IA: 1P3: Linear Circuits (AC Power) Flashcards
1
Q
What are phasors?
A
Phasors are a way to represent sinusoidal waveforms, like AC voltages and currents. They simplify the math when analyzing AC circuits. This is shown in the diagram below:
1. AC voltage’s and currents can be represented as sinusoids by the equations shown.
2. In phasor notation this can be written as “Veʲᵃ” and “Ieʲᴮ” which represents the RMS value with an associated phase shift
3. Thus it can be written as the RMS value with an associated phase shift “V ∠α” and “I ∠β”
4. The phasor diagram
The magnitude of the phasor is always the RMS value
2
Q
For an AC signal what do these symbols represent?
A
The peak voltage and current. This is the maximum value that the waveform reaches
3
Q
For an AC signal what do these symbols represent?
A
The phasor voltage and current. This encompasses the RMS value and the phase.
4
Q
For an AC signal what do these symbols represent?
A
The RMS voltage and current. This represents the effective DC equivalent.
5
Q
For an AC signal what do these symbols represent?
A
The instantaneous (time domain) voltages and current. This describes how the waveform actually varies over time.
6
Q
What is the magnitude of a phasor?
A
It is always the RMS value
7
Q
What is the equation for the instantaneous power, p(t)?
A
8
Q
What are the 2 components of instantaneous power, p(t)?
A
- An alternating component which varies at twice the supply frequency (frequency of v(t) and i(t)). This has an average value of zero.
- A constant component. This is therefore the average a.c. power consumed by the load. This quantity is denoted by P.
9
Q
In an AC circuit how do the frequencies of the voltage and current signals compare?
A
- The voltage and current signals always have the same frequency
- However, the phase can (and usually does) differ.
10
Q
What does it meant when the instantaneous power is negative?
A
When p(t) is negative, it means v(t) and i(t) are of opposite signs. Therefore the load current is flowing against the flow voltage, i.e. from a point of low potential to a point of higher potential. The load is therefore supplying voltage to the power source. This means that the load is capable of storing energy from the supply and then returning it - this is possible if the load consists in part of inductors of capacitors.
11
Q
What is the equation for the average a.c. power?
A
P = VI cos(α - β) = VI cos(φ)
Average A.C. Power = RMS Voltage x RMS Current x cos(angle of voltage w.r.t. current)
V = RMS Voltage
I = RMS Current
φ = α - β
12
Q
What is the equation for instantaneous power?
A
p(t) = VI{cos(2ωt + α + β) + cos(α + β)}
V = RMS Voltage
I = RMS Current
13
Q
What is the power factor?
A
It is the cosine of the load angle:
cos(φ) where φ = α - β
14
Q
What does the power factor indicate?
A
15
Q
What does a lagging or leading power factor mean?
A
- Lagging power factor (φ > 0): This means that the current lags the voltage
- Leading power factor (φ < 0): This means that the current leads the voltage
16
Q
How is φ measured?
A
It is the angle of the voltage with respect to the current. Therefore it is measured from the current to the voltage
17
Q
What causes a lagging power factor?
A
An inductive load
18
Q
What causes a leading power factor?
A
A capacitive load
19
Q
What is the equation for reactive power?
A
Q = VI sin(φ) where φ = α - β
20
Q
What are the units for reactive power?
A
VAR (Volt-Amps Reactive)
21
Q
How can real and reactive power be thought of in terms of phasor current?
A
The phasor current can be resolved into two components: one which is in phase with the voltage (direct/active component), and the other which is 90° out of phase with the voltage (quadrature/reactive component).
* The real power (average a.c. power) is then given by V times the direct component of current. Therefore we can deduce only the direct component of current contributes to the electrical power consumed. P = VI cosφ = VId
* The reactive power is then given by the voltage times the quadrature component of current. It can be seen by the diagram below that the product of voltage and quadrature current has zero average value. Q = VI sinφ = VIq
22
Q
What is real power?
A
Real power is the actual power consumed by an electrical device to perform useful work. It represents the average power over time in an AC circuit
23
Q
What is reactive power?
A
Reactive power is the power that oscillates between the source and the reactive components (like inductors and capacitors) in an AC system but is never consumed. It does not perform any real work but is essential to sustain the electric and magnetic fields in the system.
24
Q
Describe and explain the a.c. power consumption by a resistor:
A
Resistors only consume real power. They do not consume any reactive power. According to Ohm’s law, I = V/R. Since R is purely real, the a.c. current flowing through the resistor will be in phase with the voltage across it. Therefore φ = 0.
