Electrical Eng. fundamental Lecture 1 <Dr Hadi El-Helw> Contact details: h-elhelw@staffs.ac.uk
Introduction Electrical systems pervade our lives; they are found in home, school, workplaces, factories, and transportation vehicles-everywhere. A circuit model is used to connect our visualization to our analysis of a physical system. The challenge is to develop models that will predict the physical behaviour of real components accurately and result in mathematical equations that can be solved.
Basic Electrical Quantities Basic quantities: current, voltage and power. Electric current: Electric current in a wire is defined as the net amount of charge that passes through the wire per unit time, and is measured in amperes (A). where i dq dt i = current in amperes q = charge in coulombs t = time in sec. 1 Ampere = 1 Coulomb per second (C/s) Current in circuits physically realized by movement of electrons. Direction of current must be specified by an arrow.
By convention, current direction defined as flow of positive charge. Note that positive charge is not flowing physically. Electrons have negative charge. They move in the opposite direction of current. - - - - - - - - - - - - - - - - - - - - - - - - electron motion positive current direction In general, current can be an arbitrary function of time. Constant current is called direct current (DC). Current that can be represented as a sinusoidal function of time (or in some contexts a sum of sinusoids) is called alternating current (AC).
Voltage Voltage is the energy absorbed or expended as a unit charge moves from one point to the other. Analogous to pressure in hydraulic system. Sometimes called potential difference. Can be created by a separation of charge. Is a measure of the potential between two points. Voltage pushes charge in one direction.
We use polarity (+ and on batteries) to indicates which direction the charge is being pushed Voltage is the energy required to move a unit charge through an element, measured in volts (V) v d dq where v = voltage in volts ω = energy in Joules q = charge in coulombs i + v - A Circuit Element B
Voltage ~ Pressure Electric Current ~ Water Current Sponge ~ Resistance
Electrical Power Time rate of expending or absorbing energy and is measured by Watts. p p d dt d dq dq dt vi where p = power in watts ω = energy in Joules t = time in seconds q = charge in coulombs i = current in apperes v = voltage in volts By convention Circuit elements that absorb power have a positive value of p. Circuit elements that produce power have a negative value of p.
Active elements Elements of electrical circuits Active elements are the elements that can generate energy or power, such as voltage and current sources. Ideally, a voltage source produces V s volts regardless of the current absorbed or produced by the connected device. Ideally, a current source produces I s amps regardless of the current in the connected device. In a particular circuit, there can be active elements that absorb power for example, a battery being charged.
Passive elements passive elements are the elements that can not generate energy, such as resistors, capacitors and inductors. resistors The ability of a material to resist (impede, obstruct) the flow charge is called its resistivity. It is represented by the letter R. A resistor is a circuit element that dissipates electrical energy (usually as heat) Real-world devices that are modeled by resistors: incandescent light bulbs, heating elements, long wires Resistance is measured in Ohms (Ω) Resistor is indicated by the symbol
Resistance of a wire depends on some factors like as length (L), crosssectional area (A) and resistivity of material (ρ). R L A Where ρ resistivity in Ω.m L length in m A cross-section area in m 2 The conductance (G) of a pure resistor is the reciprocal of its resistance. The unit of conductance is the siemens (S) or mho ( ). G 1 R Ω
Ohm s Law Ohm's law states that the current through a conductor between two points is directly proportional to the potential difference or voltage across the two points, and inversely proportional to the resistance between them. The mathematical equation that describes this relationship is: i v R where v is the potential difference measured across the resistance in units of volts; i is the current through the resistance in units of amperes and R is the resistance of the conductor in units of ohms.
