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Chapter  2 Chapter 2 Presentation Transcript

  • CHAPTER 2 Basic LawsCHAPTER 2 Basic Laws To determine the values of electrical variables such as current, voltage and power in a given circuit requires understanding some fundamental laws for examples; Ohm’s law and Kirchhoff’s laws to analyze the circuit. In addition, some techniques must be used together with those fundamental laws.
  • 2.2 Ohm’s law Materials in general have a characteristic behavior of resisting the flow of electric charge which is known as resistance (R). The resistance of any material depends on cross – sectional area A and its lengthl l al with resistivity ρ R = ρ A l(Ohm) ρ – resistivity of materials (ohm – mete Good conductors low resistivity insulators high resistivity See Tab l2. 1
  • The circuit element used to model the current - resisting behavior of a material is the resistor Resistance of the resistor Ohm’s law states that the voltage V across a resistor is directly proportional to the current i flowing through the resistor iRviv =⇒α Note : AVΩ /11 =
  • If current flows from a higher potential to a lower potentialiRv = If current flows from a lower potential to a higher potential iRv −= fect conductor short - circuit ∠∞∠R0 open - circuit
  • sistor fix e dvariable Wire wound compositionlarge resistance ar resistor obey ohm’s law inear resistor does not obey
  • The reciprocal of resistance R, known as conductance and denoted by G G =I R = i v mho Or Siemens (S) G i GvpGvi 2 2 , ===
  • Example 2.2 In the circuit shown in Fig.2.8, calculate the current i , the conductance G, and the power p. Solution: The Voltage across the resistor is the same as the source voltage (30 V) because the resistor and the voltage source are connected to the same pair of terminals. Hence, the current is
  • mA xR v i 6 105 30 3 === mS xR G 2.0 105 11 3 === mWxvip 180)106(30 3 === − mWxxxRip 180105)106( 3232 === − mWxxGvp 180102.0)30( 322 === − The conductance is
  • tπsin20 tmA x t R v i π π sin4 105 sin20 3 === tmWvip π2 sin80== Example 2.3 A voltage source of Hence, V is connected ss a 5-kΩ resistor.Find the current through the r and the power dissipated. olution:
  • 2.3 Nodes, Branches and Loops A branch represents a single element such as a voltage source or a resistorA Node is the point of connection between two or more branches A Loop is any closed path in the circuit formed by starting at a node, passing through a set of nodes, and returning to starting node without passing through
  • Two or more elements are in series if they exclusively share a single node and consequent by carry the same currentTwo or more elements are in parallel if they exclusively connected to the same two nodes and consequent by have
  • Example 2.4 Determine the number of branches and node in the circuit show in Fig.2.12. Identify which elements are in series and which are in parallel. Solution: Since there are four elements in the circuit, the circuit has four branches: 10 V, 5 Ω, 6 Ω, and 2 A. The circuit has three nodes as identified in Fig. 2.13. The 5 Ω resistor is in series with the 10-V voltage source because the same current would flow in both. The 6-Ω resistor is in parallel with the 2-A current source because both are connected to the same nodes 2 and
  • 2.4 Kirchhoff’s laws Kirchhoff’s current law (KCL) states that the algebraic sum of currents entering a node is zero0 1 =∑ = N n ni 0)()( 54321 =−+++−+ iiiii 52431 iiiii +=++ enteringleaving
  • A simple application of KCL is combining current sources in parallel. The combined or equivalent current source can be found by applying KCL to node aIT = I1-I2+I3 IS = I1-I2+I3 IT b I1 I2 I3 IT a b
  • Kirchhoff’s voltage law (KVL) states that the algebraic sum of all voltages around a loop is zero -v1+v2+v3-v4+v5 = 0 v2+v3+v5 = v1+v4 When voltage sources are connected in series, KVL can be applied to obtain the total voltage
  • 5 For the circuit in Fig.2.21 (a), find voltag Solution: To find v1 and v2, we apply Ohm’s law and Kirchhoff’s voltage law. Assume that current i flow through the loop as shown in Fig. 2.21(b). From Ohm’s law,v1 =2i , v2=-3i (2.5.1)
  • Applying KVL around the loop gives -20+v1-v2=0 (2.5.2) stituting Eq. (2.5.1) into Eq. (2.5.2), we ob -20+2i+3i=0 or 5i=20 i =4 A Substituting i in Eq. (2.5.1) finally gives v1= 8 V , v2=-12 V
  • 6 Determine v0 and i in the circuit shown in Solution: We apply KVL around the loop as shown in Fig.2.23(b). The result is -12+4i+2v0-4+6i = 0 (2.6.1) Applying Ohm’s law to the 6-Ω resistor gives v0 = -6i (2.6.2)
  • bstituting Eq.(2.6.2) into Eq.(2.6.1) yields -16+10i-12i = 0 i = -8 A and v0 = 48 V.