* This means that the power factor is unity (cosφ = 1), meaning P = VI. This can be seen on the diagram where the average value of p(t) is non-zero and positive.
* However, sinφ = 0 and so the reactive power is zero (Q = 0). This can be seen on the diagram where p(t) is never negative.
25
Describe and explain the a.c. power consumption by an inductor:
**Inductors only *consume* reactive power. The average a.c. power (real power) supplied to an inductor is zero.** The impedance of an inductor is jωL = jXₗ. Therefore according to Ohm's law the current which flows is given by I = V / jXₗ = -jV / Xₗ. The -j operator rotates phasors by -90°, showing that the current lags the voltage by 90°. Therefore φ = 90°
* This means that the power factor is zero (cosφ = 0), meaning P = 0. This can be seen on the diagram where the average value of p(t) is zero.
* However, sinφ is unity, meaning the reactive power is given by Q = VI = I²Xₗ = V²/Xₗ.
26
Describe and explain the a.c. power consumption in capacitors:
**Capacitors only *generate* reactive power. The average a.c. power (real power) supplied to a capacitor is zero.** The impedance of a capacitor is 1/jωC = -jX𝒸. Therefore according to Ohm's law the current which flows is given by I = V / -jX𝒸 = jV / X𝒸. The +j operator rotates phasors by +90°, showing that the current leads the voltage by 90°. Therefore φ = -90°
* This means that the power factor is zero (cosφ = 0), meaning P = 0. This can be seen on the diagram where the average value of p(t) is zero.
* However, sinφ is -1, meaning the reactive power is given by Q = -VI = -I²X𝒸 = -V²/X𝒸 (reactive power is negative!).
27
What does it mean if the reactive power is positive or negative?
* Q > 0: The load is inductive
* Q < 0: The load is capacative
28
What are the 2 methods for determining the real and reactive power consumed by a circuit?
**Method 1:**
* Determine the impedance of the circuit and convert it to polar form
* Determine the current (in phasor form)
* Use the equations for real and reactive power (P = VI cosφ and Q = VI sinφ)
**Method 2:**
* Determine the total impedance of the circuit and convert it to polar/exponential form
* Determine the current (in phasor form)
* Determine the real power by using the fact that only a resistor consumes real power (for a series circuit use P = I²R and Q = ±I²X, for a parallel circuit use P = V²/R and Q = ±V²/X)
29
What is the polar form of impedance?
It is the magnitude-angle form
30
Why is it best to use the phasor forms of current and voltage and polar form of impedance?
When multiplying or dividing you can simply multiply/divide the magnitudes and add/subtract the angles.
31
Why must you be careful with φ?
## Footnote
Where (cosφ = power factor)
**It is always the angle of the voltage w.r.t. to the current. Therefore you must be careful which way round you have calculated (this can change whether it is positive or negative)**. For example, in the below example, φ is +53.1° NOT -53.1°. As we have calculated the current, the angle we have calculated is the current w.r.t. to the voltage, whereas φ is the voltage w.r.t. to the current!!
32
What is complex volt-amps/power?
## Footnote
Where I* is the complex conjugate of I
33
What is apparent power, S?
It is the product voltage and current without any consideration of the angle between them. It is the magnitude of the complex power/volt-amps.
34
What are the units of apparent power, S?
volt-amps (VA)
35
What is the equation for apparent power?
36
What is the power triangle?
37
How is power conserved in a circuit?
* Real power (P) is conserved: P supplied = P consumed
* Reactive power (Q) is conserved: Q supplied = Q consumed
* Apparent power (S) is **usually** not conserved: S supplied ≠ S consumed
In a parallel circuit, the total supplied P and Q equal the sum of the individual components.
Therefore, in the below circuit it can be said that Pᵢₙ = P₁ + P₂, Qᵢₙ = Q₁ + Q₂, but it cannot be said that Sᵢₙ = S₁ + S₂
38
39
How does the power factor effect power losses?
Consider two devices that consume the same amount of real power from the same voltage supply (i.e., same P, same V). One device has a high power factor, while the other has a low power factor. Since real power is given by P = V × I × cos(θ), a lower power factor (cosθ) means the device must draw more current to deliver the same amount of real power. This higher current leads to greater power losses in the transmission lines, because those losses are proportional to I². Therefore, a low power factor results in significantly higher power losses compared to a high power factor.
40
How does the reactive power lead to power losses?
Larger Q leads indirectly to larger real power losses.
41
What is power factor correction?