Resistors in Series Two elements are in series if the current that flows through one must also flow through the other. R 1 R 2 Series If we wish to replace the two series resistors with a single equivalent resistor whose voltage-current relationship is the same, the equivalent resistor has a value given by R eq R 1 R 2
For N resistors in series, the equivalent resistor has a value given by: R 1 R 2 R eq R 3 R R R R eq 1 2 3 R N
Consider two resistors in series with a voltage v across them: v 1 v 2 2 1 1 1 R R R v v 2 1 2 2 R R R v v R 1 R 2 - + + - + - v i Voltage division:
Resistors in Parallel When the terminals of two or more circuit elements are connected to the same two nodes, the circuit elements are said to be in parallel. If we wish to replace the two parallel resistors with a single equivalent resistor whose voltage-current relationship is the same, the equivalent resistor has a value given by R 1 R 2 R eq R eq R1 R2 R R 1 2
For N resistors in parallel, the equivalent resistor has a value given by R R 2 R N R eq 1 1 R 1 R 1 R 1 R 1 eq 1 2 3 R N
Consider two resistors in parallel with a voltage v across them: i Current division: + v - i 1 i 2 R 1 R 2 i 1 i 2 i i R 1 R R 1 2 R R 1 2 R 2
Electrical Eng. Fundamental Lecture 2 <Dr Hadi El-Helw> Contact details: h-elhelw@staffs.ac.uk
KCL and KVL Kirchhoff s Current Law (KCL) and Kirchhoff s Voltage Law (KVL) are the fundamental laws of circuit analysis. Gustav Kirchoff was an 18th century German mathematician
KCL The sum of all currents entering a node is zero, or The sum of currents entering a node is equal to sum of currents leaving a node. i 1 i 5 i 2 i 4 n j1 i j i 3 0 i i i i i 0 or i1 i2 i4 i3 i5 1 2 3 4 5 A node is a point where two or more circuit elements are connected together.
i 1 node i 2 i 3 i 1 flows into the node i 2 flows out of the node i 3 flows out of the node i 1 = i 2 + i 3 This equation can also be written in the following form: i 1 - i 2 - i 3 =0
Example How much are the currents i 1 and i 2? i 1 10 ma 4 ma 3 ma i 2 node + _ 4 ma + 3 ma + 7 ma = 14 ma i 2 = 10 ma 3 ma = 7 ma i 1 = 10 ma + 4 ma = 14 ma
KVL Kirchhoff s voltage law tells us how to handle voltages in an electric circuit. Kirchhoff s voltage law basically states that the algebraic sum of the voltages around any closed path (electric circuit) equal zero. n j1 v j Arrows are sometimes used to represent voltage differences; they point from low to high voltage. i i + v - 0 v
i + v 2 + v 1 +_ + v 3 + v 4 v 3 = v 4 v 1 + v 2 + v 3 = 0 v 1 + v 2 + v 4 = 0 or or v 1 = v 2 + v 3 v 1 = v 2 + v 4
Example If v 1 = 10 V and v 5 = 2 V, what are v 2, v 3, and v 4? + v 3 + v 1 = 10 V + _ + v 2 + v 4 + v 5 = 2 V v 2 = 10 V v 3 = 10 V 2 V = 8 V v 4 = 2 V
Example: The voltage v 1 and v 2 in the circuit shown are 512.41V and 330.05V respectively. Find the power supplied by each voltage source. Show that the total power supplied equals the total power dissipated in the resistors
Example: Find v 0 in the circuit shown.
Example: The current in the 12Ω resistor in the circuit shown is 1 A, find v g.
Electrical Eng. Fundamental Lecture 3 <Dr Hadi El-Helw> Contact details: h-elhelw@staffs.ac.uk
0 6 2 1 I i i i 0 5 3 1 i i i 0 2 4 3 i i i 0 ) ( 1 3 2 3 2 5 1 1 v R R i i R R i 0 ) ( 2 4 5 6 4 3 2 3 v R i R i R R i 0 6 4 7 6 5 2 R i R i R i Node b Node c Node e Applying KCL at nodes b, c, and g Applying KVL
Example Find i 1, i 2, and i 3 Solution Applying KCL at node a i i ------------------------ 1 1 2 i3 Applying KVL in the red loop 3 3 3.6 10 i 2.2 10 i 10 20 3.6 10 3 i 1 1 2.2 10 Applying KVL in the blue loop 10 3.310 3.310 3 i 2 3 i 2 3 2.2 10 i 3 3 20 3 i 30 ------------------------ 2 2.2 10 3 3 i 3 30 ------------------------ 3
Substituting from equation 1 in equation 2 3.610 3.610 3 3 ( i i 2 2 i 3 ) 2.210 5.810 3 i 3 3 i 3 30 30 ------------------------ 4 Solving equation 3 and 4 simultaneously i i 2 3 3.99mA 7.65mA Substituting in equation 1 i1 3. 66mA
The Node voltage method The Mesh current method Source transformation Superposition theorem
The Node voltage method 1-Determine the number of nodes within the network 2-Select a node as the reference node. Assign voltages v 1, v 2,,,, v n-1, to the remaining n 1 nodes. The voltages are referenced w.r.t. the reference node.
