ELECTRIC CIRCUITSCIRCUITS CIRCUITS A closed conducting path through which an electric current flows or is intended to flow Parameters The various elements of an electric circuit, like resistance, inductance, and capacitance which may be lumped or distibuted. CIRCUITS AND NETWORKS Bilateral Circuit Is one whose properties or characteristics are the same in either direction. CIRCUITS AND NETWORKS .e. They do not change with voltage and current. Non-Linear Circuit Is that circuit whose parameters change with voltage and current.CIRCUITS TYPES Linear Circuit Is one whose parameters are constant (i. Unilateral Circuit Is that circuit whose properties or characteristics change with the direction of its operation. CIRCUITS AND NETWORKS .ELECTRICAL NETWORKS ELECTRICAL NETWORK Connection of various electric elements in any manner TYPES Passive Network With no source of emf. Active Network Contains one or more than one sources of emf. ELECTRICAL NETWORKS PARTS 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 between two junctions. Mesh A loop that contains no other loop within it. Loop A closed path in a circuit in which no element or node is encountered more than once. CIRCUITS AND NETWORKS . current and resistance Developed in 1827 by German physicist Georg Simon Ohm CIRCUITS AND NETWORKS .OHM’S LAW OHM’S LAW One of the most fundamental law in electrical circuits relating voltage. OHM’S LAW According to Ohm’s Law. the current (I) flowing in an electrical circuit 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 Voltage Division for Series Circuit: E X = E T • RX RT 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 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 Voltage Division for Parallel Circuit: IX = IT • Req RT 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 SERIES/PARALLEL CIRCUITS . the algebraic sum of the current meeting at a point (or junction) is zero.” NETWORK ANALYSIS .KIRCHHOFF’S LAW 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) “In any electrical network. IA = 0 NETWORK ANALYSIS . positive current ⁻ Current away from the node. negative current IB + IC + ID = IA (IB + IC + ID) .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. ” NETWORK ANALYSIS .KIRCHHOFF’S VOLTAGE LAW Kirchhoff’s Voltage Law (KVL) “The algebraic sum of the products of currents and resistances in each of the conductors in any closed path (or mesh) in a network PLUS the algebraic sum of the emfs in the path is zero. NETWORK ANALYSIS . the sum of the voltages around the loop is equal to zero ⁻ For voltage sources. negative voltage. if the loop direction is the same as current direction. ⁻ For voltage drops. positive voltage and if loops enters on plus and goes out on minus. if loops enters on minus and goes out on plus.KIRCHHOFF’S VOLTAGE LAW In short. positive voltage drop. negative voltage drop and if the loop direction is opposite to the current direction. Current sources provide constraint equations. they are used to define branch currents. Loop Analysis Procedure: 1. a. b. Label each of the loop/mesh currents. NETWORK ANALYSIS . 2. we may write N total equations using KVL around each loop. Loop currents are those currents flowing in a loop. Solve the equations to determine the user defined loop currents. For N independent loops. Apply KVL to loops/meshes to form equations with current variables.MESH ANALYSIS MESH analysis A sophisticated application of KVL with mesh currents. 3. PRINCIPAL node – a node with three or more circuit elements joined together.NODAL ANALYSIS NODAL analysis A systematic application of KCL at a node and after simplifying the resulting KCL equation. Consist of finding the node voltages at all principal nodes with respect to the reference node. Reference node – the node from which the unknown voltages are measured. the node voltage can be calculated. NETWORK ANALYSIS . To consider effects of each source independently.SUPERPOSITION THEOREM 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. ” In general: Number of network to analyze is equal to number of independent sources. sources must be removed and replaced without affecting the final result: All voltage sources >> short circuited All current sources >> open circuited NETWORK ANALYSIS . Millman’s theorem is applicable only to those cicuits which can be re-drawn accordingly. the circuit is re-drawn as a parallel network of branches.MILLMAN’S THEOREM 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. ” In Millman’s Theorem. each branch containing a resistor or series battery/resistor combination. NETWORK ANALYSIS . MAXIMUM POWER TRANSFER THOREM 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 NETWORK ANALYSIS . 