  • Example 2.8 Find the current and voltage in the circuit show in Fig. 2.27(a) e apply Ohm’s law and Kirchhoff’s law. By v1 = 8i1, v2=3i2, v3 = 6i3 (2
  • resistor are related by Ohm’s law as show, we are really looking for three thing: (v1, v2, v3) or (i1, i2, i3). At node a, KCL gives i1- i2- i3 = 0 (2.8.2) Applying KVL to loop as in Fig. 2.27(b), -30 + v1 + v2 = 0 ess this in terms of i1 and i2 as in Fig. (2.81 -30 + 8i1 + 3i2 = 0
  • ( ) 8 i330 i 2 1 − = (2.8.3) Applying KVL to loop 2, -v2 + v3 = 0 v3 = v2 (2.8.4) as expected since the two resistors are in parallel. We express v1 and v2 in term of i1 and i2 as in Eq. (2.8.1).Eq. (2.8.4) becomes
  • 6 i3 = 3 i2 2 2 3 i i = tuting Eqs. (2.8.3) and (2.8.5) into (2.8.2) 0 28 330 2 2 2 =−− − i i i or i2 = 2 A. From the value of i2 , we now use Eqs. (2.8.1) to (2.8.5) to obtain i1 = 3 A, i3 = 1 A, v1 = 24 V, v2 = 6 V, v3 = 6 V.
  • 2.5 Series Resistors and Voltage divisionv1 = i R 1 v2 = i R 2 -v+v1+v2 = 0v = v1+v2 = i (R1+R2)v = i R eq Req = R1 + R2 ... N resistors in series then, Req = R1 + R2+ …RN = Rn Σ N n=1
  • ine the voltage across each resistor v1 v iR 1 i (R1 +R2 ) = v1 R1+R2 R1v= , v2 = R1+R2 R2v Vn= Rn R1+R2+…+Rn V
  • 2.6 Parallel resistors and current divisionv = i1R1 = i2R2 i1 = R 1 v i2 = R 2 v, at node a ;i = i1+ i2 1 Req = R 1 1 R 2 1+ R1 R1 R2 R2 + =1 Req i R 1 v R 2 v+ v R1 1 R2 1 +== v Req = R1+R2 R1R2 =Req. . .1 = R2 then Req = R1 /2 สำำหรับ R 2 ตัวต่อ ขนำน For N resistors1 R e = R 1 1 + R 2 1 R N 1+…+ If R = R1 = R2 = … = RN Req =R N
  • It is more often to use conductance rather than resistance when dealing with resistors in parallel. Geq = G1 + G2 + G3 + … + GN Given the total current i enter node a, how do use obtain current i1 and i2v = i Req = iR1R2 R1+R2 i1= R 1 v R 1 = R1R2 R1+R2 i iR2 R1+R2 = =i2 iR1 R1+R2 Current divider er current flow through the smaller resista
  • e 2.9 Find Req for the circuit shown in Fig. Solution: To get Req , we combine resistors in series and in parallel. The 6-Ω and 3-Ω resistors are in parallel, so equivalent resistance is Ω= + =ΩΩ 2 36 36 3//6 x
  • is 1Ω+5Ω = 6Ω Thus the circuit in Fig. 2.34 is reduced to that in Fig. 2.35(a). In Fig. 2.35(a), we notice that the two 2-Ω resistors are in series, so the equivalent resistance is 2Ω + 2Ω = 4Ω This 4-Ω resistor is now in parallel with the 6-Ω resistor in Fig. 2.35(a);
  • Ω= + =ΩΩ 4.2 64 64 6//4 x The circuit in Fig. 2.35(a) is now replaced with that in Fig. 2.35(b). In Fig. 2.35(b), the three resistors are in series. Hence, the equivalent resistance for the circuit is Ω=Ω+Ω+Ω= 4.1484.24Req
  • Example 2.10 Calculate the equivalent resistance Rab in the circuit in Fig. 2.37. Solution: The 3-Ω and 6-Ω resistors are in parallel because they are connected to the same two nodes c and b. Their combined resistance is Ω= + =ΩΩ 2 63 63 6//3 x
