BlockA Unit1

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  • Block A Unit 1*Block A Unit 1 OutlineConceptsCharge & Current; Voltage; Resistance; PowerTerminologyBranch, Mesh, NodeLawsKirchhoffs current and voltage laws; Ohms lawResistive networksParallel and series; current and voltage divider ruleMeasuring instruments (application)Ammeters and Voltmeters

    Block A Unit 1

  • Block A Unit 1*Electric ChargeCharge is a fundamental electric quantity, measured by the unit Coulomb (C)The smallest amount of charge that exists is the charge that is carried by 1 electron = -1.602 x 10-19 CTherefore charge quantities in real life occur in integral multiples of an electrons chargeTypically denoted by the symbol Q

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  • Block A Unit 1*Current: Free electrons on the move+-Conventional current flows from + to - (Opposite direction to electron flow)Direction of currentCurrents must run in loopsCurrent in relation to ChargeCurrents arise from flow of chargesUnit: Ampere (A)Typically denoted by the symbol, ICurrent = Rate of change of chargeMathematically, I = dQ/dt

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  • Block A Unit 1*Fundamental law for chargeCurrent has to flow in closed loopNo current flows if there is a break in the pathUnderlying physical law: Charge cannot be created or destroyedThis is the basis of Kirchhoffs Current Lawi1i4i3i2Kirchhoffs current lawSum of currents at a node must equal to zero:i1 + i2 + i3 + i4 = 0

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  • Block A Unit 1*Kirchhoffs current lawDoes the direction of current matter? i1i2i3i1i2i3i1 + i2 + i3 = 0i1 + i2 + i3 = 0 (WRONG)I2 running into nodeI2 running out of nodei1 + i3 = i2 (CORRECT)YES!!

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  • Block A Unit 1*Kirchhoffs current lawCurrents EXITING (-ve),Currents ENTERING (+ve) i1i2i3i4Entering: I1 & I2 (+ve)Exiting: I3 & I4 (-ve)i1 + i2 i3 i4 = 0Sign convention when applying KCL

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  • Block A Unit 1*KCL exampleProblem 2.14 and 2.15Find the unknown current using KCL6A - 5A + 2A - i = 0 i = 3A6A - 5A + 2A + i = 0 i = -3A

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  • Block A Unit 1*Voltage/Potential differenceABCurrent+-Energy is required to move charges between 2 pointsVoltage/potential difference is always made with reference to 2 pointsvxvx = vA - vBUnit: Volt (V)High (+)Low (-)Direction of Flow

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  • Block A Unit 1*Sign conventioniv+-In a LOADVoltage DROPS in the direction of the currentEnergy is dissipated (or consumed)iv+-In a SOURCEVoltage RISES in the direction of the currentEnergy is generated

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  • Block A Unit 1*Fundamental law on voltageEnergy is required to push electrons through a resistive elementThat same energy needs to be generated by a sourceTotal energy generated in a circuit must equal total energy consumed in the circuitEnergy cannot be created or destroyedTherefore, voltage rise = voltage drop

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  • Block A Unit 1*Kirchhoffs voltage lawsourceloadloadloadGROUNDReference voltage in a circuit set to 0VAll other nodes on the circuit can then be conveniently referenced to GROUNDGround symbol+ V1 -+ V2 -+ V3 -DEFINING signsVoltage gain (+ve)Voltage drop (-ve)v1 v2 + v3 = 0

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  • Block A Unit 1*KVL exampleProblem 2.16Apply KVL to find voltage V1 and V2Loop 1 (Clock-wise):5V - 3V - V2 = 0 V2 = 2VLoop 2 (Anti-clock-wise):V1 - 10V - V2 = 0 V1 = 12V

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  • Block A Unit 1*Resistance and Ohms LawivVI1/ROhms law: V = IRIdeal RESISTOR shows linear resistance obeying Ohms lawWhen current flows through any circuit element, there will always be a resistance to its flow which results in a voltage drop across that circuit element +_IMPORTANT: Positive current is defined here as flowing from higher to lower voltage (Remember)Unit: Ohm ()AL: resistivity (material property)A: cross-sectional area

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  • Block A Unit 1*Ohms law + KCL exampleProblem 2.17Use Ohms law + KCL to find the current through the 15 resistorKCL: I1 + I2 = 10AOhms: 15I1 = V15 (1); 30I2 = V30 (2)(KVL) V15 = V30Therefore, 2I2 = I1Solving for the variables: I2 = 3.33 A, I1 = 6.67 A

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  • Block A Unit 1*Electrical PowerIn a source, power is generatedIn a load (eg. resistor), power is dissipated/consumed Electrical power generated/dissipated in a given element is defined by the product of the voltage across that element and the current through it P = VI0.2A0.2A1.5V1.5V+_+_Source: P = 1.5V x 0.2A = 0.3WLoad: P = 1.5V x 0.2A = 0.3WUnit: Watt (W)P = I2RP = V2/RPower generated by source MUST EQUAL Power dissipated in the load

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  • Block A Unit 1*Power exampleProblem 2.22Determine which components are absorbing power and which are delivering powerIs conservation of power observed in this example?