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 . ” where: VTH – the open circuit voltage which appears across the two terminals from where the load resistance has been removed.THEVENIN’S THEOREM 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 THEVENIN’S theorem “ Any two-terminal active network containing voltage sources and resistances when viewed from its output terminals. NETWORK ANALYSIS . RN – the resistance of the network when viewed from the open circuited terminals after all voltage sources being replaced by open circuits. ” where: IN– the current which would flow in a short circuit placed across the 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 . NETWORK ANALYSIS .EQUIVALENT THREE-TERMINAL NETWORKS 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 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. NETWORK ANALYSIS . . 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.ALTERNATING CURRENT ALTERNATING CURRENT A current that is constantly changing in amplitude and direction. BASIC AC THEORY . three phase for power distribution and large motor loads or six phase for large scale AC to DC conversion. two phase for control motors. AC WAVEFORMS BASIC AC THEORY . 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 BASIC AC THEORY . 1 cycle/second (cps) = 1 Hertz (Hz) b. Proper operation of electrical equipmnent requires specific frequency c.AC WAVEFORMS Parameters of Alternating Signal f = 1 T λ = v f Period (T) – the time of one complete cycle in seconds Frequency (f) – the number of cycles per second (Hertz) a. 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: A(t) = Am sin (ωt + Φ) 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 = 2πf t – time (sec) Φ – phase shift ( + or – in degrees) BASIC AC THEORY . or RMS quantity) 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.AC WAVEFORMS Measurements of AC Magnitude AMPLITUDE It is the height of an AC waveform as depicted on a graph over time (peak. BASIC AC THEORY . Also known as the crest amplitude of a wave. average. peak-to-peak. 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 . the average value so calculated is approximately 0. For a sine wave.637 of its peak value. Technically. BASIC AC THEORY . the average amplitude of any waveform with equal-area portions above and below the “zero” line on a graph is zero.AC WAVEFORMS AVERAGE AMPLITUDE – the mathematical “mean” of all a waveform’s points over the period of one cycle. and is a way of expressing an AC quantity of voltage or current in terms functionally equivalent to DC. For a sine wave. BASIC AC THEORY .AC WAVEFORMS RMS AMPLITUDE . the RMS value is approximately 0.“RMS” stands for Root Mean Square. Also known as the “equivalent” or “DC equivalent” value of an AC voltage or current.707 of its peak value. 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 BASIC AC THEORY AC QUANTITIES Inductive Reactance (XL) • The property of the inductor to oppose the alternating current Inductive Susceptance (BL) • Reciprocal of inductive reactance XL = 2πfL BL = 1 2πfL BL = 1 XL Capacitive Reactance (XC) • The property of a capacitor to oppose alternating current XC = Capacitive Susceptance (BC) • Reciprocal of capacitive reactance 1 2πfC BL = 2πfC d BL = 1 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 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. then Z = V = Vm ∠θ = Z ∠φ I Im ∠β Z cosφ + jZsin φ = R + jX = √ R2 + X2 ∠ tan-1 X R Where: Z = Vm = √ R2 + X2 = magnitude of the impedance Im φ = θ – β = tan-1 X = phase angle of the impedance R R = Zcos φ = active or real component of the impedance X = Zsin φ = reactive or quadrature component of impedance BASIC AC THEORY . linear RLC circuit from a source voltage V = Vm ∠θ. AC QUANTITIES ADMITTANCE (Y) The reciprocal of impedance Expressed in siemens or mho (S) Y = Im ∠ β – θ = Y = Ycos φy + jYsin φy = G + jB Vm Y = √ G2 + B2 ∠ tan-1 B G 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 BASIC AC THEORY . a resistor still dissipated energy. When the current is positive (above the line). AC CIRCUITS .AC RESISTOR CIRCUIT Impedance (Z) = R With an AC circuit like this which is purely resistive. taking it from the source and releasing it in the form of heat energy. Whether the current is negative or positive. resulting in a power (p=ie) of a positive value This consistent “polarity” of a power tell us that the resistor is always dissipating power. 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. the voltage is also positive. resulting in equally frequent occurences of negative instantaneous power. while a positive power means that it is absorbing power from the circuit. there sre times when one is positive while the other is negative. AC CIRCUITS . Negative power means that the inductor is releasing power back to the circuit.AC INDUCTOR CIRCUIT Impedance (Z) = jXL 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. The inductor releases just as much power back to the circuit as it absorbs over the span of a complete cycle. 