3- Apply KCL at each node except the reference, assume that all unknown currents leave the node for each application of KCL 4-Use Ohm s law to express the branch currents in terms of node voltage. 0 0 3 2 1 2 1 1 1 1 3 2 1 R v v R v R E v i i i i i o i
ii io 0 i4 i5 I v2v1 v2 R R 3 4 I 0 4- Solve the resulting simultaneous equations to obtain the unknown node voltages.
Example: Use the node-voltage method to find v o in the circuit shown
Example: Use the node-voltage method to find v 1 and v 2 in the circuit shown
Electrical Eng. Fundamental Lecture 4 <Dr Hadi El-Helw> Contact details: h-elhelw@staffs.ac.uk
Example: Use the node-voltage method to find the branch currents i a through i e in the circuit shown below.
Example: Use the node-voltage method to find the branch currents i 1 through i 3 in the circuit shown below.
The super node when a voltage source is connected between two non-reference nodes. Problem: The current through the voltage source cannot be written as function of its two terminal voltages!
Solution: Form a supernode which is formed by enclosing the voltage source and any elements in parallel with it in a closed boundary. Since there are two nodes (two terminals of the voltage source) are enclosed in the supernode, two equations are needed for each supernode: KCL at supernode gives one equation( Write the standard node equations for the supernode) v1 2 2 v 2 4 7 0
The other equation is the relationship between the voltages of the two nodes enclosed in the supernode v 2 v1 2
Example: Use the node-voltage method to find v and i in the circuit shown below.
Electrical Eng. Fundamental Lecture 5 <Dr Hadi El-Helw> Contact details: h-elhelw@staffs.ac.uk
Mesh Analysis 1-Assign a distinct current in the clockwise direction to each independent closed loop (mesh) of the circuit. It is not absolutely necessary to choose the clockwise direction for each loop current. However, it eliminates the need to have to choose a direction for each application. Any direction can be chosen for each loop current with no loss in accuracy as long as the remaining steps are followed properly. Note:A mesh is a loop which does not contain any other loops within it. 2- Indicate the polarities within each loop for each resistance as determined by the assumed direction of loop current for that loop.
3- Apply KVL around each closed loop in the clockwise direction. Again, clockwise direction was chosen to establish uniformity and prepare us for the format approach to follow: a- If an resistance has two or more assumed currents through it, the total current through the resistance is the assumed current of the loop in which KVL is being applied, plus the assumed currents of the other loops passing through in the same direction, minus the assumed currents passing through in the opposite direction b-the polarity of a voltage source is unaffected by the direction of the assigned loop currents. 4- Solve the resulting simultaneous linear equations for the assumed loop currents
Mesh 1 E ( ) 1 i1r1 R2 i1 i2 0 Mesh 2 R2 ( i2 i1 ) i2r3 E2 0
Example Use the mesh-current method to find the branch currents i a, i b, i c in the circuit shown in the figure below
Example Use the mesh-current method to find how much power the 4-A current source delivers to the circuit shown in the figure below
Supermesh: when a current source exists between two meshes Procedure: 1. Open the current source. That is form a single mesh for the two mesh sharing the current source.
2. Write the constraint equation for the currents (Apply KCL). i 2 i1 6 3. Write the standard mesh equation for the supermesh (Apply KVL). i1 6 i2 (10 4) 20
Example Use the mesh-current method to find the current i 1 in the circuit shown in the figure below
Electrical Eng. Fundamental Lecture 6 <Dr Hadi El-Helw> Contact details: h-elhelw@staffs.ac.uk
Source Transformation Circuit 1 Circuit 2 A source transformation allows us to replace a voltage source in series with a resistance by a current source in parallel with the same resistance, or vice versa
Circuit 1 Circuit 2 Connect the same load resistor R L across terminal a-b in both circuits. If circuits 1 and 2 are equivalent i 1 = i 2 & v 1 = v 2 L s s eq s R R V R V i 1 L P p s R R R I i 2 (use the current divider role) L p p s L s s R R R I R R V i i 2 1 If we choose R p =R s V s =I s R p =I s R s
Therefore V s I s R s R s R p
Reduction of sources
Example: Use concept of source transformation to find the voltage V m in the circuit shown in the figure below
Solution
8 V m ( 2 ) 3 2V 3
Example: Use a series of source transformations to find the current i o in the circuit shown.