Power factor correction is the process of improving the power factor of an electrical system (bringing it as close to 1 as possible). This is done by reducing the reactive power that the system draws, which in turn lowers the total current required for the same real power. By connecting a capacitor/resistor in parallel, the quadrature components of input current can cancel each other out, leaving only the direct component and so the input current is now in phase with the voltage - giving a unity power factor.
42
How is power factor correction done?
* **Inductive loads:** Inductive loads cause a lagging power factor, therefore we must correct this by adding leading reactive power. To do this connect a capacitor in parallel with the load such that the combined capacitor and load reactive power is zero.
* **Capacative loads:** Capacative loads cause a leading power factor, therefore we must correct this by adding lagging reactive power. To do this connect an inductor in parallel with the load such that the combined inductor and load reactive power is zero.
43
Determine the % reduction in real power losses in the transmission line once the circuit has been power factor corrected by adding in a suitable capacitor (and determine the value of this capacitor):
19.9%
44
What's the basic operation of a transformer?
Transformers are used to convert electrical power from one voltage and current level to another via electromagnetic induction (Faraday's law) between two coils. They are composed of:
* Primary coil: Connected to the input voltage and generates a magnetic flux
* Secondary coil: Connected to the output/load
* Laminated iron core: Guides the magnetic flux
45
How is the flow of induced currents in the core of a transformer minimised?
* Laminated
* Made of many thin sheets compressed together
46
What is magnetic reluctance?
A property of a material or a magnetic circuit that opposes the creation of magnetic flux - analagous to resistance in an electrical circuit
47
What is the magnetic circuit analogy?
## Footnote
Use the transformer as an example
A magnetic system can be thought of similarly to an electrical system:
* Flux ↔ Current
* Ampere-turns (NI) (m.m.f) ↔ Voltage
* Reluctance ↔ Resistance
## Footnote
The two sources of magnetomotice force represent the two sources of flux (coils) and the single reluctance represents the core itself
48
Why in this magnetic circuit for a transformer are the two magnetomotive forces (NI) in opposite directions?
Due to Lenz's law - induced currents always flow in a manner which opposes the flux which has induced them
49
What is the equation for magnetic flux in a magnetic circuit?
**N₁i₁(t) = N₂i₂(t) + Rφ(t)**
ΣMMF = RΦ
## Footnote
This is known as Hopkinson's law and is analagous to Ohm's law
50
What are the 3 assumptions for an **ideal** transformer?
* Iron is a perfect conductor of flux: Reluctance = 0
* All the flux linking the primary coil also links the secondary coil
* Winding resistances are negligible
51
What is the significance of the ideal transformer assumption that iron is a perfect conductor of flux?
By assuming that the iron core is a perfect conductor of magnetic flux we are assuming that the reluctance of the iron core is 0. Therefore the equation "N₁i₁(t) = N₂i₂(t) + Rφ(t)" can simply become:
**N₁i₁(t) = N₂i₂(t)**
The primary and secondary m.m.f.s are exactly the same at all instants in time, therefore:
* The **ratio of I₁ to I₂ is equal to the ratio of the primary and secondary turns** (N₁ to N₂) - this is the turns ratio.
* **The primary and secondary currents are in phase**
52
What is the significance of the ideal transformer assumption that all the flux linking the primary coil also links the secondary coil?
Because we have assumed the reluctance of the iron core is zero, no magnetic flux will leave the transformer core. Therefore in the equations given by faraday's law dφ/dt is the same for both coils and so it can be seen that the winding e.m.f.s are in the ratio of the turns ratio. Therefore:
**e₁/N₁ = e₂/N₂**
53
What is the significance of the ideal transformer assumption that the winding resistances are negligible?
If we assume that the transformer windings have no resistance, then the e.m.f.s induced in them must be equal to the terminal voltage across them. Therefore:
* Primary voltage : Secondary voltage = turns ratio
* The primary and secondary voltages are in phase
* Hence, V₁I₁ = V₂I₂ (where these are the phasor voltages and currents). This means that ideal transformers only converts electrical power from one voltage and current level to another - it does not consume any power.
54
What power is conserved in an ideal transformer?
Both real and reactive power are conserved in an ideal transformer. Therefore complex volt-amps are conserved. An ideal transformer does not consume any real or reactive power.
55
What is the ideal transformer equation?
56
What is impedance referral?
When working with a transformer it is often easier to work entirely on one side of the transformer, therefore if a load Z is connected one side of an ideal transformer, this impedance can be "referred" to the otherside of the transformer. Once you have referred the impedance to the otherside you can remove the original impedance.
57
What are the impedance referral equations?
58
In a real transformer the reluctance of the iron core is NOT zero (R ≠ 0), how is this accounted for in the equivalent circuit?
