M09_BOYL3605_13_SE_C09.indd Page 359 24/11/14 1:59 PM f403 /204/PH01893/9780133923605_BOYLESTAD/BOYLESTAD_INTRO_CIRCUIT_ANALYSIS13_SE_978013 ... Network Theorems 99 • Become familiar with the superposition theorem Objectives and its unique ability to separate the impact of each source on the quantity of interest. • Be able to apply Thévenin’s theorem to reduce any two-terminal, series-parallel network with any number of sources to a single voltage source and series resistor. • Become familiar with Norton’s theorem and how it can be used to reduce any two-terminal, series- parallel network with any number of sources to a single current source and a parallel resistor. • Understand how to apply the maximum power transfer theorem to determine the maximum power to a load and to choose a load that will receive maximum power. • Become aware of the reduction powers of FPO Millman’s theorem and the powerful implications of the substitution and reciprocity theorems. 9.1 INTRODUCTION This chapter introduces a number of theorems that have application throughout the field of electricity and electronics. Not only can they be used to solve networks such as encountered in the previous chapter, but they also provide an opportunity to determine the impact of a particular source or element on the response of the entire system. In most cases, the network to be analyzed and the mathematics required to find the solution are simplified. All of the theorems appear again in the analysis of ac networks. In fact, the application of each theorem to ac networks is very similar in content to that found in this chapter. The first theorem to be introduced is the superposition theorem, followed by Thévenin’s theorem, Norton’s theorem, and the maximum power transfer theorem. The chapter concludes with a brief introduction to Millman’s theorem and the substitution and reciprocity theorems. 9.2 SUPERPOSITION THEOREM The superposition theorem is unquestionably one of the most powerful in this field. It has such widespread application that people often apply it without recognizing that their maneu- vers are valid only because of this theorem. In general, the theorem can be used to do the following: • Analyze networks such as introduced in the last chapter that have two or more sources that are not in series or parallel. • Reveal the effect of each source on a particular quantity of interest. • For sources of different types (such as dc and ac, which affect the parameters of the Th network in a different manner) and apply a separate analysis for each type, with the total result simply the algebraic sum of the results. GRIDLINE SET IN 1ST-PP TO INDICATE SAFE AREA; TO BE REMOVED AFTER 1ST-PP M09_BOYL3605_13_SE_C09.indd Page 360 24/11/14 1:59 PM f403 /204/PH01893/9780133923605_BOYLESTAD/BOYLESTAD_INTRO_CIRCUIT_ANALYSIS13_SE_978013 ... 360 NETWORK THEOREMS Th The first two areas of application are described in detail in this section. The last are covered in the discussion of the superposition theorem in the ac portion of the text. The superposition theorem states the following: The current through, or voltage across, any element of a network is equal to the algebraic sum of the currents or voltages produced independently by each source. In other words, this theorem allows us to find a solution for a current or voltage using only one source at a time. Once we have the solution for each source, we can combine the results to obtain the total solution. The term algebraic appears in the above theorem statement because the cur- rents resulting from the sources of the network can have different direc- tions, just as the resulting voltages can have opposite polarities. If we are to consider the effects of each source, the other sources obviously must be removed. Setting a voltage source to zero volts is like placing a short circuit across its terminals. Therefore, when removing a voltage source from a network schematic, replace it with a direct connection (short circuit) of zero ohms. Any internal resistance associated with the source must remain in the network. Setting a current source to zero amperes is like replacing it with an open circuit. Therefore, when removing a current source from a network schematic, replace it by an open circuit of infinite ohms. Any internal resistance associated with the source must remain in the network. The above statements are illustrated in Fig. 9.1. Rint Rint I Rint Rint E FIG. 9.1 Removing a voltage source and a current source to permit the application of the superposition theorem. Since the effect of each source will be determined