Example: Use a series of source transformations to find the current i o in the circuit shown.
Electrical Eng. Fundamental Lecture 7 <Dr Hadi El-Helw> Contact details: h-elhelw@staffs.ac.uk
Superposition Theory Whenever a linear system is excited, or driven, by more than one independent source of energy, the total response is the sum of the individual responses. An individual response is the result of an independent source acting alone. Because we are dealing with circuits made up of interconnected linearcircuit elements, we can apply the principle of superposition directly to the analysis of such circuits when they are driven by more than one independent energy source
The superposition principle states that the voltage across (or current through) an element in a linear circuits is the algebraic sum of the voltage across (or current through) that element due to each independent source acting alone. Step to apply: 1- Turn off all independent sources except one source. Find the required (voltage or current) due to that active source. Current Source open circuit(0 A) Voltage Source short circuit (0 V) 2- Repeat step 1 for each other independent sources. 3- Find the total contribution by adding algebraically all the contribution due to the independent source.
Consider only the voltage source E1
Consider only the current source I i i i i 1 2 3 4 i i 11 i i 21 31 41 i 12 i i i 22 32 42
Example: Use the principle of superposition to find v o in the circuit shown.
Example: Use the principle of superposition to find the current i o in the circuit shown.
Electrical Eng. Fundamental Lecture 8 <Dr Hadi El-Helw> Contact details: h-elhelw@staffs.ac.uk
Magnetic fields In the region surrounding a permanent magnet there exists a magnetic field, which can be represented by magnetic flux lines. Magnetic lines exist in continuous loops, as shown in fig. The symbol for magnetic flux is the Greek letter Φ (phi).
If a nonmagnetic material, such as glass or copper, is placed in the flux paths surrounding a permanent magnet, there will be an almost unnoticeable change in the flux distribution. However if a magnetic material, such as soft iron, is placed in the flux path, the flux lines will pass through the soft iron rather than the surrounding air because flux line pass with greater ease through magnetic materials than through air. as shown in fig.
A magnetic filed (represented by concentric magnetic flux line, as shown in fig.) is present around every wire that carries an electric current. The direction of magnetic flux line can be found simply by placing the thumb of the right hand in the direction of the conventional current flow and noting the direction of the fingers.(this method is commonly called the right-hand rule).
If the conductor is wound in a single-turn coil, the resulting flux will flow in a common direction through the center of the coil. A coil of more than one turn would produce a magnetic field that would exist in a continuous path through and around the coil which is quite similar to that of the permanent magnet.
The strength of the magnetic filed is determined by the density of the flux line. The filed strength of the coil can be effectively increased by placing certain materials, such as iron or steel, within the coil to increase the flux density within the coil or by increase the current in the conductor.
There are many application of the electro magnetic effect such as generator, transformer, Relay..
Magnetic filed intensity, H Assume N turns of an electric coil are wound around an iron core. If an electric current I passes through the coil, then a magnetic field is established in the core. A streamline in a magnetic field is a line so drawn that its direction is everywhere parallel to the direction of the magnetic field. The strength of the magnetic field or (The magnetic field intensity) H is given by H N. I l The magnetic field strength H is measured in ampereturns/meter of main streamline length, At/m.
Flux density In the SI system of units, magnetic flux is measured in webers and has the symbol Φ. The number of flux lines per unit area is called the flux density, is denoted by the capital letter B, and is measured in teslas. Its magnitude is determined by the following equation. B where A B=tesla (T) Φ=webers(Wb) A=square meters (m 2 )
Permeability It is easier to establish or set up the magnetic flux lines in some materials (e.g. iron) than it is in other materials (e.g. air). The magnetic lines of force, like electric current, always try to follow the path of least resistance. Permeability is the property of materials that measures its ability to permit the establishment of magnetic lines of force. It is analogous to conductivity in electric circuits. Air (or vacuum) is taken as the reference material. Its permeability is called μ o. The permeability μ of any other material is: μ =μ r μ o where μ r is called the relative permeability; it is a dimensionless quantity. Non-magnetic materials (e.g. air, glass, copper, and aluminum) are characterized by their μ r, which is approximately unity.