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. o Inductive reactance increases with increasing frequency. AC CIRCUITS . It is symbolized by the lowercase Greek letter “omega. the higher the frequency. Reactance is symbolized by the capital letter “X” and is measured in ohms just like resistance (R). In other words. the more it opposes the AC flow of electrons.” or ω. in units of electrical radians per second instead of cycles per second.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. AC CAPACITOR CIRCUIT Impedance (Z) = -jXC 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. For any given magnitude of AC voltage at a given frequency. 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 AC CIRCUITS . which is by definition always changing in instantaneous magnitude and direction. a capacitor of given size will “conduct” a certain magnitude of AC current. o Capacitive reactance can be calculated using this formula: XC = 1/(2πfC) o Capacitive reactance decreases with increasing frequency.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. AC CIRCUITS . In other words. the less it opposes (the more it “conducts”) the AC flow of electrons. the higher the frequency. Reactance is symbolized by the capital letter “X” and is measured in ohms just like resistance (R). 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-INDUCTOR CIRCUIT For a series resistor-inductor circuit. the voltage and current relation is determined in its phase shift. Thus. the total impedance will have a phase angle somewhere between 0o and +90o. AC CIRCUITS . 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.SERIES RESITOR-INDUCTOR CIRCUIT Phase shift (θ) = Arctan ( XL ) |Z| = √ R2 + jXL2 = e R i o When resistors and inductors are mixed together in circuits. 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 For a series resistor – capacitor circuit. SERIES RESISTOR-CAPACITOR CIRCUIT Phase shift (θ) = Arctan ( XC ) |Z| = √ R2 + jXC2 = e R i AC CIRCUITS . by Ohm’s Law I The basic approachwith regards to parallel circuits is using admittance because it is additive AC CIRCUITS .PARALLEL RESISTOR-INDUCTOR Y = G – jβL where: G – conductance = 1/R βL – susceptance = 1/XL Z = E . 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. The circuit current will have a phase angle somewhere between 0o and -90o. and impedances diminish (through the reciprocal formula) to form the total impedance. the total impedance will have a phase angle somewhere between 0o and +90o. AC CIRCUITS .PARALLEL RESISTOR-INDUCTOR o When resistors and inductors are mixed together in parallel circuits (just like in series cicuits). AC CIRCUITS .PARALLEL RESISTOR-CAPACITOR Y = G + jβC where: G – conductance = 1/R βC – susceptance = 1/XC o When resistors and capacitors are mixed together in circuits. the total impedance will have a phase angle somewhere between 0o and -90o. the Real Power and the Capacitive or Inductive Reactive Power S = Vrms Irms = Irms2 |Z| POWER IN 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. 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 P = Scos θ Power Factor Cosine of the power factor angle (θ) Measure of the power that is dissipated by the circuit in relation to the apparent power and is usually given as a decimal or percentage Pf = cos θ POWER IN AC CIRCUITS . 0 Pf = lagging Pf = leading Pf = 0.0 lag Pf = 0. inductive system I leads V by θ. purely inductive I leads V by 90o. purely capacitive Power factor Angle (θ) 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 . capacitive system I lags V by 90o. resistive system I lags V by θ.REAL POWER (R) Ratio of the Real Power to the Apparent Power ( P S ) when: Pf = 1.0 lead I is in phase with V. REAL POWER (R) Instantaneous Power (watts) P(t) = v(t) i(t) P(t) = ½ VmIm cos (θv – θi) + ½ VmIm cos (2ωt + θv + θi) Average Power (watts) P(t) = ½ VmIm cos (θv – θi) = VmIm cos θ Where: θ = phase shfit between v(t) and i(t) or the phase angle of the equivalent impedance POWER IN AC CIRCUITS . non-productive or wattless power Measured in volt-ampere reactive (Var) When the capacitor and inductor are both present.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. 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) Q = VmIm sin θ Reactive factor Ratio of the Reactive Power to the Apparent Power Sine of the power factor angle (θ) Rf = sin θ POWER IN AC CIRCUITS . In a series ciscuit a resistors 2200 and 4500 with an impressed voltage of 10.49 b. 6. 5 ohms d. A battery with a rating of 9 volts has an internal resistance of 20 ohms.REVIEW QUESTIONS 1.34 . 10 ohms c.34 d.67 c. 1. What is the expected resistance of the bulb that is connected across the battery to attain a maximum power transfer? a. 4. 20 ohms b. 2 ohms 2. what is the circuit current in mA? a. 1. 66.182 A 4. 264500 d.5 b. 18. What is the power in KW at the time of reading? a.REVIEW QUESTIONS 3. 0. The current needed to operate a soldering iron which has a rating of 600 watts at 110 volts is.200 A d. The ammeter reads 230 ampere while the voltmeter is 115 volts.000 A c.455 A b. a. 2645 c. 264. 