BME 205 L04 CIRCUIT THEOREMS Introduc9on • We learned how to solve circuits by applying Kirchhoff’s laws systema9cally. But it can be tedious for complex circuits. • To handle complex circuits, theorems have been developed to simplify circuit analysis, including Thevenin’s and Norton’s theorems. • These theorems are applicable to linear circuits, so we first discuss the concept of circuit linearity. We’ll also discuss the concepts of superposi4on, source transforma4on, and maximum power transfer. Linearity • Linearity is the property of an element describing a linear rela9onship between cause and effect. • Linearity combines the proper9es of homogeneity (scaling) and addi4vity. • Homogeneity: if the input (also called the excita4on) is mul9plied by a constant, then the output (also called the response) is mul9plied by the same constant. • e.g. Ohm’s law for a resistor: v = iR
• If the current is increased by a constant k, then the voltage increases correspondingly by k: kv = kiR
• Addi9vity: the response to a sum of inputs is the sum of the responses to each input applied separately. • e.g. for a linear resistor, if v1 = i1 R and v2 = i2 R then applying (i1 + i2) gives v = (i + i )R = i R + i R = v + v
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• Obviously, a resistor is a linear element. • In general, a circuit is linear if it is both addi9ve and homogeneous. • We’ll only deal with linear circuits, which consist of only linear elements, linear dependent sources, and independent sources. (Note that since p = i2R = v2/R (quadra9c func9on), the rela9onship between power and voltage (or current) is nonlinear.) • Consider the linear circuit shown. It has no independent sources inside it. It is excited by a voltage source vs, which serves as the input. The circuit is terminated by a load R. We may take the current i through R as the output. Suppose vs = 10 V gives i = 2 A. According to the linearity principle, vs = 1 V will give i = 0.2 A, and i = 1 mA must be due to vs =5 mV. Example • Find i0 when vs = 12V and vs = 24V. • Applying KVL to the two loops, we obtain add: subs9tute: Example Prac9ce Find v0 when is = 15 A and is = 30 A. Current division: io=is (2/12) vo=4is (2/12)=2is/3 vo=10 V when is=15 A vo=20 V when is=30 A v1 If vo=1, io=1/8 A v1=20io=2.5 V But v1=10 V, so vo=4 V Superposi9on • If a circuit has two or more independent sources, one way to determine the value of a specific variable (voltage or current) is to use nodal or mesh analysis as in Lecture 3. • Another way is to determine the contribu9on of each independent source to the variable and then add them up. The la`er approach is known as superposi4on and it relies on linearity. The voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltages across (or currents through) that element due to each independent source ac9ng alone. • The principle of superposi9on helps us to analyze a linear circuit with more than one independent source by calcula9ng the contribu9on of each independent source separately. • To apply the superposi9on principle, we must keep two things in mind: 1. We consider one independent source at a 9me while all other independent sources are turned off. This implies that we replace every voltage source by 0 V (or a short circuit), and every current source by 0 A (or an open circuit). This way we obtain a simpler and more manageable circuit. 