The permeability is the ratio between the magnetic flux density, B, and the field intensity, H. B= H μ =H (μ o μ r ) The permeability of a material is determined from its magnetization characteristic or (B-H) curve. The magnetization characteristic is obtained experimentally.
Material that have lower permeability than air (μ r is a fraction) are called diamagnetic materials. Those material that have slightly higher permeability than air (i.e. μ r 1 to 10) are called paramagnetic materials. On the other hand, magnetic material such as iron, steel, nickel, and alloys of such materials are called ferromagnetic materials, being characterized by their high value of μ r (from 100, to 100000). Note μ o =4π x 10-7 Wb/A.m
Reluctance The resistance of a material to the flow of charge (current) is determined by the following equation: L R (ohms,ω) A The reluctance of a material to the setting up of magnetic flux lines in the material is determined by the following equation. l A (At/Wb) Where is the reluctance, l is the length of the magnetic path, and A is its cross- section area. The t in the unit At/Wb is the number of turns of the applied winding.
Ohm s Law for magnetic circuits Effect cause opposition For magnetic circuits, the effect desired is the flux Φ. The cause is the magnetomotive (mmf), which is the external force required to set up the magnetic flux lines within the magnetic material. The opposition to the setting up of the flux Φ is the reluctance. mmf
The magnetomotive force mmf is proportional to the product of the number of turns around the core ( in which the flux is to be established) and the current through the turns of wire. mmf NI (ampere-turns, At) This equation clearly indicates that an increase in the number of turns or the current through the wire will result in an increased on the system to establish flux line through the core.
l c A c mmf c
Ampere's law for magnetic circuits Ampere's law states that the mmf in a magnetic circuit is equal to the electric current enclosed by the magnetic circuit. The mmf creates a magnetic field in core having an intensity of H ampereturns / meter along the length of the magnetic path. Upon integrating the magnetic field intensity along the magnetic path, we get, H dl = N.i If the path of integration is the mean path length of the core lc, Ampere s law becomes; H.l c = N.i
Ampere's lawfor magnetic circuits Ampere's law states that the mmf in a magnetic circuit is equal to the electric current enclosed by the magnetic circuit. The mmf creates a magnetic field in core having an intensity of H ampereturns / meter along the length of the magnetic path. Upon integrating the magnetic field intensity along the magnetic path, we get, H dl = N.i If the path of integration is the mean path length of the core lc, Ampere s law becomes; H.l c = N.i
Electrical Eng. Fundamental Lecture 9 <Dr Hadi El-Helw> Contact details: h-elhelw@staffs.ac.uk
Analysis of magnetic circuits mmf H L NI B A B H o r H
Electric circuit Magnetic circuit Driving force (cause) emf mmf Response (effect) current flux opposition resistance reluctance Equivalent circuit I emf R
Example 1: Find the current necessary to establish a flux φ=3 10-4 Wb in the series magnetic circuit shown in figure L steel core =0.3 m L iron core = 0.3m Area (throughout)= 5 10-4 m 2 N= 100 turns
Solution Φ =3 10-4 Wb, L steel core =0.3 m, L iron core = 0.3m, A = 5 10-4 m 2, N = 100 turns, I =?. NI=H 1 L 1 + H 2 L 2 100 I = H 1 L 1 + H 2 L 2 B= Φ/A = 3 10-4 / 5 10-4 = 0.6 tesla From chart H steel = 340 AT/m H iron = 2500 AT/m I= 8.52 A
Example 2: For the series parallel magnetic circuit shown in the figure, find the value of current I to establish a flux in the air gap φ=2 10-4 Wb. L ab = L bc = L fe = L ed =0.2 m L af = L be = L ed =0.1m N= 200 turns
Solution Φ g =2 10-4 Wb, I =?. B g = B 2 = Φ g / A = 2 10-4 / 5 10-4 = 0.4 tesla H 3 L 3 = H g L g + H 2 L 2 From chart H 2 =245 AT/m Hg= Bg/μ = Bg/(μo μr ) = 0.4/ (4π 10-7 1)= 318309.88 AT/m