**Place a magnetising reactance of jX₀ (an inductive impedance) in parallel with the primary winding.**
Consider an unloaded transformer (no load on the output), just enough current has to flow in the primary winding to provide the m.m.f (NI) required to produce the flux to oppose the input voltage, this flux is known as the "magnetising flux" (φₘ) and the current that produces it the "magnetising current" (Iₘ). N₁Iₘ = Rφₘ.
The back-e.m.f. induced by the magnetising flux in the primary winding is given by Faraday's law and comes out to be E₁ = jX₀Iₘ.
To model this behavior in the equivalent circuit, we place the magnetizing reactance jX₀ in parallel with the primary winding. This reactance represents the inductive impedance that the magnetizing current flows through.
Since there’s no current in the secondary coil (because it is an open circuit), there’s no current flowing through the primary coil. Therefore the current flowing through this parallel impedance jX₀ is the magnetising current, Iₘ.
59
What is back e.m.f?
Back EMF in a transformer is the voltage induced in the primary winding due to the changing magnetic flux in the core, which is created by the primary current itself
60
Explain how there are 2 distinct effect causing power losses in the iron core of a real transformer:
1. Hysteresis: Cyclical reversal of the magnetic domains in the iron core as the magnetic field reverses direction leads the a hysteresis loop. The area within the hysteresis loop is the energy lost per cycle of magnetisation per unit volume.
2. Eddy current loss: The time-varying magnetic flux in the transformer core induces e.m.f.s in them and hence a current to flow. Since the core has resistance, power loss occurs.
**Both of these losses are proportional to φₘ² and hence E₁²**
61
In a real transformer there is power loss in the iron core due to hysteresis and eddy currents, how is this accounted for in the equivalent circuit?
Since both of these losses are proportional to φₘ² and hence E₁², this can be modelled by placing a resistor (R₀) in parallel with the primary winding of the ideal transformer - in parallel with X₀. This resistor is referred to as the iron loss resistance. The parallel branch containing X₀ and R₀ is known as the magnetising branchof the transformer.
62
In a real transformer windings do have resistance, how is this accounted for in the equivalent circuit?
Windings do have resistance R₁ and R₂. They result in power losses IᵢₙR₁² and I₂R₂² respectively. These can be modelled as resistors placed in series with the respective windings.
63
In a real transformer not all the primary flux links the secondary coil (resulting in leakage flux), how is this accounted for in the equivalent circuit?
**Connect inductive reactances X₁ and X₂ in series with the primary and secondary coils respectively. **
In magnetic circuits, unless the reluctance of the core is zero there will always be some flux which takes a path through the air rather than through the core (leakage flux).
The primary leakage flux will induce a voltage in the primary winding, since it links it, but not in the second. This voltage is given by Faraday's law and since the leakage flux is proportional to the primary m.m.f (and hence the primary current) comes out to be equal to "jX₁Iᵢₙ". Where X₁ is a reactance known as the primary leakage reactance. Therefore you can connect an inductive reactance (X₁) in series with the input current. By a similar argument the secondary leakage reactance X₂ can be placed in series with the secondary coil.
64
How can the complete equivalent circuit for a transformer be simplified?
* Because the magnetising branch impedance is far greater than the impedance of the series components, the primary winding back e.m.f. (E₁) and terminal voltage (V₁) are roughly the same. Therefore, R₀ and X₀ can be moved to the input terminals with little loss of accuracy
* R₂ and X₂ can be referred from the secondary winding to the primary winding and then combined with R₁ and X₁ to give Rₜ₁ and Xₜ₁
* Finally, if the secondary coil of the transformer is connected to a load, the load can be referred to the primary
65
What are transformers limited by?
* **Winding currents:** The primary and secondary windings dissipate electrical power due to their resistive losses, I²R. This power is converted to heat, which in turn causes the temperature of the transformer to rise. This therefore places a limit on the maximum transformer current.
* **Winding voltages:** The iron core similarly gets hot due to iron losses. This is roughtly proportional to the input voltage squared, and so this places a limit on the transformer voltage.
66
What is the volt-amp rating of a transformer?
Transformers are limited by their winding currents and voltages. These limitations combine to give a volt-amp rating rather than a power rating because as far as the transformer is concerned these heating effects depend only on the current and the voltage, not the angle between them. The volt amp rating is given by:
67
How can you determine the value of the parameters R₀, X₀, Rₜ₁, and Xₜ₁ (and the turns ratio N₁:N₂)?
