Chapter 33Alternating Current (AC)R, L, C in AC circuitsAC, the descriptionA DC power source, like the one from a battery, provides a potential difference (a voltage) that does not change its polarity with respect to a reference point (often the ground)An AC power source is sinusoidal voltage source which is described as Here ()maxv V sin t∆ ∆ ω=maxvV∆∆ωis the instantaneous voltage with respect to a reference (often not the ground).is the maximum voltage or amplitude. is the angular frequency, related to frequency f and period T as 22 fTπω π= =V∆tV∆Symbol in a circuit diagram:orv∆v∆The US AC system is 110V/60Hz. Many European and Asian countries use 220V/50Hz.Resistors in an AC Circuit, Ohm’s LawThe voltage over the resistor:()R maxv v V sin t∆ ∆ ∆ ω= =Apply Ohm’s Law, the current through the resistor:( ) ( )maxRR maxVvi sin t I sin tR R∆∆ω ω= = ≡The current is also a sinusoidal function of time t. The current through and the voltage over the resistor are in phase: both reach their maximum and minimum values at the same time.PLAYACTIVE FIGUREThe power consumed by the resistor is ( )222 2maxRR R R RVvp v i i R sin tR R∆∆∆ ω= ⋅ = = =We will come back to the power discussion later.Phasor Diagram, a useful tool.The projection of a circular motion with a constant angular velocity on the y-axis is a sinusoidal function. To simplify the analysis of AC circuits, a graphical constructor called a phasor diagram is used. A phasor is a vector whose length is proportional to the maximum value of the variable it representsThe phasor diagram of a resistor in AC is shown here. The vectors representing current and voltage overlap each other, because they are in phase. xyORtωThe projection on the y-axis is()yR R sin tω=The power for a resistive AC circuit and the rms current and voltage()2R maxp p sin tω=()R maxv V sin t∆ ∆ ω=()R maxi I sin tω=When the AC voltage source is applied on the resistor, the voltage over and current through the resistor are:Both average to zero.But the power over the resistor isAnd it does not average to zero. The averaged power is:PmaxPavPmaxP( )( )22220012122 4 2TTav maxTmax maxp p sin t dtp ptsin tTωωωω= = − = ∫( )2122 4xsin xdx sin x= −∫The power for a resistive AC circuit and the rms current and voltage22rmsav rms rms rmsVp V I I RR∆∆= ⋅ = =So the averaged power the resistor consumes isIf the power were averaged to zero, like the current and voltage, could we use AC power source here?The averaged power can also be written as:PmaxPavPmaxP12 2maxav max maxpp V I∆= = ⋅221 12 2maxav maxVp I RR∆= =Define a root mean square for the voltage and current:2 2 2 21 1 and 2 2rms max rms maxV V , I I∆ ∆= =or2 and 2max rms max rmsV V , I I∆ ∆= =One get back to the DC formula equivalent:Resistors in an AC Circuit, summaryOhm’s Law applies. Te current through and voltage over the resistor are in phase.The average power consumed by the resistor isFrom this we define the root mean square current and voltage. AC meters (V or I) read these values.The US AC system of 110V/60Hz, here the 110 V is the rmsvoltage, and the 60 Hz is the frequency f, so ()R maxv V sin t∆ ∆ ω=()R maxi I sin tω=22rmsav rms rms rmsVp V I I RR∆∆= ⋅ = = and 2 2max maxrms rmsV IV , I∆∆= =2 156 Vmax rmsV V∆ ∆= =12 377 secfω π−= =Inductors in an AC circuit, voltage and currentThe voltage over the inductor is()L maxv v V sin t∆ ∆ ∆ ω= =To find the current i through the inductor, we start with Kirchhoff’s loop rule:0Lv v∆ ∆+ =( )0maxdiV sin t Ldt∆ ω− =orSolve the equation for i ( )maxVdi sin t dtL∆ω=( ) ( )2max maxmaxV Vi di sin t dt cos t I sin tL L∆ ∆πω ω ωω = = = − = − ∫ ∫or with 2maxmax maxVi I sin t , IL∆πωω = − = Inductors in an AC circuit, voltage leads currentExamining the formulas for voltage over and current through the inductor:()L maxv v V sin t∆ ∆ ∆ ω= =Voltage leads the current by ¼ of a period (T/4 or 90°or π/2) . Or in a phasor diagram, the rotating current vector is 90°behind the voltage vector.2maxi I sin tπω = − PLAYACTIVE FIGUREInductive Reactance, the “resistance” the inductor offers in the circuit.Examining the formulas for voltage over and current through the inductor again:()L maxv V sin t∆ ∆ ω= with 2maxmax maxVi I sin t , IL∆πωω = − = This time pay attention to the relationship between the maximum values of the current and the voltage: maxmaxVIL∆ω=This could be Ohm’s Law if we define a “resistance” for the inductor to be: LX Lω=And this is called the inductive reactance. Remember, it is the product of the inductance, and the angular frequency of the AC source. I guess that this is the reason for it to be called a “reactance” instead of a passive “resistance”. The following formulas may be useful: and max L max rms L rmsV X I , V X I∆ ∆= =()L max Lv I X sin t∆ ω=Capacitors in an AC circuit, voltage and currentThe voltage over the capacitor is()C maxv v V sin t∆ ∆ ∆ ω= =To find the current i through the capacitor, we start with Kirchhoff’s loop rule:0Cv v∆ ∆+ =( )0 with maxq dqV sin t , iC dt∆ ω− = =orSolve the first equation for q, and take the derivative for i ()maxq C V sin t∆ ω=( )2max maxdqi C V cos t I sin tdtπω ∆ ω ω = = = + or( )1 with 2maxmax maxVi I sin t , IC∆πωω− = + = Here I still like to keep the Ohm’s Law type of formula for voltage, current and a type of resistance.Capacitors in an AC circuit, current leads the voltageExamining the formulas for voltage over and current through the capacitor:()C maxv V sin t∆ ∆ ω=2maxi I sin tπω = + Current leads the voltage by ¼ of a period (T/4 or 90°or π/2) . Or in a phasor diagram, the rotating voltage vector is 90°behind the current vector.PLAYACTIVE FIGURECapacitive Reactance, the “resistance” the capacitor offers in the circuit.Examining the formulas for voltage over and current through the capacitor again:This time pay attention to the relationship between the maximum values of the current and the voltage: ( )1maxmaxVIC∆ω−=This could be Ohm’s Law if we define a “resistance” for the capacitor to be: 1CXCω=And this is called the capacitive reactance. It is the inverse of the
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