ac&dc circuits
DESCRIPTION
ac and dc circuitsTRANSCRIPT
DC CIRCUITS
CIRCUITS AND NETWORKS
CIRCUITS
Parameters The various elements of an electric circuit, like resistance,
inductance, and capacitance which may be lumped or distibuted.
CIRCUITS A closed conducting path through which an electric
current flows or is intended to flow
CIRCUITS
Linear Circuit Is one whose parameters are constant (i.e. They do not change with
voltage and current.
Non-Linear Circuit Is that circuit whose parameters change with voltage and current.
Bilateral Circuit Is one whose properties or characteristics are the same in either
direction.
Unilateral Circuit Is that circuit whose properties or characteristics change with the
direction of its operation.
TYPES
CIRCUITS AND NETWORKS
ELECTRICAL NETWORKS
Passive NetworkWith no source of emf.
Active NetworkContains one or more than one sources of emf.
ELECTRICAL NETWORK Connection of various electric elements in any manner
TYPES
CIRCUITS AND NETWORKS
ELECTRICAL NETWORKS
Node A junction in a circuit where two or more circuit elements and/or
branches are connected together.
Branch Part of a network which lies netween two junctions.
Loop A closed path in a circuit in which no element or node is encountered
more than once.
Mesh A loop that contains no other loop within it.
PARTS
CIRCUITS AND NETWORKS
OHM’S LAW
OHM’S LAW One of the most fundamental law in electrical circuits relating voltage,
current and resistance Developed in 1827 by German physicist Georg Simon Ohm
CIRCUITS AND NETWORKS
OHM’S LAW According to Ohm’s Law, the current (I) flowing in an
electrical circuit is directly is directly proportional to the applied voltage (E) and inversely proportional to the equivalent resistance (R) of the circuit and mathematically expressed as:
CIRCUITS AND NETWORKS
SERIES CIRCUITS
SERIES circuits A circuit connection in which the components are
connected to form one conducting path
SERIES/PARALLEL CIRCUITS
SERIES CIRCUITS
Where: EX – voltage across the resistor concerned
ET – total voltage across the circuit
RX – the resistor concerned
RT – the sum of all resistances in the circuit
Voltage Division for Series Circuit:
EX = ET • RX
RT
SERIES/PARALLEL CIRCUITS
PARALLEL CIRCUITS
PARALLEL circuits A circuit connection in which the components are
connected to form more than 1 conducting path
SERIES/PARALLEL CIRCUITS
PARALLEL CIRCUITS
Where: IX – current concerned flowing through resistor Rx
IT – total current of the circuit
Req – equivalent resistance of the parallel circuit except Rx
RT – the sum of all resistances in the circuit
Voltage Division for Parallel Circuit:
IX = IT • Req
RT
SERIES/PARALLEL CIRCUITS
KIRCHHOFF’S LAW
“In any electrical network, the algebraic sum of the current meeting at a point (or junction) is zero.”
KIRCHHOFF’S LAW More comprehensive than Ohm’s Law and is used in solving electrical Termed as “Laws of Electric Networks” Formulated by German physicist Gustav Robert Kirchhoff
Kirchhoff’s Current Law (KCL)
NETWORK ANALYSIS
KIRCHHOFF’S CURRENT LAW In short the sum of currents entering a node equals the sum
of currents leaving the node⁻ Current towards the node, positive current⁻ Current away from the node, negative current
IB + IC + ID = IA
(IB + IC + ID) - IA = 0
NETWORK ANALYSIS
KIRCHHOFF’S VOLTAGE LAW
“The algebraic sum of the products of currents and resistances in each of thr conductors in any closed path (or mesh) in a network PLUS the algebraic sum of the emfs in the path is zero.”
Kirchhoff’s Voltage Law (KVL)
NETWORK ANALYSIS
KIRCHHOFF’S VOLTAGE LAW In short, the sum of the voltages around the loop is equal to
zero⁻ For voltage sources, if loops enters on minus and goes out on plus,
positive voltage and if loops enters on plus and goes out on minus, negative voltage.
⁻ For voltage drops, if the loop direction is the same as current direction, negative voltage drop and if the loop direction is opposite to the current direction, positive voltage drop.
NETWORK ANALYSIS
MESH ANALYSIS
Loop Analysis Procedure:
1. Label each of the loop/mesh currents. 2. Apply KVL to loops/meshes to form
equations with current variables.a. For N independent loops, we may write N
total equations using KVL around each loop. Loop currents are those currents flowing in a loop; they are used to define branch currents.
b. Current sources provide constraint equations.
3. Solve the equations to determine the user defined loop currents.
NETWORK ANALYSIS
MESH analysis A sophisticated application of KVL with mesh currents.
NODAL ANALYSIS
Consist of finding the node voltages at all principal nodes with respect to the reference node.
PRINCIPAL node – a node with three or more circuit elements joined together.
Reference node – the node from which the unknown voltages are measured.
NETWORK ANALYSIS
NODAL analysis A systematic application of KCL at a node and after simplifying the
resulting KCL equation, the node voltage can be calculated.
SUPERPOSITION THEOREM
In general: Number of network to analyze is equal to number of independent
sources. To consider effects of each source independently, sources must be
removed and replaced without affecting the final result:
All voltage sources >> short circuited All current sources >> open circuited
NETWORK ANALYSIS
SUPERPOSITION theorem “ The current through or voltage across, an element in a linear
bilateral network is equal to the algebraic sum of the current or voltages produced independently in each source. ”
COMPENSATION THEOREM
NETWORK ANALYSIS
COMPENSATION theorem If the impedance Z of a branch in a network in which a current I
flows is changed by a finite amount dZ, then the change in the currents in all other branches of the network may be calculated by inserting a voltage source of -IdZ into that branch with all other voltage sources replaced by their internal impedances.
RECIPROCITY THEOREM
Simply mean, E and I are mutually transferable, or The receiving point and the sending point in a network are
interchangeable, or Interchange of an IDEAL voltage source and an IDEAL ammeter in any
network will not change the ammeter reading, Interchange of an IDEAL current source and an IDEAL voltmeter in any
network will not change the voltmeter reading
NETWORK ANALYSIS
RECIPROCITY theorem “If a voltage source E acting in one branch of a network causes a current I to flow in another branch of the network, then the same voltage source E acting in the second branch would cause an identical current I to flow in the first branch. ”
MILLMAN’S THEOREM
In Millman’s Theorem, the circuit is re-drawn as a parallel network of branches, each branch containing a resistor or series battery/resistor combination.
