current direction
current typefromto
Positive charges+-
Negative charges-+
Conventional direction+-

Current in series circuits

Current that flows through resistors in series is equal in all resistors - just like water flow through a single pipe.
ITotal = I= I= I=...
ITotal - the equivalent current in amps (A).
I1 - current of load #1 in amps (A).
I2 - current of load #2 in amps (A).
I3 - current of load #3 in amps (A).

Current in parallel circuits

Current that flows through loads in parallel - just like water flow through parallel pipes.
The total current ITotal is the sum of the parallel currents of each load:
ITotal = I1 + I+ I+...
ITotal - the equivalent current in amps (A).
I1 - current of load #1 in amps (A).
I2 - current of load #2 in amps (A).
I3 - current of load #3 in amps (A).

Current divider

The current division of resistors in parallel is
RT = 1 / (1/R2 + 1/R3)
or
I1 = IT × RT / (R1+RT)

Kirchhoff's current law (KCL)

The junction of several electrical components is called a node.
The algebraic sum of currents entering a node is zero.
∑ Ik = 0

Alternating Current (AC)

Alternating current is generated by a sinusoidal voltage source.
Ohm's law
IZ = VZ / Z
IZ  - current flow through the load measured in amperes (A)
VZ - voltage drop on the load measured in volts (V)
Z  - impedance of the load measured in ohms (Ω)
Angular frequency
ω = 2π f
ω - angular velocity measured in radians per second (rad/s)
f  - frequency measured in hertz (Hz).
Momentary current
i(t) = Ipeak sin(ωt+θ)
i(t)      - momentary current at time t, measured in amps (A).
Ipeak - maximal current (=amplitude of sine), measured in amps (A).
ω      - angular frequency measured in radians per second (rad/s).
t        - time, measured in seconds (s).
θ       - phase of sine wave in radians (rad).
RMS (effective) current
Irms =  Ieff =  Ipeak / √2 ≈ 0.707 Ipeak
Peak-to-peak current
Ip-p = 2Ipeak

Current measurement

Current measurement is done by connecting the ammeter in series to the measured object, so all the measured current will flow through the ammeter.
The ammeter has very low resistance, so it almost does not affect the measured circuit.


Electrical Resistance

Electrical resistance definition and calculations.

Resistance definition

Resistance is an electrical quantity that measures how the device or material reduces the electric current flow through it.
The resistance is measured in units of ohms (Ω).
If we make an analogy to water flow in pipes, the resistance is bigger when the pipe is thinner, so the water flow is decreased.

Resistance calculation

The resistance of a conductor is resistivity of the conductor's material times the conductor's length divided by the conductor's cross sectional area.
R=\rho \times \frac{l}{A}
R is the resistance in ohms (Ω).
ρ is the resistivity in ohms-meter (Ω×m)
l is the length of the conductor in meter (m)
A is the cross sectional area of the conductor in square meters (m2)

It is easy to understand this formula with water pipes analogy:
  • when the pipe is longer, the length is bigger and the resistance will increase.
  • when the pipe is wider, the cross sectional area is bigger and the resistance will decrease.

Resistance calculation with ohm's law

R is the resistance of the resistor in ohms (Ω).
V is the voltage drop on the resistor in volts (V).
I is the current of the resistor in amperes (A).

Temperature effects of resistance

The resistance of a resistor increases when temperature of the resistor increases.
R2 = R× ( 1 + α(T- T1) )
R2 is the resistance at temperature T2 in ohms (Ω).
R1 is the resistance at temperature T1 in ohms (Ω).
α is the temperature coefficient.

Resistance of resistors in series

The total equivalent resistance of resistors in series is the sum of the resistance values:
RTotal = R1R2R3+...

Resistance of resistors in parallel

The total equivalent resistance of resistors in parallel is given by:

Measuring electrical resistance

Electrical resistance is measured with ohmmeter instrument.
In order to measure the resistance of a resistor or a circuit, the circuit should have the power supply turned off.
The ohmmeter should be connected to the two ends of the circuit so the resistance can be read.

Superconductivity

Superconductivity is the drop of resistance to zero at very low temperatures near 0ºK.



Electric Power

Electric power is the rate of energy consumption in an electrical circuit.
The electric power is measured in units of watts.

Electric power definition

The electric power P is equal to the energy consumption E divided by the consumption time t:
P=\frac{E}{t}
P is the electric power in watt (W).
E is the energy consumption in joule (J).
t is the time in seconds (s).

Example

Find the electric power of an electrical circuit that consumes 120 joules for 20 seconds.
Solution:
E = 120J
t = 20s
P = E / t = 120J / 20s = 6W

Electric power calculation

P =  I
or
P = I  R
or
P = V 2 R
P is the electric power in watt (W).
V is the voltage in volts (V).
I is the current in amps (A).
R is the resistance in ohms (Ω).

Power of AC circuits

The formulas are for single phase AC power.
For 3 phase AC power:
 When line to line voltage (VL-L) is used in the formula, multiply the single phase power by square root of 3 (√3=1.73).
When line to zero voltage (VL-0) is used in the formula, multiply the single phase power by 3.

Real power

Real or true power is the power that is used to do the work on the load.
P = Vrms Irms cos φ

P      is the real power in watts [W]
Vrms  is the rms voltage = Vpeak/√2 in Volts [V]
Irms   is the rms current = Ipeak/√2 in Amperes [A]
φ      is the impedance phase angle = phase difference between voltage and current.
 