2. Dependent sources are lec intact because they are controlled by circuit variables. Superposi9on is applied in 3 steps: NOTES • Analyzing a circuit using superposi9on may involve more work -­‐ If the circuit has three independent sources, we may have to analyze three simpler circuits each providing the contribu9on due to the respec9ve individual source. However, superposi9on does help reduce a complex circuit to simpler circuits through replacement of voltage sources by short circuits and of current sources by open circuits. • Keep in mind that superposi9on is based on linearity. For this reason, it is not applicable to the effect on power due to each source, because the power absorbed by a resistor depends on the square of the voltage or current. If the power value is needed, the current through (or voltage across) the element must be calculated first using superposi9on. Example Solu9on: Since there are two sources let v = v1 + v2 where v1 and v2 are the contribu9ons due to the 6-­‐V voltage source and the 3-­‐A current source, respec9vely. Get v1 from (a): (voltage division) Get v2 from (b)... (current division) Then: Examples 1) Voltage division: vo=20(2/10)=4 V current division: io=8(5/10)=4 A Ohm’s law: vo=8 V Total: vo=4 V + 8 V =12 V 2) Mesh analysis at mesh 1: 10i1 – i2 = 20 – 5io i2 io=-­‐i1 à 5i1-­‐i2=20 Mesh analysis at mesh 2: 6i2 – i1 = 5io i1 6i2 + 4i1 = 0 6(5i1-­‐20)+ 4i1 = 0 Solving: i1=60/17 à io’=-­‐60/17 i1=4 Mesh analysis at mesh 2: 6i2 – 3i1 – i3 -­‐ 5io=0 io=4-­‐i3 6i2 + 4i3 = 32 Mesh analysis at mesh 3: (i3 – i2) +5io +4i3+5(i3-­‐i1) =0 – i2 +5i3 =0 i2 Solving: i3=16/17 à io’’=4-­‐16/17=52/17 i1 io=io’+io’’=-­‐60/17+52/17=-­‐8/17 i3 Example • use the superposi9on theorem to find i. i1=12/6=2 A Mesh a: 16ia-­‐4ib=-­‐24 Node 1: v1/4+v1/3=(v2-­‐v1)/4 Mesh b: -­‐4ia+7ib=0 Node 2: (v2-­‐v1)/4+v2/8=3 Solving: i2=ib=-­‐1 Solving: v1=3 à i3=1 Prac9ce Problem • Find i in this circuit using the superposi9on principle. i=16/(6+2+8)=1 A Current division: i=4 ( 2/ (2+14) ) = 0.5 A i=-­‐12/(6+2+8)=-­‐0.75 A Total: i= 1 A + 0.5 A – 0.75 A = 0.75 A Source transforma9on • Source transforma4on is another tool for simplifying circuits. Basic to these tools is the concept of equivalence. • Recall that an equivalent circuit is one whose v-­‐i characteris4cs are iden9cal with the original circuit. • In Lecture 3, we saw that node-­‐voltage (or mesh-­‐current) equa9ons can be obtained by mere inspec9on of a circuit when the sources are all independent current (or all independent voltage) sources. • It is therefore expedient in circuit analysis to be able to subs9tute a voltage source in series with a resistor for a current source in parallel with a resistor, or vice versa. • This is known as a source transforma4on. • These two circuits are equivalent – i.e. they have the same voltage-­‐current rela9on at terminals a-­‐b. This is easy to show: • If the sources are turned off, the equivalent resistance at terminals a-­‐b in both circuits is R. • Also, when terminals a-­‐b are short circuited, the short-­‐circuit current flowing from a to b is isc = vs /R in the circuit on the lec-­‐hand side and isc = is for the circuit on the right hand side. • Thus, vs /R = is in order for the two circuits to be equivalent. • Hence, source transforma9on requires that • Source transforma9on also applies to dependent sources, provided we carefully handle the dependent variable. A dependent voltage source in series with a resistor can be transformed to a dependent current source in parallel with the resistor or vice versa: • A source transforma9on does not affect the remaining part of the circuit. When applicable, source transforma9on is a powerful tool that allows circuit manipula9ons to ease circuit analysis. However, we should keep the following points in mind: 1. Note that the arrow of the current source in the last two figures is directed toward the posi9ve terminal of the voltage source. 