Millman’s theorem is applicable only to those cicuits which can be re-drawn accordingly.
NETWORK ANALYSIS
MILLMAN’S theorem “ A special case of the application of Thevenin’s Theorem/or Norton’s Theorem used for finding the COMMON voltage (VAB) across any network which contains a number of parallel voltage sources. ”
MAXIMUM POWER TRANSFER THOREM
NETWORK ANALYSIS
MAXIMUM POWER TRANSFER theorem For loads connected directly to a DC voltage supply, maximum power
will be delivered to the load when the resistance is equal to the internal resistance of the source.
For maximum power transfer: RS = RL
THEVENIN’S THEOREM
where: VTH – the open circuit voltage which appears across the two terminals from where the load resistance has been removed.RTH – the resistance looking back into the network across the two terminals with all the voltage sources shorted and replaced by their internal resistances (if any) and all current sources by infinite resistance.
NETWORK ANALYSIS
THEVENIN’S theorem “ Any two-terminal of a linear, active bilateral network of a fixed resistances and voltage source/s may be replaced by a single voltage source (VTH) and a series of internal resistance (RTH). ”
NORTON’S THEOREM
where: IN– the current which would flow in a short circuit placed across the output terminals.
RN – the resistance of the network when viewed from the open circuited terminals after all voltage sources being replaced by open circuits. NETWORK ANALYSIS
THEVENIN’S theorem “ Any two-terminal active network containing voltage sources and resistances when viewed from its output terminals, is equivalent to a constant-current source (IN) and a parallel internal resistance (RN). ”
THEVENIN-NORTON TRANSFORMATION
NETWORK ANALYSIS
NORTON-THEVENIN TRANSFORMATION
NETWORK ANALYSIS
EQUIVALENT THREE-TERMINAL NETWORKS
NETWORK ANALYSIS
DELTA to WYE The equivalent resistance of each arm to the wye is given by the PRODUCT of the
two delta sides that meet at its end divided by the SUM of the three delta resistances.
EQUIVALENT THREE-TERMINAL NETWORKS
NETWORK ANALYSIS
WYE to DELTA The equivalent delta resistance between any two twrminals is given by the
SUM of a star resistance between those terminals PLUS the PRODUCT of these two star resistances DIVIDED by the third resistance.
REVIEW QUESTIONS1. A battery with a rating of 9 volts has an internal resistance of 20 ohms.
What is the expected resistance of the bulb that is connected across the battery to attain a maximum power transfer?a. 20 ohmsb. 10 ohmsc. 5 ohmsd. 2 ohms
2. In a sireis ciscuit a resistors 2200 and 4500 with an impressed voltage of 10, what is the circuit current in mA?a. 1.49b. 6.67c. 4.34d. 1.34
REVIEW QUESTIONS3. The current needed to operate a soldering iron which has a rating of 600
watts at 110 volts is.a. 5.455 Ab. 66,000 Ac. 18,200 Ad. 0.182 A
4. The ammeter reads 230 ampere while the voltmeter is 115 volts. What is the power inKW at the time of readinga. 264.5b. 2645c. 264500d.26.45
REVIEW QUESTIONS
5. What is the type of circuit whose parameters are constant which do not change with voltage or current?a. Lumpedb. Tunedc. Reactived. Linear
6. What is the resistance of two equal valued resistor series?a. Twice as oneb. The sum of their reciprocal c. The difference of bothb. The product of both
REVIEW QUESTIONS7. What do you expect when you use two 20 kohms, 1 watts resistors in
parallel instead of one 10 kohms, 1 watt?a. Provide more powerb. Provide lighter currentc. Provide less powerd. Provide wider tolerance
8. The voltage applied in DC circuit having a power of 36 watts and a total resistance of 4 ohms.a. 6 Vb. 9Vc. 12 Vd. 24 V
REVIEW QUESTIONS9. When resistor are connected in series, what happens?
a. The effective resistanceb. Nothingc. The toleranced. The effective resistance is increased
10. Find the thevenin’s impedance equivalent across R2 of a linear close circuit having 10-V supply in series with the resistors (R1=100 ohms and R2=200 ohms)a. 666 ohmsb. 6.66 ohmsc. 66.6 ohmsd. 6666 ohms
REVIEW QUESTIONS11. How much power does electronic equipment consume, assuming a 5.5A
current flowing and a 120-V power source.a. 60 Wb. 66 Wc. 660 Wd. 125.5 W
12. How many nodes are needed to completely analyze a circuit according to Kirchoffs Current Law.a. Oneb. Two c. All nodes in the circuitd. One less than the total number of nodes in the circuit
REVIEW QUESTIONS
13. A common connection between circuit elements or conductors from different branches.a. Nodeb. Junctionc. Groundd. Mesh
14. It is used to denote a common electrical point of zero potential.a. Short circuitb. Reference pointc. Open circuitd.ground
REVIEW QUESTIONS
15. Loop currents should be assumed to flow in which direction?a. Straightb. Clockwisec. Counterclockwised. Either b or c
16. In mesh analysis, we apply:a. KCLb. KVLc. Sourced. Millman’s theorem
REVIEW QUESTIONS
17. Which of the following is not a valid expression of Ohms Lawa. R = PIb. E = IRc. I = E/Rd. R = E/I
18. Using ohms Law, what happens to the circuit current if the applied voltage increases?a. Current doublesb. Current increasesc. Current remians constantd. Current decreases
REVIEW QUESTIONS19. According to ohms law, what happen to the circuit current if the circuit
resistance increases?a. Current doublesb. Current decreasesc. Current increasesd. Current remains constant
20. If the resistance of a circuit is doubled and the applied voltage is kept constant, the current will be _______ .a. Be quaddrupledb. Remainsc. Be cut in halfd. Be doubled
REVIEW QUESTIONS
21. It is an electrical current that flows in one direction only?a. Normal currentb. Alternating currentc. Direct currentd. Eddy current
22. In Ohms Law, what is E/R?a. Amperageb. Voltagec. Resistanced. Power
REVIEW QUESTIONS23. A 33-Kohm resistor is connected in series with a parallel combination made
up of 56-Kohm resistor and a 7.8-kohm resistor. What is the total combined reistance of the three resistors?a. 390667 ohmsb. 49069 ohmsc. 63769 ohmsd. 95000 ohms