independently, the number of networks to be analyzed will equal the number of sources. If a particular current of a network is to be determined, the contribution to that current must be determined for each source. When the effect of each source has been determined, those currents in the same direction are added, and those having the opposite direction are subtracted; the algebraic sum is being determined. The total result is the direction of the larger sum and the magnitude of the difference. Similarly, if a particular voltage of a network is to be determined, the contribution to that voltage must be determined for each source. When the effect of each source has been determined, those voltages with the same polarity are added, and those with the opposite polarity are sub- tracted; the algebraic sum is being determined. The total result has the polarity of the larger sum and the magnitude of the difference. GRIDLINE SET IN 1ST-PP TO INDICATE SAFE AREA; TO BE REMOVED AFTER 1ST-PP M09_BOYL3605_13_SE_C09.indd Page 361 24/11/14 1:59 PM f403 /204/PH01893/9780133923605_BOYLESTAD/BOYLESTAD_INTRO_CIRCUIT_ANALYSIS13_SE_978013 ... Th SUPERPOSITION THEOREM 361 Superposition cannot be applied to power effects because the power R1 is related to the square of the voltage across a resistor or the current 12 ⍀ through a resistor. The squared term results in a nonlinear (a curve, not a I straight line) relationship between the power and the determining current 2 or voltage. For example, doubling the current through a resistor does not E 36 V R2 6 ⍀ I 9 A double the power to the resistor (as defined by a linear relationship) but, in fact, increases it by a factor of 4 (due to the squared term). Tripling the current increases the power level by a factor of 9. Example 9.1 demon- strates the differences between a linear and a nonlinear relationship. A few examples clarify how sources are removed and total solutions FIG. 9.2 Network to be analyzed in Example 9.1 using the obtained. superposition theorem. EXAMPLE 9.1 Current source replaced by open circuit a. Using the superposition theorem, determine the current through R resistor R2 for the network in Fig. 9.2. 1 b. Demonstrate that the superposition theorem is not applicable to 12 ⍀ power levels. Solutions: E 36 V R2 6 ⍀ IЈ2 a. In order to determine the effect of the 36 V voltage source, the cur- rent source must be replaced by an open-circuit equivalent as shown in Fig. 9.3. The result is a simple series circuit with a current equal to E E 36 V 36 V FIG. 9.3 I2′ = = = = = 2 A Replacing the 9 A current source in Fig. 9.2 by an R R R 12 Ω 6 Ω 18 Ω T 1 + 2 + open circuit to determine the effect of the 36 V Examining the effect of the 9 A current source requires replacing voltage source on current I2. the 36 V voltage source by a short-circuit equivalent as shown in Fig. 9.4. The result is a parallel combination of resistors R and R . 1 2 R Applying the current divider rule results in 1 I 12 ⍀ R1(I) (12 Ω)(9 A) I2″ = = = 6 A R1 + R2 12 Ω + 6 Ω R2 6 ⍀ I = 9 A Since the contribution to current I2 has the same direction for IЈЈ 2 each source, as shown in Fig. 9.5, the total solution for current I2 is the sum of the currents established by the two sources. That is, I2 = I2′ + I2″ = 2 A + 6 A = 8 A FIG. 9.4 b. Using Fig. 9.3 and the results obtained, we find the power delivered Replacing the 36 V voltage source by a short-circuit to the 6 Ω resistor equivalent to determine the effect of the 9 A current source on current I2. 2 2 P1 = (I2′) (R2) = (2 A) (6 Ω) = 24 W Using Fig. 9.4 and the results obtained, we find the power delivered to the 6 Ω resistor IЈ2 = 2 A I2 = 8 A P (I″)2(R ) (6 A)2(6 Ω) 216 W 2 = 2 2 = = R2 6 ⍀ R2 6 ⍀ Using the total results of Fig. 9.5, we obtain the power delivered to the 6 Ω resistor IЈЈ 2 = 6 A 2 2 PT = I 2R2 = (8 A) (6 Ω) = 384 W FIG. 9.5 It is now quite clear that the power delivered to the 6 Ω resistor Using the results of Figs. 9.3 and 9.4 to determine using the total current of 8 A is not equal to the sum of the power current I2 for the network in Fig. 9.2. levels due to each source independently. That is, P1 + P2 = 24 W + 216 W = 240 W PT = 384 W GRIDLINE SET IN 1ST-PP TO INDICATE SAFE AREA; TO BE REMOVED AFTER 1ST-PP M09_BOYL3605_13_SE_C09.indd Page 362 24/11/14 1:59 PM f403 /204/PH01893/9780133923605_BOYLESTAD/BOYLESTAD_INTRO_CIRCUIT_ANALYSIS13_SE_978013 ... 362 NETWORK THEOREMS Th To expand on the above conclusion and further demonstrate what is meant by a nonlinear relationship, the power to the 6 Ω resistor versus current through the 6 Ω resistor is plotted in Fig.