5.26.45 . What is the type of circuit whose parameters are constant which do not change with voltage or current? a. Reactive d. The sum of their reciprocal c.REVIEW QUESTIONS 5. What is the resistance of two equal valued resistor in series? a. Lumped b. The product of both . Tuned c. Linear 6. The difference of both d. Twice as one b. Provide lighter current c. 1 watt? a. Provide less power d. Provide wider tolerance 8. 12 V d. a. Provide more power b. 1 watt resistors in parallel instead of one 10 kohms. 9V c. The voltage applied in DC circuit having a power of 36 watts and a total resistance of 4 ohms. 24 V .REVIEW QUESTIONS 7. What do you expect when you use two 20 kohms. 6 V b. 666 ohms b. When resistor are connected in series. The effective resistance is increased 10.6 ohms d. Nothing c. 66. The effective resistance is decreased b. what happens? a.REVIEW QUESTIONS 9.66 ohms c. The tolerance is d. 6666 ohms . 6. 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. Two c. 66 W c.REVIEW QUESTIONS 11. How many nodes are needed to completely analyze a circuit according to Kirchhoff’s Current Law.5 W 12. a. One b. How much power does electronic equipment consume. All nodes in the circuit d.5A current flowing and a 120-V power source. assuming a 5. One less than the total number of nodes in the circuit . 60 W b. 125. a. 660 W d. Mesh 14. It is used to denote a common electrical point of zero potential. Ground d. a. A common connection between circuit elements or conductors from different branches.REVIEW QUESTIONS 13.ground . Node b. Junction c. a. Reference point c. Open circuit d. Short circuit b. Millman’s theorem .REVIEW QUESTIONS 15. KCL b. KVL c. Either b or c 16. we apply: a. Counterclockwise d. In mesh analysis. Loop currents should be assumed to flow in which direction? a. Straight b. Source d. Clockwise c. R = E/I 18.REVIEW QUESTIONS 17. E = IR c. I = E/R d. Which of the following is not a valid expression of Ohms Law a. Using ohms Law. Current increases c. R = PI b. what happens to the circuit current if the applied voltage increases? a. Current decreases . Current doubles b. Current remians constant d. Be quadrupled b. Remains c. Be doubled . According to ohms law. Be cut in half d. Current increases d. If the resistance of a circuit is doubled and the applied voltage is kept constant. Current remains constant 20.REVIEW QUESTIONS 19. Current decreases c. a. what happen to the circuit current if the circuit resistance increases? a. the current will be _______ . Current doubles b. Power . In Ohms Law. Direct current d. Normal current b. Eddy current 22. Amperage b. what is E/R? a.REVIEW QUESTIONS 21. It is an electrical current that flows in one direction only? a. Voltage c. Alternating current c. Resistance d. What is the total combined resistance of the three resistors? a. Resistance d. 95000 ohms 24. Reactance . A 33-Kohm resistor is connected in series with a parallel combination made up of 56-Kohm resistor and a 7. 49069 ohms c.REVIEW QUESTIONS 23. Current source b. Which of the following cannot be included in a loop of Kirchhoff’s Voltage Law a. 63769 ohms d.8-kohm resistor. 390667 ohms b. Voltage source c. None of the above . 25 watts c. A conductor d. 75 watts d. It was reconfigured such that 2 are in parallel and the other load is in series with a combination. An insulator c. Semiconductor b.REVIEW QUESTIONS 25. 50 watts b. the substance is usually a. What is the applied expected powers to be consumed then? a. If the number of valence electrons of an atom is less than 4. A series connected circuit consists of 3 loads and consume a total power of 50 Watts. 45 watts 26. Electric current in a wire is the flow of a. Atoms 28. Energy c. Valence electrons c. Power b. none . Charge d. Free electrons b. EMF in a circuit is a form a.REVIEW QUESTIONS 27. Bound electrons d. -m d. The SI unit of specific-resistance is a. Directly proportional b. Independent d. Ohm-m c. None of these . Inversely proportional c. The resistance of a material is ___ its area of cross-section a.REVIEW QUESTIONS 29. Mho b. Ohm-cm 30. Ohm-sq. a. Volume c. 1/265 C .REVIEW QUESTIONS 31. i. 1/272 C c. X-sectional area d.. Length b. the temperature coefficient of resistance depends upon the ____ of the material. 1/254 C d. Nature and temperature 32. The value of α.e. The value of α of a conductor is 1/236 C. 1/218 C b. The value of a α is a. Makes the operation of the appliances independent from each other . Electrical appliances are connected in parallel because it a. Results in reduced power loss c. Electrical appliances are not connected in series because a. Appliances have different current rating d.REVIEW QUESTIONS 33. Series circuit is complicated b. None of these 34. Power loss is greater c. Is a simple circuit b. Draw less current d. 625 ohms c. When a number of resistances are connected in parallel.5 ohms b. None of these . the total resistance is a. Greater than the smallest resistance c. None of these 36. 