24. Which of the following cannot be included in a loop of Kirchoff’s Volatage Lawa. Current sourceb. Voltage sourcec. Resistanced. Reactance
REVIEW QUESTIONS25. A series connected circuit consists of 3 loads and consume a total power of 50
Watts. It was reconfigured such that 2 are in parallel and the other load is in series with a combination. What is the applied expected powers to be consumed them?a. 50 wattsb. 25 wattsc. 75 wattsd. 45 watts
26. If the number of valence electrons of an atom is less than 4, the substance is usuallya. Semiconductorb. An insulatorc. A conductord. None of the above
REVIEW QUESTIONS
27. Electric current in a wire is the flow ofa. Free electronsb. Valence electronsc. Bound electronsd. Atoms
28. EMF in a circuit is a forma. Powerb. Energyc. Charged. none
REVIEW QUESTIONS
29. The SI unit of specific-resistance isa. Mhob. Ohm-nc. Ohm-sq.-md. Ohm-cm
30. The resistance of a material is ___ its area of cross-sectiona. Directly proportionalb. Inversely proportionalc. Independentd. None of these
REVIEW QUESTIONS
31. The value of α, i.e., the temperarure coefficient of resistance depends upon the ____ of the material.a. Lengthb. Volumec. X-sectional aread. Nature and temperature
32. The value of α of a conductor is 1/236 C. The value of a α is a. 1/218 Cb. 1/272 Cc. 1/254 Cd. 1/265 C
REVIEW QUESTIONS
33. Electrical appliances are not connected in series becausea. Series circuit is complicatedb. Power loss is greaterc. Appliances have different current ratingd. None of these
34. Electrical appliances are connected in parallel because ita. Is a simple circuitb. Results in reduced power lossc. Draw less currentd. Makes the operation of the appliances independent from each other
REVIEW QUESTIONS
35. The hot resistance of a 100W, 250V incandecent lamp isa. 2.5 ohmsb. 625 ohmsc. 25 ohmsd. None of these
36. When a number of resistances are connecte in parallel, the total resistance isa. Less than the smallest resistanceb. Greater than the smallest resistancec. Between the smallest and greater resistanced. None of these
REVIEW QUESTIONS37. If the resistances, each of value 36 ohms are connected in parallel, the
total resistance isa. 2 ohmsb. 54 ohmsc. 36 ohmsd. None of these
38. Two incandecent lamps of 100 W, 200V are in parallel across the 200 V. The total resistance will bea. 800 ohmsb. 200 ohmsc. 400 ohmsd. 600 ohms
REVIEW QUESTIONS39. Three resistors are connected in parallel and draws 1A, 2.5A, and 3.5A,
rspectively. If the applied voltage is 210V, what is the total resistance of the circuit?a. 5 ohmsb. 147 ohmsc. 3 ohmsd. 73.5 ohms
40. An ordinary dry cell can deliver about ____ continuously.a. 3 Ab. 2 Ac. 1/8 Ad. None of these
REVIEW QUESTIONS41. Four cells of internal resistance 1 ohms, are connected in parallel. The
battery resistance will bea. 4 ohmsb. 0.25 ohmsc. 2 ohmsd. 1 ohms
42. Of the following combination of units, the one that is not equal to the watt isa. Joule/secb. Ampere-voltc. Ampere-ohmd. Ohm/volt
REVIEW QUESTIONS
43. The power dissipated in a circuit is not equal toa. VIb. IRc. V/Rd. IR/V
44. An electric iron draws a current of 15A when connected to 120V power source. Its resistance isa. 0.125 ohmsb. 8 ohmsc. 16 ohmsd. 1,800 ohms
REVIEW QUESTIONS45. The power rating of an electric motor which draws a current of 3 A when
operated at 120 V isa. 40 Wb. 360 Wc. 540 Wd. 1,080 W
46. When a 100W, 240V, light bulb is operated at 200V, the current that flows in it isa. 0.35 Ab. 0.42 Ac. 0.5 Ad. 0.58 A
REVIEW QUESTIONS47. The equivalent resistance of a network of three 2 ohm resistors cannot
bea. 0.67 ohmsb. 1.5 ohmsc. 3 ohmsd. 6 ohms
48. A 12V potential difference is applied across a series combination of four six-ohms resistors. The current in each six-ohm resistor will bea. 0.5 Ab. 2 Ac. 8 Ad. 6 A
REVIEW QUESTIONS49. A 12V potential difference is applied across a parallel combination of
four six-ohms resistors. The current in each six-ohm resistor will be a. 0.5 Ab. 2 Ac. 8 Ad. 6 A
50. The dissipation of energy can cause burns because it proceducesa. Heatb. Firec. Frictiond. Overload
REVIEW QUESTIONS
51. The rate of expenditure of energy isa. Voltageb. Powerc. Currentd. Energy
52. In a simple DC power line, the wire that carries the current from the generator to the load is calleda. Return wireb. Feederc. Outgoing wired. Conductor
REVIEW QUESTIONS53. A circuit in which the resistance are connected in a continuous run, i.e.,
end-to-end is a _____ circuit.a. Saries b. Parallelc. Series-paralleld. None of these
54. A battery is connected to an external circuit. The potential drop with the battery is proportional to a. The EMF of the batteryb. The current of the circuitc. The equivqlent circuit resistanced. Power dissipated in the circuit
REVIEW QUESTIONS55. Two wires A and B have the same cross-sectional area and are made of the
same material. Ra = 600 ohms and Rb = 100 ohms. The number of times A is longer tahn B isa. 6b. 2c. 4d. 5
56. A coil has a resistance of 100 ohms at 90 C. At 100 C, its resistance is 101 ohms. The temperature coefficient of the wire isa. 0.01b. 0.1c. 0.0001d. 0.001
REVIEW QUESTIONS
57. The resistance of a conductor does not depend on itsa. Resistivityb. Lengthc. Cross-sectiond. Mass
58. A material which has a negative temperature coefficient of resistance is usually a/ana. Insulatorb. Conductorc. Semi-conductord. All of these
REVIEW QUESTIONS
59. Which of the following statements is true both for a series and a parallel dc circuit?a. Power additiveb. Current are additivec. Voltage are additived. All of these
60. Two resistors are said to be in series when a. Both carry the same value of currentb. Total current equals the sum of the branch currentc. Sum of IR drops equal to EMFd. Same current phases through both
REVIEW QUESTIONS61. According to KCL as applied to a juction in a network of conductors.