2. Note that source transforma9on is not possible when R = 0, which is the case with an ideal voltage source. However, for a prac9cal, nonideal voltage source, R ≠ 0. Similarly, an ideal current source with R =∞ cannot be replaced by a finite voltage source. Example three transforming steps: i=2(2/10)=0.4 A à vo=3.2 V Prac9ce examples 1) Find i0 using source transforma9on. 5 -­‐ + 6 30 + -­‐ " 9 −3 0 %" i1 % " 30 %
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3 i1 5 7 i2 i3 + -­‐ 15 io = i2 − i3 = 1.78 A
Prac9ce examples 2) Find vx 4 3 4 3 1 vx 2 vx + -­‐ 18 3 + -­‐ + -­‐ + -­‐ 18 2 1 + -­‐ vx 4 vx + -­‐ vx -­‐3 + 5i + vx + 18=0 -­‐3 + i + vx = 0 i=-­‐18/4 A à vx = 7.5 V + -­‐ 18 Prac9ce examples 3) Find ix v 4 10 5 2ix/5 KCL: 4=v/10 + v/5 + 2/5 ix Ohm’s Law: v=10ix Solving: ix=1.176 A Thevenin’s Theorem • It ocen occurs in prac9ce that a par9cular element in a circuit is variable (usually called the load) while other elements are fixed. • As a typical example, a household outlet terminal may be connected to different appliances cons9tu9ng a variable load. Each 9me the variable element is changed, the en9re circuit has to be analyzed all over again. To avoid this problem, Thevenin’s theorem provides a technique by which the fixed part of the circuit is replaced by an equivalent circuit. • According to Thevenin’s theorem, the linear circuit (a) can be replaced by that in (b). (a) (b) (The load may be a single resistor or another circuit.) The circuit to the lec of the terminals a-­‐b is known as the Thevenin equivalent circuit. Thevenin’s theorem states that a linear two-­‐terminal circuit can be replaced by an equivalent circuit consis9ng of a voltage source VTh in series with a resistor RTh, where VTh is the open-­‐circuit voltage at the terminals and RTh is the input or equivalent resistance at the terminals when the independent sources are turned off. •
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(a) Two circuits are said to be equivalent if they have the same voltage-­‐current rela9on at their terminals. What will make the last two circuits (a) and (b) equivalent? If the terminals a-­‐b are made open-­‐circuited (by removing the load), no current flows, so that the open-­‐circuit voltage across the terminals a-­‐b in (a) must be equal to the voltage source VTh in (b) since the two circuits are equivalent. Thus VTh is the open-­‐circuit voltage across the terminals as shown below; that is, (b) •
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Again, with the load disconnected and terminals a-­‐b open-­‐circuited, we turn off all independent sources. The input resistance (or equivalent resistance) of the dead circuit at the terminals a-­‐b in (a) must be equal to RTh in (b) because the two circuits are equivalent. Thus, RTh is the input resistance at the terminals when the independent sources are turned off, as shown below; that is, To apply this idea in finding RTh, we need to consider two cases: CASE 1 If the network has no dependent sources, we turn off all independent sources. RTh is the input resistance of the network looking between terminals a and b, as shown above. CASE 2 If the network has dependent sources, we turn off all independent sources. As with superposi9on, dependent sources are not to be turned off because they are controlled by circuit variables. (a) (b) Case 2 cont’d: We apply a voltage source v0 at terminals a and b and determine the resul9ng current