2. The hot resistance of a 100W. Less than the smallest resistance b.REVIEW QUESTIONS 35. 250V incandecent lamp is a. Between the smallest and greater resistance d. 25 ohms d. 54 ohms c. 600 ohms . 200 ohms c. None of these 38.REVIEW QUESTIONS 37. 36 ohms d. each of value 36 ohms are connected in parallel. The total resistance will be a. the total resistance is a. If the resistances. Two incandecent lamps of 100 W. 2 ohms b. 800 ohms b. 200V are in parallel across the 200 V. 400 ohms d. 5A. 73. 2 A c. If the applied voltage is 210V. what is the total resistance of the circuit? a. 1/8 A d. An ordinary dry cell can deliver about ____ continuously. None of these . 2.REVIEW QUESTIONS 39. 3 A b. respectively. a. 5 ohms b.5A. 147 ohms c. and 3.5 ohms 40. 3 ohms d. Three resistors are connected in parallel and draws 1A. REVIEW QUESTIONS 41. 4 ohms b. Ampere-ohm d. 2 ohms d. 1 ohms 42. Joule/sec b. the one that is not equal to the watt is a. are connected in parallel. Ohm/volt . Ampere-volt c. Four cells of internal resistance 1 ohms.25 ohms c. The battery resistance will be a. Of the following combination of units. 0. 800 ohms . An electric iron draws a current of 15A when connected to 120V power source. IR/V 44.REVIEW QUESTIONS 43. V/R d. VI b. 16 ohms d.125 ohms b. Its resistance is a. The power dissipated in a circuit is not equal to a. IR c. 0. 8 ohms c. 1. 0. 0.5 A d. 240V.080 W 46. 540 W d. 0.42 A c. the current that flows in it is a. 0. 1. When a 100W.35 A b.58 A . light bulb is operated at 200V.REVIEW QUESTIONS 45. 40 W b. 360 W c. The power rating of an electric motor which draws a current of 3 A when operated at 120 V is a. A 12V potential difference is applied across a series combination of four six-ohms resistors. 3 ohms d.REVIEW QUESTIONS 47. 1.5 A b. 8 A d. 0. 6 ohms 48.67 ohms b. 2 A c. The current in each six-ohm resistor will be a.5 ohms c. 0. 6 A . The equivalent resistance of a network of three 2 ohm resistors cannot be a. The current in each six-ohm resistor will be a. 0. Friction d. 6 A 50. Heat b.REVIEW QUESTIONS 49.5 A b. 2 A c. 8 A d. Fire c. Overload . The dissipation of energy can cause burns because it produces a. A 12V potential difference is applied across a parallel combination of four six-ohms resistors. the wire that carries the current from the generator to the load is called a. Current d. Voltage b. The rate of expenditure of energy is a. Energy 52. Outgoing wire d. In a simple DC power line. Power c. Conductor . Return wire b.REVIEW QUESTIONS 51. Feeder c. e. i.. a. The current of the circuit c. The EMF of the battery b. A battery is connected to an external circuit. A circuit in which the resistance are connected in a continuous run.REVIEW QUESTIONS 53. The equivqlent circuit resistance d. None of these 54. Series b. The potential drop with the battery is proportional to a. Power dissipated in the circuit . Series-parallel d. Parallel c. end-to-end is a _____ circuit. At 100 C.1 c. 0. The number of times A is longer than B is a. 0. Ra = 600 ohms and Rb = 100 ohms. 4 d. 6 b.001 . its resistance is 101 ohms.0001 d. 0. A coil has a resistance of 100 ohms at 90 C.REVIEW QUESTIONS 55. 5 56. The temperature coefficient of the wire is a. 2 c. 0.01 b. Two wires A and B have the same cross-sectional area and are made of the same material. Conductor c. Semi-conductor d. Mass 58. A material which has a negative temperature coefficient of resistance is usually a/an a. Length c.REVIEW QUESTIONS 57. The resistance of a conductor does not depend on its a. Insulator b. Cross-section d. All of these . Resistivity b. Voltage are additive d. Which of the following statements is true both for a series and a parallel dc circuit? a. Two resistors are said to be in series when a. Currents are additive c. Sum of IR drops equal to EMF d. Both carry the same value of current b. Same current phases through both . Powers are additive b. All of these 60.REVIEW QUESTIONS 59. Total current equals the sum of the branch current c. Electric circuit .REVIEW QUESTIONS 61. Electronic circuits c. Algebraic sum of the currents meeting at the juction is zero 62. According to KCL as applied to a juction in a network of conductors. a. Total sum of currents meeting at the juction is Zero b. Net current flow at he juction is positive d. No current can leave the juction without same current passing throuhg it c. Closed-loop circuit b. Juctions in a network d. Kirchoff’s Current Law is applicable only to a. According to KVL.REVIEW QUESTIONS 63. Determined by battery EMF . Kirchoff’s Voltage Law is concerned with a. the algebraic sum of all IR drops and EMF’s in any closed loop of a network is always a. Zero b. Battery EMF’s c. Juction voltages d. Positive d. a and b 64. IR drops b. Negative c. Choose the wrong statement. Direction of current c. Affect the