a. Total sum of currents meeting at the juction is Zerob. No current can leave the juction without same current passing throuhg itc. Net current flow at he juction is positived. Algebraic sum of the currents meeting at the juction is zero
62. Kirchoff’s Current Law is applicable only toa. Closed-loop circuitb. Electronic circuitsc. Juctions in a networkd. Electric circuit
REVIEW QUESTIONS
63. Kirchoff’s Voltage Las\w is concerned witha. IR dropsb. Battery EMF’sc. Juction voltagesd. A and b
64. According to KVL, the algebraic sum of all IR drops and EMF’s in any closed loop of a network is alwaysa. Zerob. Negativec. Positived. Determined by battery EMF
REVIEW QUESTIONS65. The algebraic sign of an IR drops primarily dependent upon
a. The amount of current flowing through itb. Direction of currentc. The value of the resistanced. The battery connection
66. Choose the wrong statement. In the node voltage technique of solvingnetwork parameters, the choice of the reference node does not a. Affect the operation of the circuitb. Change the voltage across the element c. Alter the potential difference between any pair of nodesd. Affect the volatge of various nodes
REVIEW QUESTIONS
67. The nodal analysis is primarily based on the application ofa. KVLb. KCL.c. Ohms Lawd. b and c
68. Superposition theorem can be applied only to circuits having ____ elementsa. Non-linearb. Passivec. Linear bilaterald. Resistive
REVIEW QUESTIONS
69. The superposition theorem is essentially based on the concept ofa. Reciprocityb. Linearityc. non-linearityd. Duality
70. What are the electrons in motion called?a. Current variationb. Electric currentc. Electron velocityd. Dynamic electricity
REVIEW QUESTIONS
71. An active element in a circuit is one which _____ .a. Receives energyb. Supplies energyc. a or bd. None of these
72. The siperposition theorem is used when the circuit containsa. A single voltage sourceb. A number of voltage sourcec. Passive element onlyd. None of these
REVIEW QUESTIONS
73. Thevenin’s theorem is ____ form of equivalent circuit.a. Voltageb. Currentc. Both a and bd. None of these
74. Norton’s theorem is ____ form of an equivalent circuit.a. Voltageb. Currentc. Both a and bd. None of these
REVIEW QUESTIONS75. In the analysis of vacuum tube circuit, we can generally use ____
theorem.a. Norton’sb. Thevenin’sc. Superpositiond. Reciprocity
76. In the analysis of transistor circuits, we generally use _____ theorem.a. Voltageb. Currentc. Both a and bd. None of these
REVIEW QUESTIONS
77. Under the condition of Maximum Power Transfer, the efficiency isa. 75 %b. 100 %c. 50 %d. 25 %
78. The maximum power transfer theorem is used ina. Electronic circuitsb. Home lightningc. Power sytemd. None of the above
REVIEW QUESTIONS
79. delta/star or star/delta transformation technique is applied toa. One terminalb. Two terminal c. Three terminald. None of these
80. _____ will be used under elctrostatics.a. Incandecent lampb. Electric motorc. Electric irond. Lightning rod
REVIEW QUESTIONS
81. The value of the absulute permitivity of air is ______ F/m.a. 9 x 10 b. 8,854 x 10 ^ -12c. 5 x 10d. 9 X 10
82. When two similar charges eaach of 1 coulumb each are placed 1m apart in air, then the force of repusion isa. 8 x10 Nb. 10 Nc. 9 x 10^9 Nd. 5 x 10 N
REVIEW QUESTIONS83. Another name for dielectric strength is
a. Potetial gradientb. Breakdown voltagec. Dielectric constantd. Electric intensity
84. A heater connected to a 100 V supply, generates 10,000 J of heat energy is 10 sec. How much time is needed to generate the same amount of heat when it is used on 220V line?a. 5 secb. 2.5 secc. 7.5 secd. 4 sec.
REVIEW QUESTIONS
85. Kirchoff’s current law is applied in what type of circuit analysis?a. Meshb. Thevenin’sc. Superpositiond. Nodal
86. Inductance and capacitance are not relevant in a dc circuit becausea. Frequency of DC is zerob. It is a simple circuitc. They do not exist in dc circuitd. None of the above
REVIEW QUESTIONS
87. Three resistors of 3 ohms resistance each are connected in delta, the equivalent wye-connected resistors will bea. 1 ohmb. 3 ohmsc. 9 ohmsd. 0.111 ohm
88. Cells are conneted in series when _____ is requireda. High currentb. High voltagec. High powerd. All of these
AC CIRCUITS
ALTERNATING CURRENT
ALTERNATING CURRENT A current that is constantly changing in amplitude and direction.
Advantages of AC: Magnitude can easily be changed (stepped-up or stepped down) with the use of a
transformer Can be produced either single phase for light loads, two phase for control motors, three
phase for power distribution and large motor loads or six phase for large scale AC to DC conversion. BASIC AC THEORY
AC WAVEFORMS
BASIC AC THEORY
AC WAVEFORMS
Period (T) – the time of one complete cycle in seconds Frequency (f) – the number of cycles per second (Hertz)
a. 1 cycle/second (cps) = 1 Hertz (Hz)b. Proper operation of electrical equipmnent requires specific frequencyc. Frequencies lower than 60 Hz would cause flicker when used in lightning
Wavelength (λ) – the length of one complete cycle Propagation Velocity (v) – the speed of the signal Phase (Φ) – an angilar measurement that specifies the position of a sine wave relative to
reference
f = 1 T
λ = v f
Parameters of Alternating Signal
BASIC AC THEORY
AC WAVEFORMS
THE SINUSOIDAL WAVE Is the most common AC waveform that is practically
generated by generators used in household and industries General equation for sine wave:
Where:a(t) – instantaneous amplitude of voltage or current at a given time (t)Am – maximum voltage or current amplitude of the signal
ω – angular velocity in rad/sec; ω = 2πft – time (sec)Φ – phase shift ( + or – in degrees)
A(t) = Am sin (ωt + Φ)
BASIC AC THEORY
AC WAVEFORMS
PEAK AMPLITUDE – the height of an AC waveform as measured from the zero mark to the highest positive or lowest negative point on a graph. Also known as the crest amplitude of a wave.