i0. Then RTh = v0 /i0 , as shown below Alterna9vely, we may insert a current source i0 at terminals a-­‐b as shown below and find the terminal voltage v0 . Again RTh = v0 /i0 . Either of the two approaches will give the same result. We may assume any value of v0 and i0 . For example, we may use v0 = 1 V or i0 = 1 A, or even use unspecified values of v0 or i0 . • Thevenin’s theorem is very important in circuit analysis. A large circuit may be replaced by a single independent voltage source and a single resistor. This is a powerful tool in circuit design. • As men9oned earlier, a linear circuit with a variable load can be replaced by the Thevenin equivalent, exclusive of the load. The equivalent network behaves the same way externally as the original circuit. Consider a linear circuit terminated by a load RL : (a) (b) • The current IL through the load and the voltage VL across the load are easily determined once the Thevenin equivalent of the circuit at the load’s terminals is obtained, as shown in (b). We obtain Example • Find the Thevenin equivalent circuit of the circuit shown, to the lec of the terminals a-­‐b. Then find the current through RL = 6, 16, and 36 ohms. 16i1-­‐12i2=32 i2=-­‐2 i1=0.5 Aà Vth =12(i1-­‐i2)=30 V Examples/Prac9ce 1) Find Equivalent circuit to the lec of terminals a-­‐b To find Rth, disconnect all independent sources. 6 6 4 To find Vth, open-­‐circuit a-­‐b and solve the circuit. 6 6 12 V + -­‐ i1 i2 4 Rth=12||4=3 Ohms Mesh analysis (supermesh): -­‐12 + 6i1 + 6i2 + 4i2=0 KCL at current source: i1 +2= i2 Solving: i2 =1.5 A Vth=4i2=6 V 2) To find Rth, disconnect all independent sources and apply a unit voltage source at a-­‐b. i1 io i2 i3 + -­‐ Mesh analysis at mesh 1: -­‐2vx+2(i1-­‐i2)=0à -­‐2vx+2i1-­‐2i2=0à2i1+6i2=0 Mesh analysis at mesh 2: 4i2+2(i2-­‐i1)+6(i2-­‐i3)=0à-­‐2i1+12i2-­‐6i3=0 Mesh analysis at mesh 3: 2i3+1+6(i3-­‐i2)=0à-­‐6i2+8i3=-­‐1 Solving: i3=-­‐1/6àio=-­‐i3=1/6àRth=1/(1/6)=6 Ohms 1 V (vx=-­‐4i2) 2) To find Vth, open-­‐circuit a-­‐b and solve the circuit. i3 Mesh 1: i1=5 Mesh 2: 12i2-­‐4i1-­‐2i3=0à12i2-­‐2i3=20 Mesh 3: -­‐2vx+2(i3-­‐i2)=0à vx=i3-­‐i2 vx=4(i1-­‐i2)à20-­‐4i2=i3-­‐i2à3i2-­‐i3=20 Solving: i2=10/3 A ; i3=10 A Vth=6(10/3)=20 V i1 i2 3) To find Rth, disconnect all independent sources and apply a unit current source at a-­‐b. v1 v2 1 A Nodal analysis at node 1: 1.5Ix = v1/5 + Ixà Ix =2v1/5 Ix = (v1-­‐v2)/3=2v1/5àv1+5v2=0 Nodal analysis at node 2: Ix +1=v2/4 (v1-­‐v2)/3+1=v2/4à4v1-­‐7v2=-­‐12 Solving: v1=-­‐2.22; v2=0.44 Rth=v2/1=0.44 Ohms 3) v1 v2 To find Vth, open-­‐circuit a-­‐b and solve the circuit. Nodal analysis at node 1: (v1-­‐6)/5+Ix=1.5Ix Nodal analysis at node 2: Ix=v2/4 Ix = 2(v1-­‐6)/5=(v1-­‐v2)/3 v1+5v2=36 (v1-­‐v2)/3=v2/4à4v1-­‐7v2=0 Solving: v1=9.33; v2=5.33 V Vth=v2=5.33 Example Find the Thevenin equivalent. Since it has no independent sources, VTh=0. To find RTh apply a current source at the terminals as in (b). Nodal analysis gives: but so Thus Nega9ve resistance tells us the circuit is supplying power. This must come from the dependent source – we’ve used it to simulate nega9ve resistance. Prac9ce Problem • Obtain the Thevenin equivalent of this circuit: 1 A i1 Vth=0 (no independent sources) Mesh analysis: 5i1+10i1+4vx+15(i1-­‐i2)=0 Vx=-­‐5i1 i2=-­‐1 10i1=-­‐15ài1=-­‐1.5 A vo=15(i1-­‐i2)=15(-­‐1.5+1)=-­‐7.5 V Rth=-­‐7.5 V / 1 A = -­‐7.5 Ohms i2 Norton’s theorem • Very similar to Thevenin: can be replaced by: • We find RN in the same way we find RTh. • In fact, from what we know about source transforma9on, the Thevenin and Norton resistances are equal; that is, • To