H 3 L 3 = H g L g + H 2 L 2 H 3 0.1 = 318309.88 0.002 + 245 (0.2 + 0.2 + (0.1-0.002)) H 3 =758.62 AT/m B 2 = H 2 (μ o μ r ) 0.4 = 245 4π 10-7 μ r μ r = 1300 B 3 = H 3 (μ o μ r ) = 758.62 4π 10-7 1300 =1.239 Tesla Φ 3 = B 3 A = 1.239 2 10-4 = 0.0002478 wb Φ 1 = Φ 2 + Φ 3 = 0.0002478 + 2 10-4 = 0.0004478 wb B 1 = Φ 1 / A = 0.0004478/ 5 10-4 = 0.8956 tesla From chart H 1 =620 AT/m
N I = H 1 L 1 + H 3 L 3 200 I = 620 (0.2+0.2+0.1) + 758.62 0.1 I= 0.518 A
Electrical Eng. Fundamental Lecture 10 <Dr Hadi El-Helw> Contact details: h-elhelw@staffs.ac.uk
Direct Current (DC) Direct Current (DC) always flows in the same direction, but it may increase and decrease. A DC voltage is always positive (or always negative), but it may increase and decrease. Electronic circuits normally require a steady DC supply which is constant at one value or a smooth DC supply which has a small variation called ripple. Steady DC Smooth DC Varying DC
Alternating Current (AC) All the industrial power-distribution networks operate with alternating currents.
The magnitude of the emf and current varies with time: Maximum when coil is perpendicular to the field Zero when the coil is parallel to the field
Alternating Current (AC) flows one way, then the other way, continually reversing direction. An AC voltage is continually changing between positive (+) and negative (-). The rate of changing direction is called the frequency of the AC and it is measured in hertz (Hz) which is the number of forwards-backwards cycles per second.
General format for the Sinusoidal voltage or current The basic mathematical format for the sinusoidal waveform is: A m A m θ sin The peak value of the waveform is the unit of the horizontal axis Where θ =ωt ω t ω=2 π f The angular velocity of the rotating vector Time
V ( t) Vm sint Thus, we have consider only sine waves that have maxima value at π/2 and 3 π/2. With a zero value at 0, π, and 2 π as shown in the previous fig. if the waveform is shifted to the right or left of θ o, the expression becomes V ( t) V sin( t ) Where θ is the angel in degrees that the waveform has been shifted m
V ( t) V sin( t ) m V ( t) V sin( t ) m
If the wave form crosses the horizontal axis with positive going slope 90 o, it is called a cosine wave. sin( t 90) cost
The average (mean) value of a wave The average value of an alternating current is equal to the steady (DC) value which transfers same charge to the circuit within a specific period. I av 1 T 0 T i( t ) dt I I av av 1 2 I m 2 2 0 I m sin d 2 cos 0 0 In the case of a symmetrical alternating current (i.e. One whose two-half cycle are exactly similar, whether sinusoidal or non-sinusoidal), the average value over a complete cycle is zero.
average value over one (or more) cycles of a sine wave is clearly zero. however, it is often useful to know the average magnitude of the waveform independent of its polarity: We can think of this as the average value over half a cycle or as the average value of the rectified signal I I I av av av 1 I m sin 0 I m cos 0.637 I m 0 d 2 I m
Example 1: Determine the average value of the voltage waveform shown below.
Example 2: Determine the average value of the voltage waveform shown below.
Example 3: Determine the average value of the current waveform shown below.
Electrical Eng. Fundamental Lecture 11 <Dr Hadi El-Helw> Contact details: h-elhelw@staffs.ac.uk
Root Mean Square (RMS) value Most AC instruments are calibrated to show the RMS value of the voltage or current and not the peak value When the value of an AC voltage or current is given it is understood that it is the RMS value. The RMS value is sometimes called the effective value of the AC voltage. The root mean square value of an alternating current is given by that steady (d.c.) current which when flowing through a given circuit for a given time produces the same heat as produced by the alternating current when flowing through the same circuit for the same time.