AMPLITUDE It is the height of an AC waveform as depicted on a graph over time (peak,
peak-to-peak, average, or RMS quantity)
Measurements of AC Magnitude
BASIC AC THEORY
AC WAVEFORMS
PEAK-TO-PEAK AMPLITUDE – the total height of an AC waveform as measured from maximum positive to maximum negative peaks on a graph. Often abbreviated as “P-P”
BASIC AC THEORY
AC WAVEFORMS
AVERAGE AMPLITUDE – the mathematical “mean” of all a waveform’s points over the period of one cycle. Technically, the average amplitude of any waveform with equal-area portions above and below the “zero” line on a graph is zero.
For a sine wave, the average value so calculated is approximately 0.637 of its peak value.
BASIC AC THEORY
AC WAVEFORMS
RMS AMPLITUDE - “RMS” stands for Root Mean Square, and is a way of expressing an AC quantity of voltage or current in terms functionally equivalent to DC. Also known as the “equivalent” or “DC equivalent” value of an AC voltage or current.
For a sine wave, the RMS value is approximately 0.707 of its peak value.
BASIC AC THEORY
AC WAVEFORMS
The crest factor of an AC waveform is the ratio of its peak (crest) to its RMS value.
The form factor of an AC waveform is the ratio of its RMS value to its average value.
BASIC AC THEORY
AC QUANTITIES
RESISTANCE (R) Opposes the AC current similar to DC circuits Opposition offered by resistors
REACTANCE (X) Depends on the AC frequency of the AC source which is the
opposition to current due to inductance and capacitance
BASIC AC THEORY
AC QUANTITIESInductive Reactance (XL)
• The property of the inductor to oppose the alternating current
Inductive Susceptance (BL)• Reciprocal of inductive reactance
Capacitive Reactance (XC)• The property of a capacitor to oppose alternating current
Capacitive Susceptance (BC)• Reciprocal of capacitive reactance
d
XL = 2πfL
BL = 1 BL = 1 XL 2πfL
XC = 1 2πfC
BL = 1 BL = 2πfC XC
BASIC AC THEORY
AC QUANTITIES
IMPEDANCE (Z) Total opposition to the flow of Alternating current Combination of the resistance in a circuit and the reactances involved
Phasor Diagram of Impedance
Z = R + jXeq Z = |Z| ∠φ
Where: |Z| = √ R2 + X2 φ = Arctan Xeq
R
BASIC AC THEORY
AC QUANTITIES
If I = Im ∠β is the resulting current drawn by a passive, linear RLC circuit from a source voltage V = Vm ∠θ, then
Where: Z = Vm = √ R2 + X2 = magnitude of the impedance Imφ = θ – β = tan-1 X = phase angle of the impedance
RR = Zcos φ = active or real component of the impedance
X = Zsin φ = reactive or quadrature component of impedance
Z = V = Vm ∠θ = Z ∠φ I Im ∠β Z cosφ + jZsin φ = R + jX = √ R2 + X2 ∠ tan-1 X
R
BASIC AC THEORY
AC QUANTITIES
ADMITTANCE (Y) The reciprocal of impedance Expressed in siemens or mho (S)
Where: Y = Im = √ G2 + B2 = 1 = magnitude of the admittance Z
φy = β – θ = φ = tan-1 B = phase angle of the admittance G
G = Ycos φy = conductive/conductance component
B = Ysin φy = susceptive/susceptance component
Y = Im ∠ β – θ = Y = Ycos φy + jYsin φy = G + jB Vm
Y = √ G2 + B2 ∠ tan-1 B G
BASIC AC THEORY
AC RESISTOR CIRCUIT
With an AC circuit like this which is purely resistive, the relationship of the voltage and current is as shown:
Voltage (e) is in phase with the current (i) Power is never a negative value. When the current is positive (above the line), the
voltage is also positive, resulting in a power (p=ie) of a positve value This consistent “polarity” of a power tell us that the resistor is always dissipating
power, taking it from the source and releasing it in the form of heat energy. Whether the current is negative or positive, a resistor still dissipated energy.AC CIRCUITS
Impedance (Z) = R
AC INDUCTOR CIRCUIT
The most distinguishing electrical characteristics of an L circuit is that current lags voltage by 90 electrical degrees
Because the current and voltage waves arae 90o out of phase, there sre times when one is positive while the other is negative, resulting in equally frequent occurences of negative instantaneous power.
Negative power means that the inductor is releasing power back to the circuit, while a positive power means that it is absorbing power from the circuit.
The inductor releases just as much power back to the circuit as it absorbs over the span of a complete cycle.
Impedance (Z) = jXL
AC CIRCUITS
AC INDUCTOR CIRCUIT
o Inductive reactance is the opposition that an inductor offers to alternating current due to its phase-shifted storage and release of energy in its magnetic field. Reactance is symbolized by the capital letter “X” and is measured in ohms just like resistance (R).
o Inductive reactance can be calculated using this formula: XL = 2πfL
o The angular velocity of an AC circuit is another way of expressing its frequency, in units of electrical radians per second instead of cycles per second. It is symbolized by the lowercase Greek letter “omega,” or ω.
o Inductive reactance increases with increasing frequency. In other words, the higher the frequency, the more it opposes the AC flow of electrons.
AC CIRCUITS
AC CAPACITOR CIRCUIT
The most distinguishing electrical characteristics of an C circuit is that leads the voltage by 90 electrical degrees
The current through a capacitor is a reaction against the change in voltage across it A capacitor’s opposition to change in voltage translates to an opposition to alternating voltage in
general, which is by definition always changing in instantaneous magnitude and direction. For any given magnitude of AC voltage at a given frequency, a capacitor of given size will “conduct” a certain magnitude of AC current.
The phase angle of a capacitor’s opposition to current is -90o,meaning that a capacitor’s opposition to current is a negative imaginary quantity
Impedance (Z) = -jXC
AC CIRCUITS
AC CAPACITOR CIRCUIT
o Capacitive reactance is the opposition that a capacitor offers to alternating current due to its phase-shifted storage and release of energy in its electric field. Reactance is symbolized by the capital letter “X” and is measured in ohms just like resistance (R).
o Capacitive reactance can be calculated using this formula: XC = 1/(2πfC)
o Capacitive reactance decreases with increasing frequency. In other words, the higher the frequency, the less it opposes (the more it “conducts”) the AC flow of electrons.