find the Norton current IN, we determine the short-­‐circuit current flowing from terminal a to b. The short-­‐circuit current in the Norton equivalent is IN , which must be the same as in the original circuit since the two circuits are equivalent. Thus: • This is essen9ally source transforma9on. For this reason, source transforma9on is ocen called Thevenin-­‐Norton transforma9on. • Dependent and independent sources are treated the same way as in Thevenin’s theorem. So to determine the Thevenin or Norton equivalent circuit requires that we find: • The open-­‐circuit voltage voc across terminals a and b. • The short-­‐circuit current isc at terminals a and b. • The equivalent or input resistance Rin at terminals a and b when all independent sources are turned off. We can calculate any two of the three using the method that takes the least effort and use them to get the third using Ohm’s law. Also since the open-­‐circuit and short-­‐circuit tests are sufficient to find any Thevenin or Norton equivalent. Example and the Norton equivalent cct is: Thus, Prac9ce Problem • Find the Norton Equivalent circuit To find RN, disconnect independent sources. 3 3 To find IN, short circuit a-­‐b and solve. v 6 RN=6||6=3 Nodal analysis: (v-­‐15)/3+v/3=4 Note: no current through 6 Ω resistor v=13.5 V IN=v/3 Ω=13.5 V / 3 Ω=4.5 A Examples 1) Using Norton’s theorem, find RN and IN of this circuit between terminals a & b To find RN, disconnect independent sources, apply a unit voltage to terminals, and solve for io. io + -­‐ vo=1 V Ohm’s law: Ix=(0-­‐1) V / 5 Ω =-­‐0.2 A KCL at node a: 2Ix + Ix +io=0àio=0.6 A RN=vo/io=1 V/ 0.6 A = 1.67 Ohms Examples 1) Using Norton’s theorem, find RN and IN of this circuit between terminals a & b To find iN, short circuit the terminals and solve for the short-­‐circuit current: i3 iN i1 i2 Mesh analysis at mesh 1: 4i1+10=0à i1=2.5 A Mesh analysis at mesh 2: -­‐10+5(i2-­‐i3)=0 ix=i2-­‐i3 -­‐10+5ix=0àix=2 A KCL at terminal a: iN=ix+2Ix=6 A Examples 2) Using Norton’s theorem, find RN and IN of this circuit between terminals a & b To find RN, disconnect independent sources, apply a unit current to terminals, and solve for vo. vo=vx i1 To find IN, short circuit a-­‐b and solve for the short-­‐circuit current. i2 To find IN, short circuit a-­‐b io=1 A 6i1+2vx+2(i1-­‐i2)=0 vx=2(i1-­‐i2) i2=-­‐1 i=0 Subs9tu9ng: 6i1+4(i1-­‐i2) +2(i1-­‐i2)=0ài1=i2/2=-­‐1/2 6 vo=vx=2(i1-­‐i2)=2(-­‐1/2+1)=1 V RN=vo/io=1 Ohm IN 60 V + -­‐ IN vx=0 V IN=I=60 V /6 Ω=10 A Maximum Power Transfer • In many prac9cal situa9ons, a circuit is designed to provide power to a load. While for electric u9li9es, minimizing power losses in the process of transmission and distribu9on is cri9cal for efficiency and economic reasons, there are other applica9ons in areas such as communica9ons where it is desirable to maximize the power delivered to a load. • How do we deliver the maximum power to a load when given a system with known internal losses? • The Thevenin equivalent is useful in finding the maximum power a linear circuit can deliver to a load. We assume that we can adjust the load resistance RL. • If the en9re circuit is replaced by its Thevenin equivalent except for the load as shown, the power delivered to the load is • For a given circuit, VTh and RTh are fixed. By varying the load resistance RL, the power delivered to the load varies as sketched above. The power is small for small or large values of RL but maximum for some value of RL between 0 and ∞. • It’s easy to show that this maximum power occurs when RL = RTh. This is known as the maximum power theorem.