volatge of various nodes . In the node voltage technique of solvingnetwork parameters. The value of the resistance d. Alter the potential difference between any pair of nodes d. The battery connection 66. The algebraic sign of an IR drops primarily dependent upon a. Change the voltage across the element c. Affect the operation of the circuit b. the choice of the reference node does not a. The amount of current flowing through it b.REVIEW QUESTIONS 65. Linear bilateral d. Non-linear b. KCL. b and c 68. Ohms Law d.REVIEW QUESTIONS 67. KVL b. Resistive . The nodal analysis is primarily based on the application of a. Passive c. Superposition theorem can be applied only to circuits having ____ elements a. c. The superposition theorem is essentially based on the concept of a. Electric current c. Electron velocity d. Dynamic electricity . What are the electrons in motion called? a.REVIEW QUESTIONS 69. Reciprocity b. Duality 70. non-linearity d. Linearity c. Current variation b. A number of voltage source c. Passive element only d. None of these 72. An active element in a circuit is one which _____ . None of these .REVIEW QUESTIONS 71. a. Supplies energy c. Receives energy b. The superposition theorem is used when the circuit contains a. A single voltage source b. a or b d. Both a and b d. Current c. Both a and b d. a. None of these 74. Thevenin’s theorem is ____ form of equivalent circuit. Current c. None of these . Norton’s theorem is ____ form of an equivalent circuit. Voltage b.REVIEW QUESTIONS 73. a. Voltage b. None of these . Thevenin’s c. In the analysis of vacuum tube circuit. Norton’s b. Both a and b d. a. Voltage b.REVIEW QUESTIONS 75. a. we generally use _____ theorem. Reciprocity 76. we can generally use ____ theorem. Superposition d. Current c. In the analysis of transistor circuits. Under the condition of Maximum Power Transfer. 75 % b. Power sytem d. None of the above .REVIEW QUESTIONS 77. Home lighting c. 100 % c. Electronic circuits b. 25 % 78. the efficiency is a. 50 % d. The maximum power transfer theorem is used in a. Three terminal d. Lightning rod . None of these 80. a. _____ will be used under elctrostatics. Electric iron d. delta/star or star/delta transformation technique is applied to a. Electric motor c. Two terminal c.REVIEW QUESTIONS 79. Incandecent lamp b. One terminal b. 8 x10 N b. 9 X 10 82.854 x 10 e -12 c. a. 5 x 10 d. 10 N c.REVIEW QUESTIONS 81. When two similar charges each of 1 coulumb each are placed 1m apart in air. 9 x 10^9 N d. 5 x 10 N . then the force of repulsion is a. 9 x 10 b. The value of the absolute permitivity of air is ______ F/m. 8. 5 sec b. Breakdown voltage c.5 sec c. 7.000 J of heat energy is 10 sec. generates 10. Another name for dielectric strength is a. Potetial gradient b. Electric intensity 84. . 4 sec.REVIEW QUESTIONS 83. 2. Dielectric constant d. A heater connected to a 100 V supply. How much time is needed to generate the same amount of heat when it is used on 220V line? a.5 sec d. Nodal 86. Kirchoff’s current law is applied in what type of circuit analysis? a. Superposition d. It is a simple circuit c. None of the above .REVIEW QUESTIONS 85. Thevenin’s c. Mesh b. Inductance and capacitance are not relevant in a dc circuit because a. Frequency of DC is zero b. They do not exist in dc circuit d. All of these . High power d. 3 ohms c. Cells are conneted in series when _____ is required a.REVIEW QUESTIONS 87. 0. 1 ohm b. High current b.111 ohm 88. High voltage c. 9 ohms d. Three resistors of 3 ohms resistance each are connected in delta. the equivalent wye-connected resistors will be a. Reactive power d. a. True power c. Sine waves in phase b. Sine waves in coordination 2. Peak envelope power . Effective power b. Term used for the out of phase.REVIEW QUESTIONS 1. Stepped sine waves c. non-productive power associated with inductors and capacitors. a. Phased sine waves d. The description of two sine waves that are in step with each other going through their maximum and minimum points are the same time and in the same direction. REVIEW QUESTIONS 3. Peak envelope power d. Refers to a reactive power. Wattless. Power consumed in wire resistance in an inductor 4. Real power c. a. Power loss because of capacitor leakage d. True power . Effective power b. Power consumed in circuit Q c. It is equal to the product of the apparent power and the power factor. a. non productive power b. The distance covered or traveled by a waveform during the time interval of one complete cycle. The product of current and voltage in an AC circuit refers to the a. Time slot d. a. Wavelength c. Useful power c. Wave time .REVIEW QUESTIONS 5. Frequency b. Apparent power d. DC power 6. Real power b. Reactive power c. The power dissipated accross the resistance in an AC circuit. a. Frequency c. It is the number of complete cycles of alternating voltage or current complete each second a. Period b. Real power b. Phase .REVIEW QUESTIONS 7. Amplitude d. Apparent power d. True power 8. The impedance in the study of electronics is represented by resistance and ___ . Reactance b.REVIEW QUESTIONS 9. 