AC CIRCUITS
SERIES RESITOR-INDCUTOR CIRCUIT
For a series resistor-inductor circuit, the voltage and current relation is determined in its phase shift. Thus, current lags voltage by a phase shift (θ)
Impedance (Z) = R + jXL
Admittance (Y) = 1 = R – jXL
R + jXL R2 + jXL2
AC CIRCUITS
SERIES RESITOR-INDCUTOR CIRCUIT
o When resistors and inductors are mixed together in circuits, the total impedance will have a phase angle somewhere between 0o and +90o. The circuit current will have a phase angle somewhere between 0o and -90o. Series AC circuits exhibit the same fundamental properties as series DC circuits: current is uniform throughout.
Phase shift (θ) = Arctan ( XL ) |Z| = √ R2 + jXL2 = e
R i
AC CIRCUITS
SERIES RESISTOR-CAPACITOR CIRCUIT
For a series resistor – capacitor circuit, the voltage and current relation is determined by the phase shift. Thus the current leads the voltage by an angle less than 90 degrees but greater than 0 degrees.
Impedance (Z) = R – jXC
Admittance (Y) = 1 = R + jXC
R – jXC R2 + jXC2
AC CIRCUITS
SERIES RESISTOR-CAPACITOR CIRCUIT
Phase shift (θ) = Arctan ( XC ) |Z| = √ R2 + jXC2 = e
R i
AC CIRCUITS
PARALLEL RESISTOR-INDUCTOR
Y = G – jβL where: G – conductance = 1/R
βL – susceptance = 1/XL
Z = E , by Ohm’s Law I
The basic approachwith regarda to parallel circuits is using admittance because it is additive
Z parallel = 1 Admittance (Y)
AC CIRCUITS
PARALLEL RESISTOR-INDUCTOR
o When resistors and inductors are mixed together in parallel circuits (just like in series cicuits), the total impedance will have a phase angle somewhere between 0o and +90o. The circuit current will have a phase angle somewhere between 0o and -90o.
o Parallel AC circuits exhibit the same fundamental properties as parallel DC circuits: voltage is uniform throughour the circuit, brach currents add to form the total current, and impedances diminish (through the reciprocal formula) to form the total impedance.
AC CIRCUITS
PARALLEL RESISTOR-CAPACITOR
Y = G + jβC where: G – conductance = 1/R
βC – susceptance = 1/XC
o When resistors and capacitors are mixxed together in circuits, the total impedance will have a phase angle somewhere between 0o and -90o.
AC CIRCUITS
APPARENT POWER (S)
APPARENT POWER Represents the rate at which the total energy is supplied to the system Measured in volt-amperes (VA) It has two components, the Real Power and the Capacitive or Inductive
Reactive Power
POWER IN AC CIRCUITS
S = Vrms Irms = Irms2 |Z|
APPARENT POWER (S)
Power Triangle
Complex Power
S = P ± jQ
POWER IN AC CIRCUITS
REAL POWER (R)REAL POWER
The power consumed by the resistive component Also called True Power, Useful Power and Productive Power Measured in Watts (W) It is equal to the product of the apparent power and the power factor
Cosine of the power factor angle (θ) Measure of the power that is dissipated by the cicuit in relation to the
apparent power and is usually given as a decimal or percentage
POWER IN AC CIRCUITS
Pf = cos θ
P = Scos θPower Factor
REAL POWER (R) Ratio of the Real Power to the Apparent Power ( P )
S
when:Pf = 1.0 I is in phase with V; resistive systemPf = lagging I lags V by θ; inductive systemPf = leading I leads V by θ; capacitive systemPf = 0.0 lag I lags V by 90o; purely inductivePf = 0.0 lead I leads V by 90o; purely capacitive
The angle between the apparent power and the real poweer in the power triangle
Let v(t) = Vm cos(ωt + θv) volts
V = Vrms ∠θv
i(t) = Im cos(ωt + θi) A
I = Irms ∠θi
POWER IN AC CIRCUITS
Power factor Angle (θ)
REAL POWER (R)
Where: θ = phase shfit between v(t) and i(t) or the phase angle of the equivalent impedance
POWER IN AC CIRCUITS
Instantaneous Power (watts)
Average Power (watts)
P(t) = v(t) i(t)
P(t) = ½ VmIm cos (θv – θi) + ½ VmIm cos (2ωt + θv + θi)
P(t) = ½ VmIm cos (θv – θi) = VmIm cos θ
REACTIVE POWER (QL or QC)
REACTIVE POWER Represents the rate at which energy is stored or released in any of the
energy storing elements (the inductor or the capacitor) Also called the imaginary power, non-productive or wattless power Measured in volt-ampere reactive (Var) When the capacitor and inductor are both present, the reactive power
associated with them take opposite signs since they do not store or release energy at the same time
It is positive for inductive power (QL) and negative for capacitive power (QC)
Ratio of the Reactive Power to the Apparent Power Sine of the power factor angle (θ)
POWER IN AC CIRCUITS
Q = VmIm sin θ
Reactive factor
Rf = sin θ
BALANCED THREE PHASE SYSTEMSBALANCED 3-PHASE SYSTEM
Comprises of three identical single-phase systems operating at a 120o phase displacement from one another. This means that a balance three-phase system provides three voltages(and currents) that are equal in magnitude and separated by 120o from each other
Three-Phase, 3-wire systems Provide only one type of voltage(line to line) both single phase and
three phase loads
Three-Phase, 4-wire systems Provide two types of voltages(line to line and line to neutral) to both
single phase and three phase loads
BALANCED THREE PHASE SYSTEM
CLASSIFICATION
BALANCED THREE PHASE SYSTEMS
and
VLL and VLN are out of phase by 30o
BALANCED Y-system
VLL = √3 VLN IL = IP
and
IL and IP are out of phase by 30o
Where: VLL or VL - line to line or line voltage
VLN or VP - line to neutral or phase voltage
IL - line current
IP - phase current
BALANCED ∆-system
IL = √3 IP VLL = VLN
BALANCED THREE PHASE SYSTEM
ALTERNATING CURRENT
VJH
watts vars va
BALANCED THREE PHASE SYSTEM
Note: for balanced 3-phase systems:
IA + IB + IC = 0
VAN + VBN + VCN = 0
VAB + VBC + VCA = 0
P = 3VPIPcos θ = √3 VLIL cos θ
Q = 3VPIPsin θ = √3 VLIL sin θ
S = 3VPIP = √3 VLIL
THREE-PHASE POWER
REVIEW QUESTIONS1. The description of two sine waves that are in step with each other going
through their maximum and minimum points ate the same time and in the same direction.
