Chap 1.ppt - 1st chapter concepts of circuit 1.1 function of circuit 1.2 circuit model 1.3 reference direction of current 1.4 ohm\u2019s law 1.5 power and

Chap 1.ppt - 1st chapter concepts of circuit 1.1 function...

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Unformatted text preview: 1st chapter concepts of circuit 1.1 function of circuit 1.2 circuit model 1.3 reference direction of current 1.4 ohm’s law 1.5 power and load 、 open circuit 1.6 Kirchhof’s law 1.7 concept of potential 上一页 下一页 返回 退出 1.1 function and components of circuit 1. Function of circuit Power system Power transmission 、 distribution Electronic system Signal transmission 、 control 上一页 下一页 返回 退出 1.1 function and components of circuit 2. Components of circuit consists of power source, load and transmission load Power source generator transfor mer line transfor mer applia nces transmission 上一页 下一页 返回 退出 1.1 function and components of circuit 2. Components of circuit signal process amplifier Signal Source microphone speaker load 上一页 下一页 返回 退出 Power sources load: change the electrical energy to other energy 上一页 下一页 返回 退出 1. 2 circuit model use ideal component to represent real components e.g. a light bulb Resistor i Inductor R ignore L L 上一页 下一页 返回 R 退出 Ideal components resistor 、 inductor 、 capacitor 、 power sources + R resistor C L inductor US – IS capacitor 上一页 下一页 返回 退出 Circuit model diagram of ideal components e.g. flashlight circuit I power load switch E + + – U RO line Real circuit S R – power line load Ideal model 上一页 下一页 返回 退出 Switch S Q L 220V R C Fluorescent lamp s us + R C L Ideal model 上一页 下一页 返回 退出 1.3 reference direction of current and voltage 1. Real directions of current and voltage Definition: Direction of current :Moving direction of positive charges ; Direction of voltage :From high potential to low potential ; 上一页 下一页 返回 退出 2. Reference directions I Reference direction can be assumed in analysis current : a + – U R0 b R – voltage : I R E + +U – a b Uab Note : positive or negative directions are reference 上一页 下一页 directions 返回 退出 3. Relationship between real and reference directions Same direction: current or voltage is positive ; Opposite direction: current or voltage is negative ; e.g. 1 I : a R if I = 5A , from a to b ; b + U – a R if I = –5A , from b to a 。 if U = 5V , from a to b ; b if U= –5V , from b to a 。 跳转 上一页 下一页 返回 退出 e.g. 2 I = 0.28A I = – 0.28A voltage U = 2.8V, from to ; E 3V + R0 + U U´ 2.8V – 2.8V + e.m.f. E = 3V the reference direction of voltage U´ U´= –2.8V reference direction of current I I = 0.28A from to 上一页 下一页 返回 退出 1.4 Ohm’s Law U 、 I in same direction , in opposite direction + + U = – IR U=IR U – I R U I R – two signs of I : (1) sign of I is determined by the reference direction (2) real direction is determined by the sign of I 上一页 下一页 返回 退出 e.g. : calculate resistance R by using ohm’s law + U 6V – I 2A (a) Sol: (a) U = IR (b) U = – IR + R I U R 6V –2A – (b) U 6 so : R 3Ω I 2 U 6 so : R 3Ω I 2 上一页 下一页 返回 退出 Concept of linear resistor : a resistor is called linear resistor when its resistance obeys ohm’s law U R Const. I The relationship between voltage and current is called volt-ampere characteristic I/A volt-ampere characteristic of linear resistor o U/V 上一页 下一页 返回 退出 1.5 open circuit and closed I circuit 1.5.1 circuit with power and load E 1. current E I R0 R U E + + – U R0 U = IR voltage R – or U = E – IR0 With inner resistance: I U 。 