V rms 1 T 0 T v 2 ( t ) dt Square Mean Root
The Root Square value of a sine wave m m m rms m rms o m rms m rms m m V V V V V V d V V d V V V t V t v 0.707 2 sin 0) (0 ) 2 sin 4 (2 4 2 sin 2 4 ) cos 2 (1 2 1 2 ) sin ( 2 1 sin sin ) ( 2 2 0 2 2 2 2 0 2 r.m.s value of current or voltage= 0.707 x max. value of current or voltage
Form factor for any waveform the form factor is defined as Form factor average r.m.s. value value for a sine wave this gives Form factor 0.707 0.637 V V m m 1.11
Peak factor for any waveform the peak factor is defined as Peak factor peak value r.m.s. value for a sine wave this gives Peak V factor m 0.707V m 1.414
Example 1: Determine the form factor and the peak factor of the voltage waveform shown below.
Example 2: Determine the form factor and the peak factor of the voltage waveform shown below.
Example 3: Determine the form factor and the peak factor of the voltage waveform shown below.
Electrical Eng. Fundamental Lecture 12 <Dr Hadi El-Helw> Contact details: h-elhelw@staffs.ac.uk
The response of the basic elements R,L, and C to a sinusoidal voltage or current Resistance The resistance is unaffected by the frequency of the applied sinusoidal voltage or current For purely resistive circuit V and I were in phase Vm I m or Vm I m R R
Inductance For the inductor the voltage across an inductor is directly related to the rate of change of current through the coil. Consequently, the higher the frequency, the greater will be the rate of change of current through the coil, and the greater the magnitude of the voltage
di L vl L dt di L d ( I m dt dt di L vl L dt or sin t) L( I m I m cos t) cos t LI m cos t vl Vm sin( t 90) Where V m LI m Therefore, for the inductor, v L leads i L by 90 o, or i L lags v L by 90 o The quantity ωl, called the reactance of an inductor, is symbolically represented by X L and is measured in ohms
X L L 2fL (ohms) In an ohm s law format, its magnitude can be determined from: X L V I m m Inductive reactance is opposition to the flow of current, which results in the continual interchange of energy between the source and the magnetic field of the inductor
Capacitance The voltage across the capacitor is limited by the rate at which charge can be deposited on, or released by, the plates of the capacitor during the charging and discharging phases, respectively. In other words, an instantaneous change in voltage across a capacitor is opposed by the fact that there is an element of time required to deposit charge on the plates of a capacitor
The fundamental equation relating the voltage across a capacitor to the current of a capacitor is: applying differentiation Therefore i c dv dt c dvc c dt i d dt c c dv dt ( v sin t) V m m c m cost c( V cost) cv or m cost i I sin( t 90) c m where I m CV m
For a capacitor i c leads v c by 90 o or v c lags i c by 90 o The quantity 1/ωc called the reactance of a capacitor, is symbolically by X c and is measured in ohm 1 X c c Capacitor reactance is the opposition to the flow of charge, which results in the continual interchange of energy between the source and the electric field of a capacitor.
Series RL circuit V V m v I I m i Where I m R 2 V m ( L) 2 i v L tan 1 R
Series RC circuit V V m v I I m i Where I m R 2 V m 1 ( ) C 2 i v tan 1 1 RC
Series RLC circuit v V m V i I m I Where 2 2 ) 1 ( C L R V I m m R c L v i 1 tan 1
Example 1 A coil is connected to supply voltage as shown in the figure below calculate the following 1.the inductive reactance(x L ) 2.the total impedance (Z) 3.the current(i) 4.the phase angle Ө between the current and the supply voltage
Example 2 Two coils are connected in series across a voltage supply (V) as shown in the figure below. Calculate the value of the voltage V 1,V 2 across each coil. each coil has its own resistance
Electrical Eng. Fundamental Lecture 13 <Dr Hadi El-Helw> Contact details: h-elhelw@staffs.ac.uk
Series RC circuit V V m v I I m i Where I m R 2 V m 1 ( ) C 2 i v tan 1 1 RC
Example 1 Capacitor (C)of 100µf and resistor(r) of 10 Ω are connected in series across a supply voltage (V) as shown in the figure below. calculate the current (I) flowing in the circuit calculated the phase angle (Ө) between the current and voltage
Series RLC circuit v V m V i I m I Where 2 2 ) 1 ( C L R V I m m R c L v i 1 tan 1
Example 2 A circuit consists of R.L.C connected in series as shown in the figure below, calculated the value of the current (I) showing if it is leading or lagging in the following two cases frequency F=50HZ frequency F=150HZ