Intro Electrical Engineering
Intro Electrical Engineering EE 221
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This 18 page Class Notes was uploaded by Maudie Quitzon on Saturday September 12, 2015. The Class Notes belongs to EE 221 at West Virginia University taught by Staff in Fall. Since its upload, it has received 15 views. For similar materials see /class/202857/ee-221-west-virginia-university in Electrical Engineering at West Virginia University.
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Date Created: 09/12/15
1 ri E 1 quotI I quotW iquot a a 339 pi r a u 7 quotfr Lt Lt 3 Instantaneous and average power is RMS value amp Apparent power and power factor amp Complex power Product of timedomain voltage and timedomain current W vt ilttgt vow o Determine maximum value Transients Steadystate POW steadystate Power supplied by source As the transient dies out the circuit returns to steadystate operation Since the only source remaining in the circuit is dc the inductor eventually acts as a short circuit absorbing zero power vt Vm COSoatG and it Im COSoat p 39 P Vm m COSoat6 COS03t p trigonometric 12 Vm lm 0089 p 12 Vm Im COS2ot e p Identlty pt 12 Vm m cos p Lz Vm Irm oos2mt 6 a V constant periodic two parts independent oft period is 12T average the quotaveragequot average is zero periodic wanted unwanted r l r I m pt 12 V cosm p 12 Vn m COS2nt H p H4 W J PviWV xx 39 y 1 VI y I T I I ILI II ms V2 1 2 3 i l 12 4 T g voltage V 440 V impedance Z 2460 C a 1 6 rads I2z60 A pt24cos7tl360 W time s V R2 2 V 2445 Example 1 Average power delivered to resistor time s 45 V X V 2 lt3 3 as O 12 Vm m cos190 12 Vm m 0036 p ll llll II III Illll A Example 2 Average power delivered to p ure T 4 Iy reactive elements Example 3 Voltage across impedance V 1004250 V and Z 504550 Q Determine active power absorbed VIZ 242555 2430 17jA lt30 z 287 j419 o P 12 22 287 12 17212 287 574 W Example 4 Chapter 11 Problem 8 In the circuit shown in Fig 1127 find the average power being a dissipated in the 39 resistor b generated by the source 1 Figm1127 31j34j3 2 z 3 01 03 W R Ignore30 onVNIR52j531Rlz 6j8 10 miit a 2 T x3 a b 4 540 W 1346345194 v 1 0 PW Egtlt13463gtlt500s 5194 2075 W II 7 ffquot til 7 u i m a 39llf39lill quot 7 A simple loop circuit used to illustrate the derivation of the maximum power transfer theorem as it applies to circuits operating in the sinusoidal steady state Zth Rththh ZL RLjXL 30 u Rm and XL Xt Example a what value of ZL will absorb a maximum average power b What is the value of this maximum power Vm 2120101 2107334 11657 V 105 000jw m N ZL 107334 11657 m quot ma a396 1 10733 2 P x8 mow A 16 8 4Q Figure 1 1 30 Hi I IL Chapter 11 Problem 12 For the circuit of Fig 1130 l W m 1 T HU L Measure for sinusoidal voltages and currents Power outlets 60 Hz quotvoltage of 115Vquot Not the mean of T T2 Not the Amplitude M 115V Measure of effectiveness of a source in delivering power to a resistive load Effective value of periodic current is equal to the DC value that delivers the same average power to resistor it R gt pt gt PR and compare to lDC R Mathematical expression P 131sz zli2dz 12 R 1 T 0 DC 0 I iYIiszz e TO Square root of the mean of the square current rms value Defined for all periodic signals 2 e R j u ll J f i w y 39 l r 4 l i is H 7 i r i l i 39 w er l 7 a l l 7 7w n 1739 7 7 7 7 7 l7 7 7 7 7 7 47 7 7 7 it Im COSoat p with a period of T 27503 1 T 2 2 a 27ra 1 167 OJ Im cos mtodt ImE 3quot cos2at2gp dt real quantity independent of phase angle equal to 0707 the amplitude example g30 A delivers the same as lDC 1A In general P 2Vmmcos6p For resistors Vet39f Iet f COS6 39 p PVe le Ve 2R Ie ZR Note We can use amplitude or rms value Use V and V rms to designate voltages ll iiiquot tiWillilvilli Example Chapter11 Problem 30 The series combination of a 1kQ resistor and a 2H inductor must not dissipate more than 250 mW of power at any instant Assuming a sinusoidal current with 0 500 rads what is the largest rms current that can be tolerated The peak instantaneous power is 250 mW The combination of elements yields Z 1000 j1000Q 1414445 Q szo sz 45 1414 Arbitrarily designate V Vm 40 so that I A and Vm 1414 Im We may write pt 12 Vm m cos 4 12 Vm m cos 2at 4 where 4 the angle of the current 45 This function has a maximum value of 12 lem cos 4 12 lem Thus 0250 12 lem 1 cos 4 12 1414 Im2 1707 and Im 1439 mA In terms of rms current the largest rms current permitted is l1439m N5 1018 mA rms We had P Vm Im COS6 39 p Veff Ieff COS6 39 p In case of direct current we would use voltage times current 8 Veff Iet f This is not the average power Is the quotapparentquot power 8 or AP Measured in voltampere or VA rather than Wto avoid confusion Magnitude of S is always greater or equal to magnitude of P S 2 P Defined as the ratio of average real power to apparent power PF PS Pve le l In the sinusoidal case the power factor is PF cos6 p 6 p is the angle the voltage leads the current PF angle note Purely resistive load has PF 1 Purely reactive load has PF O PF 05 means a phase angle of i60 Resolve ambiguity PF leading or lagging for capacitive or inductive load
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