when R0<<R , then U E O I 上一页 下一页 返回 退出 e.g. : a light bulb of 60 W, working in 220 V circuit , what is the resistor of the light bulb? If it works 3h every day , how much energy it uses in one month? sol: the current I is: P 60 I A 0.273 A U 220 the resistance is: U 220 R 806 I 0.273 energy use in one month W = P t = 60W (3 30) h = 0.06 kW 90 h = 5.4k W. h 上一页 下一页 返回 退出 1.5.2 open circuit I=0 U = U0 = E P= 0 E + – R0 IS 1.5.3 short circuit E I IS R0 U= 0 load P= 0 PE = P = I²R0 R E + – R R0 上一页 下一页 返回 退出 1. 6 Kirchhof’s law I1 R1 a R2 I2 3 R3 2 I3 E1 1 E2 b branch circuit : every branch in a circuit junction : the point of three or more braches join together loop : closed circuit 上一页 下一页 返回 退出 e.g.1 : R1 a I1 I2 R2 IG G d R3 R4 I3 I4 b I + E branch : ab 、 bc 、 ca 、 … (6) junction : a 、 b 、 c c 、d 4 )、 adbca loop : abda 、(abca … (7 ) – 上一页 下一页 返回 退出 1.6.1 Kirchhof’s 1st law (KCL) At any instant the algebraic sum of the currents at a 1 . junction in a network is zero. I1 E1 R1 a I3 R2 R3 I=0 I2 E2 a: I1+I2 = I3 or I1+I2–I3= 0 b Continuity of current 上一页 下一页 返回 退出 e.g. 1 : if I1 = 2A , I2 = 3A , I3 = 0.5A , I5 = 1A, what is IR and I4 ? I2 A I1 sol : for A: for B: R I3 B IR I4 I5 IR = I1 + I2 = 5 A I 4 + I 3 + IR = I 5 I 4 = + I5 IR I3 = 3.5 A = + 1 5 ( 0.5) 上一页 下一页 返回 退出 1.6.2 Kirchhof’s 2nd law ( KVL) 1. At any instant in a closed loop, the algebraic sum of the e.m.f.s acting round the loop is equal to the algebraic sum of the p.d.s round the loop. E = IR E1 = I1R1 + I3R3 E1 – E2 = I1R1 – I2R2 I1 R1 + E1 – R2 I2 a I3 R3 + – E2 b 上一页 下一页 返回 退出 1.6.2 Kirchhof’s 2nd law ( KVL) I2 I1 R1 R 2 a E1 = I1R1 + I3R3 I3 + + I1R1 + I3R3 – E1 = 0 E2 E1 1 R3 2 – I2R2 + I3R3 – E2 = 0 – b U=0 Note : the signs of potential. Recall the definition of positive potential. 上一页 下一页 返回 退出 a e.g. : I1 R1 I2 apply U = 0 I6 R 2 R6 d R3 I3 I b + I4 – c R4 for abda : I6 R6 – I3 R3 +I1 R1 = 0 for acba : I2 R2 – I4 R4 – I6 R6 = 0 for bcdb : I4 R4 + I3 R3 –E = 0 E for adbca , in counter-clockwise direction :– I1 R1 + I3 R3 + I4 R4 – I2 R2 = 0 for cadc , in counter-clockwise direction :– I2 R2 – I1 R1 + E = 0 上一页 下一页 返回 退出 e.g. 2 : if E1 =7V , E2=16V , E3=14V , R1=16 , R2=3 , R3=9. when S is open , what is Uab= ? when S is closed , what is I3= ? sol : (1) S open , I3 = 0 ∴ UR3 = 0 E1+Uab+E3 E2= 0 a S b + E3 + E1 R1 Uab = 714 +16 = 9V (2) S close , Uab = 0 E1+E3 E2 + I3R3= 0 I3 = (E1 E3 + E2) / R3 = 9 / 9 = 1A R2 R3 I3 上一页 下一页 返回 + E2 退出 1.7 concept of potential 1. Concept of potential potential : voltage to the reference point “VX” calculation of potential: (1) select a reference point, its potential is 0 (2) draw the reference direction of current (3) calculate potential to the reference point 上一页 下一页 返回 退出 2. e.g. c 20 a 5 d 6A 4A potential: E1 E2 6 10A Va 、 Vb 、 Vc 、 Vd 140V 90V 。 Sol. : b let a ref. point , Va=0V let b ref. point , Vb=0V Vb=Uba= –10×6= 60V Va = Uab=10×6 = 60 V Vc=Uca = 4×20 = 80 V Vc = Ucb = E1 = 140 V Vd =Uda= 6×5 = 30 V   Vd = Udb =E2 = 90 V   Uab = 10×6 = 60 V Uab = 10×6 = 60 V Ucb = E1 = 140 V Ucb = E1 = 140 V Udb = E2 = 90 V   Udb = E2 = 90 V   上一页 下一页 返回 退出 Simplify the diagram of circuit c 20 4A E1 140V 6 b R1 a + US1 5 a 6A 10A d +140V 20 a 5 +90V c d E2 90V c R2 d R4 R5 b + US2 + US1 a R3 6 R1 R5 b + US2 c R2 d R3 R4 上一页 下一页 返回 退出 e.g. 1: determine the potential VA +6V Sol: (1) when S is open I1 2k 2k I1 = I2 = 0 , VA = 6V S (2) when S is closed as ( b ) I2 = 0 , VA = 0V 2k + 6V – I1 A I2 (a) 2k I2 A (b) 上一页 下一页 返回 退出 ...
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