a. Sine waves in phaseb. Stepped sine wavesc. Phased sine wavesd. Sine waves in coordination
2. Term used for the out of phase, non-productive power associated with inductors and capacitors.a. Effective powerb. True powerc. Reactive powerd. Peak envelope power
3. Refers to a reactive power.a. Wattles, non productive powerb. Power consumed in circuit Qc. Power loss because of capacitor leakaged. Power consumed in wire resistance in an inductor
4. Term used for an out-of-phase, non-productive power associated with inductors and capacitors.a. Effective powerb. Reactive powerc. Peak envelope powerd. True power
REVIEW QUESTIONS
5. The product of current and voltage in an AC circuit refers to thea. Real powerb. Useful powerc. Apparent powerd. DC power
6. The distance covered or traveled by a waveform during the time interval of one complete cycle.a. Frequencyb. Wavelengthc. Time slotd. Wave time
REVIEW QUESTIONS
7. The power dissipated accross the resistance in an AC circuit.a. Real powerb. Reactive powerc. Apparent powerd. True power
8. It is the number of complete cycles of alternating voltage or current complete each seconda. Periodb. Frequencyc. Amplituded. Phase
REVIEW QUESTIONS
9. How many degrees are there in one complete cycle?a. 720 degb. 360 degc. 180 degd. 90 deg
10. The impedance in the study of electronics is represented by resistance and ___ .a. Reactanceb. Inductance and capacitancec. Inductanced. capacitance
REVIEW QUESTIONS
11. It is the current that is eliminated by a synchro capacitor?a. Magnetizing statorb. Lossc. Statord. Rotor
12. It is a rotaing sector that represent either current or voltage in an AC circuit.a. Resistanceb. Phasorc. Solar diagramd. velocity
REVIEW QUESTIONS
13. The relationship of the voltage accros an inductor to its current is described asa. Leading the current by 90 degb. Lagging the current by 90 degc. Leading the current by 180 degd. In phase with the current
14. Find the phase angle between the voltage across through the cicuit when Xc is 25ohms, R is 100 ohms and XL is 50 ohms.a. 76 deg with the voltage leading the currentb. 24 deg with the voltage lagging the currentc. 14 deg with the voltage lagging the currentd. 76 deg with the voltage lagging the current
REVIEW QUESTIONS
15. Calculate the period of an alternating current having a equation of I=20sin 120πta. 4.167 msb. 8.33 msc. 16.67 msd. 33.33 ms
16. What do you mean by root-mean-square (rms) value?a. It is the average valueb. It is the effective valuec. It is the value that causes the same heating effect as the DC voltaged. b or c
REVIEW QUESTIONS
17. The maximum instances value of a vrying current, voltage or power equal to 1.414 times the effective value of a sine wave.a. RMS valueb. Peak valuec. Effective valued. Peak to Peak value
18. If an AC signal has a peak voltage of 55V, what is the average value?a. 34.98 Vb. 61.05Vc. 86.34 Vd. 38.89 V
REVIEW QUESTIONS
19. If an AC signal has an average voltage of 18V, what is the rms voltage?a. 12.726 Vb. 19.980 Vc. 25.380 Vd. 16.213 V
20. A 220-V, 60Hz is driving a series RL circuit. Determine the current if R=100 ohms and 20 mH inductancea. 2.2 A (lagging)b. 2.0 A (lagging)c. 2.2 A (leading)d. 2.0 A (leading)
REVIEW QUESTIONS
21. Ignoring any inductive effects, what is the impedance of RC series capacitor made up of a 56K ohm resistor and a 0.33uF capacitor at a signal frequency of 4650 Hz.a. 66730 ohmsb. 57019 ohmsc. 45270 ohmsd. 10730 ohms
22.What is the time constant of a 500mH coil and a 3300 ohm resistor in series?a. 0.00015 secb. 6.6 secc. 0.0015 secd. 0.0000015 sec
REVIEW QUESTIONS
23. What is the realtionship between frequency and the value of XC ?
a. Frequency has no effectb. XC varies inversely with frequencyc. XC varies indirectly with frequencyd. XC varies directly with frequency
24. The reactance of a 25mH coil at 5000Hz which of the following?a. 785 ohmsb. 785000 ohmsc. 13 ohmsd. 0.0012 ohms
REVIEW QUESTIONS
25. There are no transients in pure resistive circuites becaus theya. Offer high resistanceb. Obey ohm’s Lawc. Are linear circuitsd. Have no stored energy
26. The reciprocal of capacitance is called a. Elastanceb. Conductancec. Permitivityd. permeability
REVIEW QUESTIONS
REVIEW QUESTIONS
27. The AC system is prefered over DC system becausea. Ac voltages can easily changed in amgnitudeb. Dc motors do not have fine speed controlc. High voltage AC transmission is less efficientd. DC voltage cannot be used for domestic aplliences
28. An altenating voltage is given by v = 20 sin 157 t. The frequency of the alternating voltage isa. 50 Hzb. 25 HZc. 100 Hzd. 75 Hz
REVIEW QUESTIONS29. An alternating current given by i = 10 sin 314 t. The time taken to generate two
cycles of current isa. 20 msb. 10 msc. 40 msd. 50 ms
30. In a pure resistive circuit, the instantaneous voltage and are current are given by:v=250 sin 314t i=10sin314t
The peak power in the circuit isa.1250 Wb. 25 Wc. 2500 Wd. 250 w
REVIEW QUESTIONS31. An average value of 6.63 A is _____ the effective value of 7.07 A.