360 deg c. Inductance d. Inductance and capacitance c. 720 deg b. 180 deg d. 90 deg 10. a. How many degrees are there in one complete cycle? a. capacitance . Stator d. a. velocity . Solar diagram d. Loss c. Resistance b. Rotor 12.REVIEW QUESTIONS 11. Magnetizing stator b. It is a rotaing sector that represent either current or voltage in an AC circuit. It is the current that is eliminated by a synchro capacitor? a. Phasor c. Leading the current by 180 deg d.REVIEW QUESTIONS 13. Find the phase angle between the voltage across through the cicuit when Xc is 25ohms. Lagging the current by 90 deg c. 76 deg with the voltage lagging the current . Leading the current by 90 deg b. In phase with the current 14. 76 deg with the voltage leading the current b. 14 deg with the voltage lagging the current d. R is 100 ohms and XL is 50 ohms. The relationship of the voltage accros an inductor to its current is described as a. 24 deg with the voltage lagging the current c. a. 33 ms c.33 ms 16. b or c .REVIEW QUESTIONS 15. 33. 4. It is the effective value c. It is the value that causes the same heating effect as the DC voltage d. It is the average value b. What do you mean by root-mean-square (rms) value? a.67 ms d. Calculate the period of an alternating current having a equation of I=20sin 120πt a. 16.167 ms b. 8. 38. Peak to Peak value 18. The maximum instances value of a varying current. If an AC signal has a peak voltage of 55V.REVIEW QUESTIONS 17. what is the average value? a.34 V d.414 times the effective value of a sine wave.98 V b.05V c. voltage or power equal to 1.89 V . 34. Peak value c. a. 86. Effective value d. 61. RMS value b. 19. A 220-V. 2. If an AC signal has an average voltage of 18V.2 A (leading) d. 12.0 A (leading) . what is the rms voltage? a.REVIEW QUESTIONS 19.380 V d. 2. 2. 25.980 V c. 16.2 A (lagging) b.726 V b. 2.213 V 20. Determine the current if R=100 ohms and 20 mH inductance a. 60Hz is driving a series RL circuit.0 A (lagging) c. 57019 ohms c. 0. what is the impedance of RC series capacitor made up of a 56K ohm resistor and a 0. 0.REVIEW QUESTIONS 21. Ignoring any inductive effects. a. 45270 ohms d.33uF capacitor at a signal frequency of 4650 Hz.0015 sec d. 0.What is the time constant of a 500mH coil and a 3300 ohm resistor in series? a.6 sec c. 10730 ohms 22.00015 sec b. 66730 ohms b. 6.0000015 sec . 0012 ohms . What is the relationship between frequency and the value of XC ? a. XC varies indirectly with frequency d. 785000 ohms c.REVIEW QUESTIONS 23. XC varies inversely with frequency c. 13 ohms d. XC varies directly with frequency 24. 0. Frequency has no effect b. Which of the following is the reactance of a 25mH coil at 5000Hz? a. 785 ohms b. Permitivity d. permeability . The reciprocal of capacitance is called a. Offer high resistance b. Conductance c.REVIEW QUESTIONS 25. Are linear circuits d. Elastance b. Obey ohm’s Law c. Have no stored energy 26. There are no transients in pure resistive circuites because they a. DC voltage cannot be used for domestic aplliences 28. Ac voltages can easily changed in magnitude b. 25 HZ c. 75 Hz . 50 Hz b. An alternating voltage is given by v = 20 sin 157 t. Dc motors do not have fine speed control c. The AC system is preferred over DC system because a. The frequency of the alternating voltage is a. 100 Hz d.REVIEW QUESTIONS 27. High voltage AC transmission is less efficient d. REVIEW QUESTIONS 29.1250 W b. The time taken to generate two cycles of current is a. 50 ms 30. 40 ms d. 10 ms c. 2500 W d. 25 W c. In a pure resistive circuit. 250 w . An alternating current given by i = 10 sin 314 t. 20 ms b. the instantaneous voltage and current are given by: v=250 sin 314t i=10sin314t The peak power in the circuit is a. REVIEW QUESTIONS 31. An average value of 6.63 A is _____ as the effective value of 7.07 A. a. The same area b. Less than c. 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 be a. 45 deg b. 30 deg c. 60 deg d. 36.8 deg REVIEW QUESTIONS 33.An R-L series ac circuit has 15V across the resistor and 20V across the inductor. The supply volatge is a. 35 V b. 5 V c. 25 V d. 175 V 34. The active and reactive powers of an inductive circuit are equal. The power factor of the circuit is a. 0.8 lagging b. 0.707 lagging c. 0.6 lagging d. 0.5 lagging REVIEW QUESTIONS 35. 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 is a. 10 ohms b. 8.66 ohms c. 20 ohms d. 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 be a. 0.8 lagging b. 0.8 leading c. 0.6 lagging d. 0.6 leading REVIEW QUESTIONS 37. 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 deg b. 50 deg c. 75 deg d. 100 deg 38. A current wave is represented by the equation i = 10 sin 251t. The average and RMS value of current are a. 7.07 A; 6.63A b. 6.36A; 7.07A c. 10A; 7.07A d. 6.36A; 10A 1174 V c. 0.14H takes a current of 25 A. when the frequency is 50Hz.1H.0092 c.95 40.4 V b. 0. 