  • Similarly, the 12-Ω and 4- Ω resistors are in parallel since they are connected to the same two nodes d and b. Hence Ω= + =ΩΩ 3 412 412 4//12 x Also the 1-Ω and 5-Ω resistors are in series; hence, their equivalent resistance is Ω=Ω+Ω 651With these three combinations, we can replace the circuit in Fig. 2.37 with that in Fig. 2.38(a). In Fig.2.38(a),3-Ω in parallel with 6-Ω gives 2-Ω, as calculated in Eq. (2.10.1). This 2-Ω
  • equivalent resistance is now in series with the 1-Ω resistance to give a combined resistance of 1Ω + 2Ω =3Ω . Thus, we replace the circuit in Fig. 2.38(a) with that in Fig. 2.38(b). In Fig. 2.38(b), we combine the 2-Ω and 3-Ω resistors in parallel to getΩ= + =ΩΩ 2.1 32 32 3//2 x Ω resistor is in series with the 10-Ω resistor Rab = 10 + 1.2 = 11.2 Ω
  • Example 2.12 Find iO and vO in the circuit shown in Fig. 2.42(a). Calculate the power dissipated in the 3-Ω resistor. Solution: The 6-Ω and 3-Ω resistors are parallel, so their combined resistance isΩ= + =ΩΩ 2 36 36 3//6 xThus our circuit reduces to that shown in Fig. 2.42(b). Notice that vO is not affected by the combination of the resistors because the resistors are in parallel and therefore have the same voltage vO. From Fig. 2.42(b), we can obtain vO in two ways. One way is to
  • Ai 2 24 12 = + = and hence, vO = 2i = 2x2 = 4 V. Another way is to apply voltage division, since the 12 V in Fig. 2.42(b) is divided between the 4-Ω and 2-Ω resistors. Hence, VVvO 4)12( 42 2 = + = Similarly, iO can be obtained in two way. One approach is to apply Ohm’s law to the 3-Ω resistor in Fig. 2.42(a) now that we know vO; thus,
  • 43 == OO iv AiO 3 4 = Another approach is to apply current division to the circuit in Fig. 2.42(a) now that we know I, by writing AAiiO 3 4 )2( 3 2 36 6 == + = ower dissipated in the 3-Ω resistor is Wivp OOO 333.5 3 4 4 =      ==
  • Example 2.13 For the circuit shown in Fig. 2.44(a), determine: (a) the voltage vO, (b) the power supplied by the current source, (c) the power absorbed by each resistor. Solution: (a) The 6-kΩ and 12-kΩ resistors are in series so that their combined value is 6 + 12 = 18 kΩ . Thus the circuit in Fig. 2.44(a) reduces to that shown in Fig. 2.44(b). We now apply the current division
  • mAmAi 20)30( 000,18000,9 000,18 1 = + = mAmAi 10)30( 000,18000,9 000,9 2 = + = Notice that the voltage across the 9-kΩ and 18-kΩ resistors are the same, and vO = 9,000i1 = 18,000i2 = 180 V, as expected. (b) Power supplied by the source is WmWivp OOO 4.5)30(180 ===
  • Power absorbed by the 12-kΩ resistor is WxRiRiiivp 2.1)000,12()1010()( 232 222 ===== − wer absorbed by the 6-kΩ resistor is WxRip 6.0)000,6()1010( 232 2 === − wer absorbed by the 9-kΩ resistor is W R v p O 6.3 000,9 )180( 22 === WmWivp O 6.3)20(1801 === Notices that the power supplied (5.4W) equals the power absorbed (1.2 + 0.6 + 3.6) = 5.4 W). This is one way of
  • 2.7 Wye – Delta transformationsWhen the resistors are neither in parallel nor in series. For example, the bridge circuit, this circuit can be simplified by using three – terminal equivalent networks, wye (y) and delta ( ) as will be shown in Ex.2.15.
