05_Nodal_Analysis

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 (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

Physics 302/328: Nodal Analysis o R 3 and R 4 are zero The numeric answer is that V 4 = 1.056 Volts. Using the other equations, we can then get V 2 = 3.68 Volts, V 3 = 2.04 volts. Double check that these numbers work by mapping them on the circuit. 4) If we compare the solution in the last question to a solution using our previous analysis technique, it isn’t clear that the Nodal Analysis technique is any easier. So, what’s the point of Nodal Analysis? Well, Nodal Analysis is the best choice in some circumstances. Consider the same circuit as above but with R 5 = 0. Then there is only one equation for the whole circuit. At node V 2 we would have V 1 − V 2 R 1 + 0 − V 2 R 3 + R 4 = I S 2 and V 2 is the only unknown (remember V 1 =  V S1 ) ============================================ 5) Quick Summary of Nodal Analysis a) It expands the  V in  V = IR . b) It is typically used if you want to solve for voltages (and the circuit is relatively simple as illustrated in the last example) c) In terms of solving problems, i) Label currents, as usual, with passive sign convention. ii) Label node voltages; choose ground. iii) Then, even though you want to solve for voltages, you actually start with KCL….examine junctions, write down KCL, then sub in for the currents using I =  V /R. d) Note about the book’s notation: G = 1 R . Why would this be useful? Page 5 of 10

Physics 302/328: Nodal Analysis Also, because of the power supplies themselves we have: (eqn 4) V 1 – V o = 4V A. We have a lot of unknowns here (3 currents and 3 voltages). But if we recognize the following, we can reduce it. First of all, (eqn 5) V A = V S6 – V 1 . Also, KCL gives us that (eqn 6) I 4 = I 6 + I 12 Now we have enough equations for our unknowns. Remember that we’re solving for V o . Sub 1, 2, and 3 into 6: V S 6 − V 1 R 4 = V 1 − 0 R 6 + V o − 0 R 12 Try to eliminate V 1 for V o . Combine our last two equations: 5 and 4. That gives V 1 – V o = 4(V S6 – V 1 ) Solve that for V 1 and sub it into the previous equations. Then solve for V o V o = V S 6 R 12 ( R 6 − 4 R 4 ) R 4 R 12 + R 6 R 12 + 5 R 4 R 6 Evaluate (1) units (2) check our educated guess (a) If R 12 goes to zero, (b) If R 4 goes to zero or infinity (c) If R 6 and R 12 go to infinity Page 7 of 10

Physics 302/328: Nodal Analysis d) Notice how these types of problems all have many nodes that are all ground . It makes the problem much easier. For example in the problem that we actually solve above (with the current source), setting R 5 to zero greatly simplifies the problem to only 3 eqns and 3 unknowns 7) Last Example: What is the power in the 6 ohm resistor? a) 8.2 W b) 15.3W c) 4.4 W d) 13.5 W Answer: Concept Recognize that P 6 = (V x – 0) 2 /6  Use Nodal Analysis (ohm’s law, KCL, and KVL) and solve for V x . Educated Guess for V x (1) If either R 12 or R 6 = 0, then V x = 0 (2) If R 4 = 0, then V x = 12 Volts Outline solution (eqn 1) I 1 + I L = I 6 (from KCL) This can be rewritten using Ohm’s law (such as I 1 = V S 12 − V x R 4 ) as (eqn 1a) V S 12 − V x R 4 + I L = V x − 0 R 6 Need to find I L . Use the other node that has I L (eqn 2) 2I 1 = I L + I 12 (from KCL) This can be rewritten using Ohm’s law as Page 8 of 10

Physics 302/328: Nodal Analysis Page 10 of 10