a. The same areab. Less thanc. Greater than d. Any of these
32. In an R-L series circuit, the resistance is 10 ohms and the inductive reactance is 10 ohms. The phase angle between the applied voltage and circuit current will bea. 45 degb. 30 degc. 60 degd. 36.8 deg
REVIEW QUESTIONS33.An R-L series ac circuit has 15V across the resistor and 20V across the
inductor. The supply volatge isa. 35 Vb. 5 Vc. 25 Vd. 175 V
34. The active and reactive powers of an inductive circuit are equal. The power factor of the circuit isa. 0.8 lagging b. 0.707 laggingc. 0.6 laggingd. 0.5 lagging
REVIEW QUESTIONS35. A circuit when connected to 200V mains takes a current of 20 A, leading
the voltage by one-twelfth of time period. The circuit resistance isa. 10 ohmsb. 8.66 ohmsc. 20 ohmsd. 17.32 ohms
36. An AC circuit has a resistance of 6 ohms, inductive reactance of 20 ohms, and capacitive reactance of 12 ohms. The circuit power will bea. 0.8 laggingb. 0.8 leadingc. 0.6 laggingd. 0.6 leading
REVIEW QUESTIONS37. An alternating voltage of 80 + j60 V is applied to a circuit and the current
flowing is -40 + j10 A. Find the phase angle.a. 25 degb. 50 degc. 75 degd. 100 deg
38. A current wave is represented by the equation i = 10 sin 251t. The average and RMS value of current area. 7.07 A; 6.63Ab. 6.36A; 7.07Ac. 10A; 7.07Ad. 6.36A; 10A
REVIEW QUESTIONS39. Calculate the susceptance in mho of a circuit consisting of resistor of 10
ohms in series with a conductor of 0.1H, when the frequency is 50Hz.a. 0.0303b. 0.0092c. -0.029d. 32.95
40. An inductive circuit of resistance 16.5 ohms and inductive of 0.14H takes a current of 25 A. If the frequency is 50Hz, the supply voltage isa. 117.4 Vb. 1174 Vc. 1714 Vd. 1471 V
REVIEW QUESTIONS41. The current taken by a circuit is 1.2 A when the applied potential
difference is 250 V and the power taken is 135W. The power factor isa. 0.35b. 0.45c. 0.55d. 0.65
42. A capacitor has a capacitance of 20uF. The current supplied if it is placed across a 1100 V, 25 Hz supply.a. 3.554 Ab. 6.91 Ac. 3.45 Ad. 9.61 A
REVIEW QUESTIONS43. A non-inductive load takes a 100A at 100V. Calculate the inductance of the
inductor to be connected in series in order that the same current is supplied from 220 V, 50 Hz mains.a. 1.96 ohmsb. 6.91 ohmsc. 19.6 ohmsd. 9.61 ohms
44. An inductor having negligible resistance and an inductance of 0.07H is connected in series with a resiostor of 20 ohms resitance across a 200, 50 Hz supply. The maximum energy stored in the coil isa. 3.175 Jb. 1.585 Jc. 0.236 Jd. 0.33 J
REVIEW QUESTIONS45. A coil has 1200 turns and procedures 100 uWb mwhen the current
flowing is 1A. The inductance of the coil isa. 0.21 Hb. 0.12 Hc. 0.31 Hd. 0.41 H
46. A capacitor connected to a 115 V, 25 Hz supply takes 5 A. What current will it take when the capacitance and frequency are doubled?a. 2 Ab. 5 Sc. 10 Ad. 20 A
REVIEW QUESTIONS47. A resistor of 20 ohms is connected in parallel with a capacitor across a 110 V,
40 Hz supply. If the current taken is 6A, what is the capacitance?a. 88.6 uFb. 68.8 uFc. 86.8 uFd. 76.8 uF
48. What capacitance must be placed in series with an inductance of 0.05H, so that when the frequency is 100 Hz, the impedance becomes equal to the ohmic resitance?a. 70.5 uFb. 50.7 uFc. 5.7 uFd. 7.05 uF
REVIEW QUESTIONS49. A reactance of 20 ohms and inductance of 0.1 H is connected in parallel with a
capacitor. The capacitance of the capacitor required to produce a resonance when connected to a 100V, 50 Hz isa. 71.2 uFb. 1.277 uFc. 17.2 uFd. 72.1 uF
50. What is the resonant frequency of a circuit when an inductance of 1uH and capacitance of 10 picofarad are in series?a. 15.9 MHzb. 50.3 MHzc. 15.9 kHzd. 50.3 KHz
REVIEW QUESTIONS
51. The _____ the Q of a circuit, the narrower the bandwidth.a. Lowerb. Higherc. Broaderd. Selective
52. Find the half power bandwidth of a parallel resonant circuit which has a resonant frequency of 3.6MHz and Q of 218.a. 606 kHzb. 58.7 kHzc. 16.5 kHzd. 47.3 kHz
REVIEW QUESTIONS
53. At series resonance _____ .a. Circuit impedance is very largeb. Cicuit power factor is minimumc. Voltage across L or C is zerod. Circuit power factor is unity
54. At series resonance, the voltage across the inductor isa. Equal to the applied voltageb. Much more than the apllied voltagec. Equal to voltage across Rd. Less than the applied voltage
REVIEW QUESTIONS55. The Q factor of the coil is _____ the resistance of the coil.
a. Inversely proportional tob. Directly proportional toc. Independent ofd. None of these
56. An RLC circuit is connected is connected to 200V AC source. If Q factor of the coil is 10, then the voltage across the capacitor at resonance isa. 200 Vb. 2000 Vc. 20 Vd. 210 V
REVIEW QUESTIONS57. At parallel resonance
a. Circuit impedance is minimumb. Power factor is zeroc. Line current is maximumd. Power factor is unity
58. The dynamic impedance of parallel resonant circuit is 1 Mohms. If the capacitance is 1uF, and the resistance is 1ohm, then the value of the inductancea. 1 Hb. 10 Hc. 10 pHd. 10 uH
REVIEW QUESTIONS59. When supply frequency is less than the resonant frequency in a parallel
ac circuit, then the circuit isa. Resistiveb. Capacitivec. Inductived. None of these
60. At parallel resonance, the circuit drwas a current of 2mA. If the Q of the circuit is 100, then the current through the capacitor isa. 2 mAb. 1 mAc. 200 mAd. None of these
REVIEW QUESTIONS61. A circuit has an impedance of (1-j2) ohms. The susceptance of the
circuit in mho isa. 0.1 b. 0.2c. 0.4d. None of these
62. If the admittance of a parallel ac circuit increased, the circuit currenta. Remains constantb. Is increasedc. Is decreasedd. None of these
REVIEW QUESTIONS63. The resistance between any pair two terminals of a balanced wye-
connected load is 12 ohms.a. 6 ohmsb. 18 ohmsc. 24 ohmsd. None of these
64. If an AC circuit contains three nodes, the number of each mesh equations that can be formulated isa. 1b. 2c. 3d. 4
REVIEW QUESTIONS
65. The relation of the voltage across an inductor to its current is describe asa. Leading the current by 90 degb. Lagging the current by 90 degc. Leading the current by 180 degd. In phase with the current
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