  • Delta to Wye conversion Wye to delta = Ra = R1R2 + R2R3 + R3R1 R1 Rc =R1R2 + R2R3 + R3R1 R3 Rb =R1R2 + R2R3 + R3R1 R2 Y and balance wh e n R1 = R2 = R3 = Ry Ra = Rb = Rc = R Ry = R or R = 3Ry 3 Rb Rc Ra+Rb+Rc Rc Ra Ra+Rb+Rc Ra Rb Ra+Rb+Rc 2 = 3 =
  • Example 2.14 Convert the ∆ network in Fig. 2.50(a) to an valent Y network. ution: Using Eqs. (2.49) to (2.51), we obta Ω== ++ = ++ = 50 50 250 151025 1025 1 x RRR RR R cba cb
  • Ω== ++ = 5.7 50 1525 2 x RRR RR R cba ac Ω== ++ = 3 50 1015 3 x RRR RR R cba ba equivalent Y network is shown in Fig. 2.50
  • Example 2.15 Obtain the equivalent resistance Rab for the circuit in Fig. 2.52 and use it to find current i.
  • Solution: In this circuit, there are two Y network and one ∆ network. Transforming just one of these will simplify the circuit. If we convert the Y network comprising the 5-Ω, and 20-Ω resistors, we may select Ω= 101R Ω= 202R Ω= 53R s from Eqs. (2.53) to (2.55) we have 10 1055202010 1 133221 xxx R RRRRRR Ra ++ = ++ = Ω== 35 10 350
  • Ω== ++ = 5.17 20 350 2 133221 R RRRRRR Rb Ω== ++ = 70 5 350 3 133221 R RRRRRR Rc pairs of resistors in parallel, we obtain Ω= + = 21 3070 30x70 30//70 Ω= + = 2917.7 5.175.12 5.175.12 5.17//5.12 x Ω= + = 5.10 3515 3515 35//15 x
  • so that the equivalent circuit is shown in Fig. 2.53(b). Hence, we find Ω= + =+= 632.9 21292.17 21292.17 21//)5.10292.7( x Rab A R v i ab s 458.12 632.9 120 ===
  • 2.8 Application Lighting systems, such as in a house, often consist of N lamps connected either in parallel or in series 2.8.1 Lighting systems
  • Assuming that all the lamps are identical and v0 is the power – line voltage, the voltage across each lamp is v0 for parallel connection and v0/N for series connection. The series connection is easy to manufacture but is seldom used in practice for two reasons. First, it is less reliable. Second, it is harder to maintain.
