05_Nodal_Analysis

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Physics 302/328: Nodal Analysis Objectives:  Discuss how the technique of Nodal Analysis works (based on KVL, KCL and Ohm’s Law) and how it can be used to answer questions. This is section 3.1 in the book Basic Engineering Circuit Analysis, 11 th ed. 1) Review (relevant for what will come next): Ohm’s Law and all of the voltage labels we’ve been using so far are really  V . a) Here are some examples of this using a “reference point” in the circuit. I’ll call this reference point “ground” and since voltage differences are all that matter we can choose this ground to be any voltage we want. Choose something easy…zero. b) Set point d to ground and then find V a , V b , V c , V e and  V ad c) Find V s given that V eb = 18 Volts Page 1 of 10

Physics 302/328: Nodal Analysis ============================= 2) What is Nodal Analysis? a) This is a decent technique to use if you want to find voltages. It is nothing new: Just KVL, KCL and Ohm’s Law. It is built on the ideas of the last two examples. b) In this case, the “unknowns” are the node voltages. c) Choose a “ground” (i.e. a place where the voltage is zero 1 ) because we need a reference voltage d) Still draw currents, apply passive sign convention, and use KVL, KCL, and ohm’s Law e) Foreshadowing: it is only a more useful technique (compared to what we just did) if you have a circuit where there is a “common” wire…. ============================ 3) Nodal Analysis Example 1: R 1 = 50 ohms, R 3 = 100 ohms, R 4 = 60 ohms and R 5 = 40 ohms. ΔV S 1 = 5 Volts, I S2 = 10 mA. Find the value of the voltage at the node between R 5 and R 4 with the given ground. Answer:  Concepts: o Use KVL, KCL, and Ohm’s Law  Approach: o Same as always, draw the currents and use the passive sign convention. o But with the nodal analysis, we go one step further. We label the voltages at the nodes 1 How can we just choose a ground? Well, remember that all that really matters is change-in-voltage or voltage differences….So choosing a zero doesn’t really effect the laws of circuits. Page 2 of 10

Physics 302/328: Nodal Analysis There are 4 unknowns here, namely V 1 thru V 4 . We will probably need 4 equations. Concepts:  Ohm’s Law says: ΔV = IR . For us that means the following: o V 1 – V 2 = I 1 R 1 ; o V 3 – V 2 = I 3 R 3 , o V 4 – V 3 = I 3 R 4 , and o V 4 – 0 = I 1 R 5 .  KCL (I 1 + I 3 ) in = (I S2 ) out , etc.. to be used at each node.  KVL is usually not used very much in Nodal analysis but it is implicitly used (in the sense that each node has one voltage). Educated Guess for V 4  If R 5 = 0, V 4 = 0  If R 3 or R 4 is infinity, then the current is confined to the left loop. Since there is a current source I S2 , then that will be the current. Hence the V 4 would at a voltage of I S2 R 5 above ground… V 4 = I S2 R 5 . Here is how we will construct our equations: Look at each node: What circuit elements are directly connected to that node? And decide which is coming “in” and “out” by the direction of the current  If it is a source, you can use energy principles to relate the node voltage to the source voltage (see for example node V 1 below)  If it is a resistor, usually you start with KCL (at the node) and sub in using Ohm’s Law because (1) we know that KCL applies to nodes and (2) we want to get voltages not currents => So use ohm’s law to replace the currents with voltages Find V 4 : From KCL at the node of V 4 : (I 1 + I 3 ) out = (I S2 ) in becomes (using ohm’s law; V 4 – 0 = I 1 R 5 and V 4 – V 3 = I 3 R 4 ): (eqn 1) V 4 − 0 R 5 + V 4 − V 3 R 4 = I S 2 Note that we can simplify all of this a little more by using the R 4 + R 3 series combination and equation 1 would be (eqn 1b) V 4 − 0 R 5 + V 4 − V 2 R 4 + R 3 = I S 2 Find V 2 From KCL at the node of V 2 : (I 1 + I 3 ) in = (I S2 ) out becomes Page 3 of 10

Physics 302/328: Nodal Analysis (eqn 2) Find V 1 From energy at node V 1 (eqn 3) V 1 – 0 = ΔV S 1 (don’t need KCL because this node is directly connected to voltage source) Now, we’ve got 3 equations and 3 unknowns. Side note: why don’t we use the node at V 1 as V 4 − 0 R 5 = V 1 − V 2 R 1 (which is perfectly legitimate because I 1 = I 1 )? Or for that matter at V 3 ? Well the reason is these are not new equations. It is the same reason that we don’t use all the loops for KVL…(and we can solve the problem using just the 3 equations we have) Now for algebra: Let’s first get things out of the denominator by multiplying them through with the R value and then use eqn 3 to eliminate V 1 from all of our equations, the result is: Eqn 1b => V 4 ( R 4 + R 3 ) + ( V 4 − V 2 ) R 5 = R 5 ( R 4 + R 3 ) I S 2 Eqn 2 => ( ΔV S 1 − V 2 ) ( R 3 + R 4 ) + ( V 4 − V 2 ) R 1 = R 1 ( R 3 + R 4 ) I S 2 At this point, we want to collect common terms of our unknowns: Eqn 1b => V 4 ( R 4 + R 3 + R 5 ) − V 2 R 5 = R 5 ( R 4 + R 3 ) I S 2 Eqn 2 => ( ΔV S 1 ) ( R 3 + R 4 ) − V 2 ( R 3 + R 4 + R 1 ) + V 4 R 1 = R 1 ( R 3 + R 4 ) I S 2 Solution, after some algebra, is V 4 = R 5 ( I S 2 [ R 3 + R 4 ] + ΔV S 1 ) R 2 + R 3 + R 4 + R 5 Evaluate  Units  Check our educated guess: o R 5 = 0 o R 3 or R 4 is infinity. Page 4 of 10

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