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  • Block A Unit 1*Power example solution This slide is meant to be blankFinally calculate power through each element A: (12V)(5A) = 60W [generating]B: (3V)(5A) = 15W [generating]C: (-5V)(5A) = -25W [absorbing]D: (-10V)(3A) = -30W [absorbing]E: (-10V)(2A) = -20W [absorbing]Last part: Is power conserved? Generating: 60W + 15W = 75WAbsorbing: 25W + 30W + 20W = 75WYES!

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  • Block A Unit 1*Terminology: Branch and NodeBranchBRANCH: Any path of a circuit with 2 TERMINALS connected to itNodeNODE: Junction of 2 or more branches

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  • Block A Unit 1*Terminology: Loop and MeshLOOP: Any closed connection of branchesIn the above circuit, there are 6 loops in total+-Vs+-VsMesh: Loop that does not contain other loopsIn the above circuit, there are 3 meshes in total

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  • Block A Unit 1*DC vs ACDC Direct Current: Current is constant with timeAC Alternating Current: Current varies with time and reverses direction periodically

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  • Block A Unit 1*Independent Voltage Source+_Circuitii+_v+_vIndependent Voltage Source supplies a prescribed voltage across its terminals irrespective of current flowing through itCurrent supplied is determined by load circuit connected

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  • Block A Unit 1*Independent Current SourceCircuitii+_i+_vIndependent Current Source supplies a prescribed current to any load circuit connected to it Voltage supplied is determined by load circuit connected

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  • Dependent SourcesDEPENDENT source generates v or i that is a function of some other v or i in the circuit Symbol diamond shape outlineBlock A Unit 1*+-5V+vx-IndependentVoltage Controlled Voltage Source (VCVS): vs = vxCurrent Controlled Voltage Source (CCVS): vs = rix Voltage Controlled Current Source (VCCS): is = gvxCurrent Controlled Current Source (CCCS): is = ix _+isvs

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  • Block A Unit 1*Example of a VCVS+vi-Avi+-Note: This is a general model for an amplifier which shall be re-visited later in BLOCK CRiRo

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  • Block A Unit 1*Short Circuit0.1AAB0.1AABShort circuit A & BShort circuit: Connect 2 or more terminals so that the voltage between each of them is the sameTypically associated with current, eg. short-circuit currentIscShort circuit current Isc = 0.1A20 resistor is now bypassedCurrent from source flows through the short-circuit to give a short-circuit current2020

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  • Block A Unit 1*Open CircuitABOpen circuit: Leave 2 terminals unconnected externallyTypically associated with voltage, eg. open-circuit voltage0.1A0.1AAB+Voc-Ioc= 0A2020Open circuit voltageVoc = 0.1A 20 = 2V

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  • Block A Unit 1*Self-contradictory circuits+-vsABWhat is the voltage across A and B?IsABWhat is the current arriving at A?Prefixes: Memorize and apply them!Write as 2.15mA instead of 0.00215A

    teraT1012gigaG109megaM106kilok103millim10-3micro10-6nanon10-9picop10-12femtof10-15

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  • Block A Unit 1*Parallel network (Highlights)R1R2IsRNI1I2INRPIsEquivalent Resistance1/RP = 1/R1 + 1/R2 + + 1/RNCurrent divider rule

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  • Block A Unit 1*Parallel Network: ProofR1R2IsI2I1Apply Kirchhoffs current law (KCL) at X:Is = I2 + I1 This can be seen as Is is split into the 2 branchesR1R2IsI2I1XXY123456Note that points 1, 2, 3 are all at the same voltage, therefore same node (X)Points 4, 5, 6 are all at the same voltage, therefore same node (Y)Voltage across R1 = Voltage across R2 = VXY

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  • Block A Unit 1*Parallel network: ProofR1R2IsI2I1XY123456Apply Ohms to both R1 and R2:VXY = I1R1 = I2R2I1 = VXY/R1; I2 = VXY/R2Adding I2 & I1 up according to KCL:Is = VXY(1/R1 + 1/R2)We can now find how much of Is is distributed between the 2 branchesThis is referred to as the current divider rule

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  • Block A Unit 1*Parallel network: ProofR1R2IsI1I2Is = VXY(1/R1 + 1/R2)Replace the parallel network of resistors with a single equivalent resistorRPIsRemember that the voltage across RP is still VXY!VXY/RP = VXY(1/R1 + 1/R2)1/RP = 1/R1 + 1/R2XY

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  • Block A Unit 1*Parallel network (seeing it)R1R2IsRNI1I2INCurrent splits at one nodeCurrent re-combine at the other nodeSuggestion: Think about parallel resistors as the rungs on a ladder

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  • Block A Unit 1*Current divider: Example 1i1V2R1i2R2V1IFind i1 and i2 in terms of IIf R1 = R2, find the ratio between i1 and i2If R2 = 3R1, find the ratio between i1 and i2i1 = [R2 / (R1 + R2)]I; i2 = [R1 / (R1 + R2)]IIf R1 = R2, then i1 = i2 = I/2If R2 = 3R1, then i1 = 3I/4, i2 = I/4Therefore i1 = 3i2

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  • Block A Unit 1*Current divider: Example 2Find the current through each resistor in terms of IFind the current through each resistor in terms of I if a 4th resistor was added in parallelHow many resistor are required to reduce t