You can perform 2 simple tests, the open-circuit test and the short-circuit test:
* The open-circuit test determines the turns ratio, R₀, and X₀ (turns ratio, iron loss resistance, magnetising reactance)
* The short-circuit test determines Rₜ₁ and Xₜ₁ (Combined winding resistance, combined leakage reactance)
68
How is the open circuit-test in transformers used?
In the open circuit test, the primary winding is excited at its rated voltage. The secondary winding is open circuited, this means that no current can flow in the secondary, and so I₁ is also zero.
* **Turns ratio:** There is no voltage drop across the series parameters Rₜ₁ and Xₜ₁ and so the primary terminal voltage and back-e.m.f are identical, as are secondary voltage and back e.m.f. Since the ratio E₁:E₂ gives the turns ratio, we see that under-open circuit conditions, this is the same as V₁:V₂. Therefore to find the turns ratio of the transformer, the primary and secondary voltages are measured under open-circuit conditions and the ratio found. **N₁/N₂ = V₁/V₂**
* **Iron loss resistance:** Since I₁ is zero, everything to the right of the magnetising branch can be removed. If the input current and input real power are measure, then the input apparent power and input reactive power can be determined. Since resistors only consume real power, R₀ can be given by the equation: **R₀ = V₁²/P**
* **Magnetising reactance:** Similarly, reactive power is only consumed by the magnetising reactance, X₀. Therefore X₀ can be given by the equation: **X₀ = V₁²/Q**
69
What is the equation for the turns ratio?
Measured in an open circuit test:
* V₁ = Input terminal voltage
* V₂ = Output terminal voltage
70
What is the equation for the iron loss resistance, R₀?
Measured in an open circuit test:
* V₁ = Input terminal voltage
* P = Input real voltage
71
What is the equation for magnetising reactance, X₀?
Measured in an open circuit test:
* V₁ = Input terminal voltage
Calculated from other measurements:
* Q = Input reactive voltage
72
How is the short circuit-test in transformers used?
In the short circuit test the secondary is short circuited so that V₂ = 0. Since V₂ is zero, E₂ is also zero and therefore E₁ is zero. The magnetising branch impedance is far larger than the total series impedance, and so it may be ignored as the current flowing through the series branch will be far higher. This results in a series circuit with just Rₜ₁ and Xₜ₁ with I₁ flowing through them and V₁ across the input terminal.
* **Rₜ₁ (Combined winding resistance):** If input voltage, current, and real power are measured, we can determine Rₜ₁ as it will only consume real power. Therefore it can be given by the equation: **Rₜ₁ = P/I₁²**
* **Xₜ₁ (Combined leakage reactance):** As input voltage, current, and real power are measured, the input apparent power can be determined and hence the input reactive power. We can determine Xₜ₁ as it will only consume reactive power. Therefore is canbe given by the equation:**Xₜ₁ = Q/I₁²**
73
What is the equation for Rₜ₁ (Combined winding resistance)?
**Rₜ₁ = P/I₁²**
Measured in a short circuit test:
* P = Input real power
* I₁ = Input current
74
What is the equation for Xₜ₁ (Combined leakage reactance)?
**Xₜ₁ = Q/I₁²**
Measured in a short circuit test:
* I₁ = Input current
Calculated from other measurements:
* Q = Input reactive power
75
What are 3 uses of transformers?
* **The change a.c. voltage and current levels**
* **To provide circuit isolation** - primary and secondary windings are connected only by flux. This can help protect electronic componetns which are sensitive to high voltage. The energy is transferred with no physical connection.
* **Intrumentation** - Measurement of high voltages and currents. Measuring large voltages or currents can be very dangerous, therefore the current OR voltage may be stepped down to a safe level for measurement
76
What is the equation for the efficiency of a transformer?
77
What influences the construction of a transformer?
The operating frequency, losses in the iron core increase significantly with frequency. This is because hysteresis losses increase with respect to the frequency and eddy current losses increase with respect to the frequency squared.
78
What are iron/fixed losses?
These are losses occuring in the iron core - due to hysteresis and eddy currents. They are "fixed" as the input voltage is usually fixed. It is given by the equation:
79
What are copper/variable losses?
Copper losses are due to resistive heating of the windings. They are "variable" as the losses depend on the current squared and the current can vary with the load. It is given by the equation:
80
What is regulation?
Regulation measures the change in load voltage from open-circuit value due to the load current. Ideally a transformer maintains a constant secondary terminal voltage irrespective of the load current taken from it, therefore a small regulation is better than a large regulation as this indicates that the load voltage is independent of the load current. Regulation is given by:
## Footnote
Ideal regulation = 0%
81
What is the regulation in this circuit?
3.43%