Reactive power

Reactive power is the power that is wasted and not used to do work on the load.
Q = Vrms Irms sin φ

Q      is the reactive power in volt-ampere-reactive [VAR]
Vrms  is the rms voltage = Vpeak/√2 in Volts [V]
Irms   is the rms current = Ipeak/√2 in Amperes [A]
φ      is the impedance phase angle = phase difference between voltage and current.
 

Apparent power

The apparent power is the power that is supplied to the circuit.
S = Vrms Irms

S      is the apparent power in Volt-amper [VA]
Vrms  is the rms voltage = Vpeak/√2 in Volts [V]
Irms   is the rms current = Ipeak/√2 in Amperes [A]
 

Real / reactive / apparent powers relation

The real power P and reactive power Q give together the apparent power S:
P2 + Q2 = S2

P      is the real power in watts [W]
Q      is the reactive power in volt-ampere-reactive [VAR]
S      is the apparent power in Volt-amper [VA]


Electric Power Efficiency

Power efficiency

Power efficiency is defined as the ratio of the output power divided by the input power:
η = 100% ⋅ Pout / Pin
η is the efficiency in percent (%).
Pin is the input power consumption in watts (W).
Pout is the output power or actual work in watts (W).
Example
Electric motor has input power consumption of 50 watts.
The motor was activated for 60 seconds and produced work of 2970 joules.
Find the efficiency of the motor.
Solution:
Pin = 50W
E = 2970J
t = 60s
Pout = E / t  = 2970J / 60s = 49.5W
η = 100% * Pout / Pin = 100 * 49.5W / 50W = 99%

Energy efficiency

Energy efficiency is defined as the ratio of the output energy divided by the input energy:
η = 100% ⋅ Eout / Ein
η is the efficiency in percent (%).
Ein is the input energy consumed in joule (J).
Eout is the output energy or actual work in joule (J).
 
Example
Light bulb has input power consumption of 50 watts.
The light bulb was activated for 60 seconds and produced heat of 2400 joules.
Find the efficiency of the light bulb.
Solution:
Pin = 50W
Eheat = 2400J
t = 60s
Ein = Pin * t = 50W * 60s = 3000J
Since the light bulb should produce light and not heat:
Eout Ein - Eheat = 3000J - 2400J = 600J
η = 100 * Eout / Ein = 100% * 600J / 3000J = 20%


Power Factor

In AC circuits, the power factor is the ratio of the real power that is used to do work and theapparent power that is supplied to the circuit.
The power factor can get values in the range from 0 to 1.
When all the power is reactive power with no real power (usually inductive load) - the power factor is 0.
When all the power is real power with no reactive power (resistive load) - the power factor is 1.

Power factor definition

The power factor is equal to the real or true power P in watts (W) divided by the apparent power |S| in volt-ampere (VA):
PF = P(W) / |S(VA)|
PF - power factor.
P   - real power in watts (W).
|S|   - apparent power - the magnitude of the complex power in volt⋅amps (VA).

Power factor calculations

For sinusuidal current, the power factor PF is equal to the absolute value of the cosine of the apparent power phase angle φ (which is also is impedance phase angle):
PF = |cos φ|
PF is the power factor.
φ   is the apprent power phase angle.

The real power P in watts (W) is equal to the apparent power |S| in volt-ampere (VA) times the power factor PF:
P(W) = |S(VA)| × PF = |S(VA)| × |cos φ|

When the circuit has a resistive impedance load, the real power P is equal to the apparent power |S| and the power factor PF is equal to 1:
PF(resistive load) = P / |S| = 1

The reactive power Q in volt-amps reactive (VAR) is equal to the apparent power |S| in volt-ampere (VA) times the sine of the phase angle φ:
Q(VAR) = |S(VA)| × |sin φ|
Single phase circuit calculation from real power meter reading P in kilowatts (kW), voltage V in volts (V) and current I in amps (A):
PF = |cos φ| = 1000 × P(kW) / (V(V) × I(A))

Three phase circuit calculation from real power meter reading P in kilowatts (kW), line to line voltage VL-L in volts (V) and current I in amps (A):
PF = |cos φ| = 1000 × P(kW) / (3 × VL-L(V) × I(A))

Three phase circuit calculation from real power meter reading P in kilowatts (kW), line to line neutral VL-N in volts (V) and current I in amps (A):
PF = |cos φ| = 1000 × P(kW) / (3 × VL-N(V) × I(A))

Power factor correction

Power factor correction is an adjustment of the electrical circuit in order to change the power factor near 1.
Power factor near 1 will reduce the reactive power in the circuit and most of the power in the circuit will be real power. This will also reduce power lines losses.
The power factor correction is usually done by adding capacitors to the load circuit, when the circuit has inductive components, like an electric motor.

Power factor correction calculation

The apparent power |S| in volt-amps (VA) is equal to the voltage V in volts (V) times the current I in amps (A):
|S(VA)| = V(V) × I(A)
The reactive power Q in volt-amps reactive (VAR) is equal to the square root of the square of the apparent power |S| in volt-ampere (VA) minus the square of the real power P in watts (W) (pythagorean theorem):
Q(VAR) = √(|S(VA)|2 - P(W)2)
The reactive power Q in volt-amps reactive (VAR) is equal to the square of voltage V in volts (V) divided by the reactance Xc:
Q(VAR) = V(V)2 / XC = V(V)2 / (1 / (2πf(Hz)·C(F))) = 2πf(Hz)·C(F)·V(V)2
So the power factor correction capacitor in Farad (F) that should be added to the circuit in parallel is equal to the reactive power Q in volt-amps reactive (VAR) divided by 2π times the frequency f in Hertz (Hz) times the squared voltage V in volts (V):
C(F) = Q(VAR) / (2πf(Hz)·V(V)2)

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