§3: Linear Circuit Analysis§3.0 Introduction§3.1 Linear Systems§3.2 Superposition§3.3 Node-Voltage Analysis§3.4 Mesh-Current Analysis§3.5 Equivalent Circuits§3.6 Thevinin Equivalents§3.7 Norton Equivalents§3.8 Source Transformations§3.9 Maximum Power Transfer§3.10 Equivalent Resistance Revisited§3.11 Examples§3.12 Summary§3.0 IntroductionWhen a carpenter works on a house or a mechanic tunes an engine, they each have many tools and various options forcompleting their tasks. Most tasks are straightforward: the carpenter uses a hammer to pound in nails, and the mechanic usesa specially designed wrench to remove the head gasket. Other tasks are slightly more ambiguous and require more information.Consider the carpenter working on a doorframe. Whether he chooses to use a hand saw, a power saw, or a pocket knife dependson the precision required, the number of doorframes to be built, and quite possibly the tools he has immediately available. Themechanic may have even more trouble in determining whether to use a half-inch or 13mm wrench for tightening an assembly.The analysis methods presented in this chapter will provide you with a toolbox capable of tackling almost any linear DC circuit.The various methods will invariably terminate at the same answer, but some are clearly more efficient in solving particular circuits.Our goal is to construct both the toolbox and an understanding for what problems each tool is effective. More importantly, thesetools will provide you with the necessary preperation to analyze more complex circuits: those containing diodes, transistors, anddigital logic.Beginning with simple applications of Ohm’s and Kirchoff’s laws, we will use equivalent impedances and dividers to lay thefoundation for our remaining circuit analyses. Larger circuits using both current and voltage dividers will create the framework.The properties of basic linear systems will allow us to perform analyses by superposition and Node/Mesh equations, furtherfleshing out our options. The last analysis step will be to create simple circuit equivalents for any linear circuit, thus completingour toolbox. Finally, we will address the various methods in a general setting and demonstrate a few tricks that can be used toview the problems more easily.Where possible, we will demonstrate multiple methods in analyzing the circuits: you need only pick the one with which youfeel most comfortable. The benefit of solving a circuit multiple ways lies in the ability to check your answers–you can be prettywell certain that you are correct when you solve by completely different methods and obtain the same solution.We will begin with a simple example to determine an output voltage given a DC input voltage as shown in Figure 3.1.Example: Determine the current I50traveling downward in the 50Ω resistor.Figure 1We know two methods so far to solve this problem:121. We may determine the equivalent resistance looking to the right from the voltage source to then find the total currententering the collection of resistors. Application of a current divider then gives the current through the 50 Ω resistor.Method 1: Req= 25 + 50k(30 + 20) = 50 ΩIT=2V50Ω= 40 mAI50=(30 + 20)(30 + 20) + 50IT= 20 mAFigure 22. We may combine the 20, 30, and 50 Ohm resistors as an equivalent and then find the voltage across the equivalent (sameas the voltage across the 50 Ω resistor). Simple application of Ohm’s law then gives us the current.Method 2: Rcomb= 50k(30 + 20) = 25 ΩVcomb=RcombRcomb+ 25(2V ) = 1 VI50=Vcomb50Ω= 20 mAFigure 3In many ways, this is a very typical problem to solve: a circuit containing one source and perhaps four or five resistors. Let’s tryanother, more difficult, example that verifies the conservation of energy principle: all energy dissipated by a resistor is generatedby the source.Example: Verify the conservation of energy in the circuit shown in Figure 3.4.3Figure 4We use a combination of dividers and Ohm’s law applications to calculate the currents and the power dissipated.Current calculations:I10Ω=1V10Ω= 100 mA I15Ω=1V(15 + 6k30) Ω=1V20Ω= 50 mAI6Ω=3030 + 6· I15Ω= 41.67 mA I30Ω=630 + 6· I15Ω= 8.33 mA I1V= I10Ω+ I15Ω= 150 mAPower calculations:P1V= (1V )(150mA) = 150 mW P6Ω= I26(6Ω) = 10.41 mW P10Ω= I210(10Ω) = 100 mWP15Ω= I215(15Ω) = 37.5 mW P30Ω= I230(30Ω) = 2.08 mWA fully labeled circuit is shown in Figure 3.5.Figure 5To verify that the conservation of energy holds, compare the power generated by the voltage source to the sum of powersdissipated in the resisters.150 mW.= (10.41 + 100 + 37.5 + 2.08) mWNotice that there is an implicit assumption in this last step: we have shown that the powers are equal, but not the energy! Understeady-state conditions, the rate of energy into a circuit (the power) is constant and therefore the conservation of instantaneouspower implies conservation of energy over any time-scale.§3.1: Linear SystemsOne of the most common models studied in engineering is the linear system. The generic linear system is characterized by theproperty of linearity, which requires that the output is a weighted sum of the inputs. Figure 3.6 shows the block diagram of abasic linear system: for inputs x1, x2, ..., xnthere are corresponding outputs y1, y2, ..., yn.4Figure 6While this definition sounds trivial, answer the following question: is f(x) = x + 1 a linear system? Actually, no! Plotting thefunction on a set of axes as shown in Figure 3.7, you will see the answer jump out at you.Figure 7If we scale an input, say αx1, then the output is the same multiple times the corresponding output, αy1. Likewise, if we addtwo inputs and process them simultaneously, say x1+ x2, then the output is the corresponding sum y1+ y2.Now, let’s return to the question of whether f (x) = x + 1 qualifies as a linear system: consider the three values f (0) = 1,f(1) = 2, and f (2) = 3. If f(x) was linear, then f (0 + 1) = f(0) + f(1), but 2 6= 1 + 2 = 3. Further, if f (x) were linear thenf(2 · 1) = 2 · f(1) which is again contradicted. The only conclusion is that f(x) = x + 1 is nonlinear, despite being a line; a lineneed not be mathematically linear! If the line happens to pass through the origin, then it will be linear.Combining the properties discussed, we obtain the system responseyk= ck· xk∀k ∈ Nsuccinctly stated as: “the output is a weighted sum of the inputs.” Two important characteristics of the linear system tonotice are that a