1714 V d. If the frequency is 50Hz. An inductive circuit of resistance 16. 117.5 ohms and inductive of 0. the supply voltage is a. Calculate the susceptance in mho of a circuit consisting of resistor of 10 ohms in series with a conductor of 0. 1471 V . -0. a.REVIEW QUESTIONS 39. 32.029 d.0303 b. 3. The current supplied if it is placed across a 1100 V.45 c.55 d. a. The current taken by a circuit is 1. 0.35 b. 25 Hz supply. A capacitor has a capacitance of 20uF.2 A when the applied potential difference is 250 V and the power taken is 135W. 0. 3.REVIEW QUESTIONS 41. 6. 0.91 A c.65 42.554 A b.45 A d.61 A . 9. The power factor is a. 0. 1. 0. 9.585 J c.236 J d.61 ohms 44.175 J b. The maximum energy stored in the coil is a.96 ohms b. 19. A non-inductive load takes a 100A at 100V.33 J . a. 6. 50 Hz mains. 0. 3. Calculate the inductance of the inductor to be connected in series in order that the same current is supplied from 220 V. 1.91 ohms c.REVIEW QUESTIONS 43. 50 Hz supply. An inductor having negligible resistance and an inductance of 0.07H is connected in series with a resiostor of 20 ohms resitance across a 200.6 ohms d. 2 A b. 10 A d.12 H c.31 H d. The inductance of the coil is a.41 H 46. 0. 0. A coil has 1200 turns and produces 100 uWb when the current flowing is 1A.21 H b. 20 A . 0.REVIEW QUESTIONS 45. What current will it take when the capacitance and frequency are doubled? a. A capacitor connected to a 115 V. 0. 25 Hz supply takes 5 A. 5 S c. If the current taken is 6A. 76. 86. 68. what is the capacitance? a.REVIEW QUESTIONS 47. What capacitance must be placed in series with an inductance of 0.7 uF d.8 uF c. 88. 70. 40 Hz supply. A resistor of 20 ohms is connected in parallel with a capacitor across a 110 V. the impedance becomes equal to the ohmic resitance? a. so that when the frequency is 100 Hz. 5.5 uF b.05 uF .05H.6 uF b.8 uF 48.8 uF d. 50.7 uF c. 7. 15.3 MHz c.1 H is connected in parallel with a capacitor. The capacitance of the capacitor required to produce a resonance when connected to a 100V. A reactor of resistance of 20 ohms and inductance of 0. 50 Hz is a.2 uF d. 50. 17.9 kHz d.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 MHz b.3 KHz . 1.277 uF c. 50. 72.REVIEW QUESTIONS 49.2 uF b. 71. 3 kHz .6MHz and Q of 218. 47. 606 kHz b. the narrower the bandwidth. Broader d. Lower b. 58. 16.7 kHz c. a.5 kHz d. Find the half power bandwidth of a parallel resonant circuit which has a resonant frequency of 3. The _____ the Q of a circuit. a. Higher c. Selective 52.REVIEW QUESTIONS 51. Equal to the applied voltage b. Circuit power factor is minimum c. At series resonance _____ . the voltage across the inductor is a. a. Circuit power factor is unity 54. Voltage across L or C is zero d. Much more than the applied voltage c.REVIEW QUESTIONS 53. At series resonance. Less than the applied voltage . Circuit impedance is very large b. Equal to voltage across R d. None of these 56. 210 V . Independent of d. then the voltage across the capacitor at resonance is a.REVIEW QUESTIONS 55. The Q factor of the coil is _____ the resistance of the coil. 20 V d. An RLC circuit is connected is connected to 200V AC source. a. 200 V b. Inversely proportional to b. If Q factor of the coil is 10. Directly proportional to c. 2000 V c. 1 H b. 10 H c. If the capacitance is 1uF. 10 pH d. and the resistance is 1ohm.REVIEW QUESTIONS 57. Line current is maximum d. Power factor is unity 58. 10 uH . The dynamic impedance of parallel resonant circuit is 1 Mohms. Power factor is zero c. Circuit impedance is minimum b. At parallel resonance a. then the value of the inductance is a. At parallel resonance. Resistive b. If the Q of the circuit is 100. 1 mA c. When supply frequency is less than the resonant frequency in a parallel ac circuit. 200 mA d. the circuit drwas a current of 2mA. Inductive d. then the circuit is a. 2 mA b. Capacitive c. None of these . then the current through the capacitor is a. None of these 60.REVIEW QUESTIONS 59. 0. If the admittance of a parallel ac circuit increased. 0. Is increased c.4 d.REVIEW QUESTIONS 61. 0. None of these 62.1 b. None of these . The susceptance of the circuit in mho is a.2 c. Is decreased d. A circuit has an impedance of (1-j2) ohms. Remains constant b. the circuit current a. REVIEW QUESTIONS 63. 3 d. None of these 64. 6 ohms b. 24 ohms d. a. 4 . The resistance between any pair two terminals of a balanced wyeconnected load is 12 ohms. 2 c. If an AC circuit contains three nodes. 1 b. the number of each mesh equations that can be formulated is a. 18 ohms c. REVIEW QUESTIONS 65. Leading the current by 90 deg b. Leading the current by 180 deg d. Lagging the current by 90 deg c. In phase with the current . The relation of the voltage across an inductor to its current is describe as a. . ...ThANk YoU.