  • Potentiometer is a three – terminal device that operates on the principle of voltage division. It is essentially an adjustable voltage divider. As a voltage regulator, it is used as a volume or level control on radios, TVs and other devices. 2.8.2 Design of DC Meters Potentiometer controlling potential levels Vout = Vbc = Vin Rbc Rac Where Rac = Rab+Rbc Thus, vout decreases or increases as the sliding contact of the pot moves toward c
  • Another application where resistors are used to control current flow is the analog dc meters, ammeter, voltmeter and ohmmeter. Each of these meter employs the d’Arsonval meter movement. The movement consists of a movable iron – core coil mounted on a pivot between the poles of a permanent magnet. When current flows through the coil, it creates torque which
  • Voltmete rIt measures the voltage across a load and is connected in parallel with the element. It consists of a d’Arsonval movement in series with a resistor whose resistance Rm is deliberately made very large to minimize the current drawn from the circuit. ( ) m fs fs n nmfsfs R I V R RRIV −= +=
  • Ammete rIt measures the current through the load and is connected in series with it. It consists of a d’ Arsonaval movement in parallel with a resistor whose resistance Rm is deliberately made very small to minimize the voltage drop across it. m mfs m n fs mn n m R II I R I RR R I − = + =
  • Ohmm eterIt consists of d’Arsonval movement, a variable resistor and a battery
  • ( ) ( ) )1.....(....................m m x mxm RR I E R IRRRE +−= ++= The resistor R is selected such that the meter gives a full-scale deflection when Rx = 0 fsm II = ( ) )2........(....................fsm IRRE += Substitute eq. (2) in (1) ( )m m fs x RR I I R +         −= 1
  • connected to a 9-V battery as shown in Fig. 2.56(a). Calculate: (a) the total current supplied by the battery, (b) the current through each bulb, (c) the resistance of each bulb. Solution: (a) The total power supplied by battery is equal to the total power absorbed by the bulbs, that is, p = 15 + 10 + 20 = 45 W Since p = VI, then the total current
  • A V p I 5 9 45 ===(b) The bulbs can be modeled as resistors as shown in Fig. 2.56(b). Since R1 (20-W bulb) is in parallel with the battery as the series combination of R2 and R3, V1 = V2 + V3 = 9 V The current through R1 isA V p I 222.2 9 20 1 1 1 ===
  • current through the series combination of AIII 778.2222.2512 =−=−= (c) Since p = I2 R, Ω=== 05.4 222.2 20 22 1 1 1 I p R Ω=== 945.1 777.2 15 22 2 2 2 I p R Ω=== 297.1 777.2 10 22 3 3 3 I p R
  • setup of Fig. 2.60, design a voltmeter for the following multiple ranges: (a) 0-1 V (b) 0-5 V (c) 0-50 V (d) 0-100 V Assume that the internal resistance Rm = 2 kΩ and the full scale current .100 AI fs µ= Solution: We apply Eq. (2.60) and assume that R1, R2, R3, and R4 correspond with ranges 0-1 V, 0-5 V, 0-50 V, and 0-100 V, respectively. For range 0-1 V, Ω=−=−= − k x R 82000000,102000 10100 1 61
  • or range 0-5 V, Ω=−=−= − k x R 482000000,502000 10100 5 62 or range 0-50 V, Ω=−=−= − k x R 4982000000,5002000 10100 50 63 or range 0-100 V, Ω=−=−= − k x R 9982000000,000,12000 10100 100 64
  • .9 Summary1. A resistor is a passive element in which the voltage v across it is proportional to the current i through itiRv =;R is the resistance of the resistor en circuit is a resistor with infinite resistan 2. Short circuit is a resistor with zero resistance(R = 0) 3. The conductor G of a resistor is the reciprocal of its resistance R G 1 =
  • 4. A branch is a single two-terminal element in an electric circuit A node is the point of connection between two or more branches A loop is a closed path in a circuit5. KCL states that the sum of the currents entering a node equals the sum of currents leaving the node6. KVL states that the voltage around a closed path algebraically sum to zero
  • Elements are in parallel if they have the same voltage across them8. When two resistors are in series 7. Elements are in series if the same current flows through them 21 RRReq += 21 21 GG GG Geq + = 9. The voltage division principle for two resistors in series is v RR R vv RR R v 21 2 2 21 1 1 , + = + =
  • 10.When two resistors R1 and R2 are in parallel 21 21 21 , GGG RR RR R eqeq += + = 11.The current division principle for two resistors in parallel is i RR R ii RR R i 21 1 2 21 2 1 , + = + = 12.Delta – to – Wye conversion cba ba cba ac cba cb RRR RR R RRR RR R RRR RR R ++ = ++ = ++ = 321 ,,
  • 13.Wye – to – Delta conversion 3 133221 2 133221 1 133221 R RRRRRR R R RRRRRR R R RRRRRR R c b a ++ = ++ = ++ =