Unit 2 Circuit Analysis Techniques In this unit we apply our knowledge of KVL, KCL and Ohm s Law to develop further techniques for circuit analysis. The material is based on Chapter 4 of the text and that chapter should be read in conjunction with the notes. For the most part we will study the new techniques by way of examples. Again, see the definitions in Table 4.1 of the text. In particular we note the following which will be directly relevant in this Unit: (1) An essential node is a point in a circuit where three or more circuit elements join. (2) An essential branch is a path that connects two essential nodes without passing through an essential node. In solving the circuits of this Unit that is, in finding any unknown parameters (voltages, currents or resistances) of course, we must analyze the circuit to obtain the same number of independent equations as there are unknowns. We have already seen that if there are n nodes in a circuit, we may obtain n 1 equations by using KCL. If there are b unknown currents in total, then clearly b equations are required so that b (n 1) equations must be found by some other means KVL is a good candidate. Remember that the b equations must be independent. Deciding on which techniques to use in developing the minimum number of equations required takes PRACTICE. 2.1 The Node-Voltage Method In circuit analysis it is often useful to consider one node in a circuit to be a reference node relative to which the voltages at the other (non-reference) nodes are measured. It also usually a good idea to choose the reference node as that node connected to 1
the most branches. If there are n e essential nodes in a circuit, the circuit may be described by n e 1 equations. A node voltage is defined as the voltage rise from a reference node to a nonreference node. In the circuit shown below there are only 3 essential nodes. This is emphasized by labelling one node twice eg., electrically, the two cs are the same point. It may be noted that the following node-voltage technique is really a result of KCL. We will now use the node-voltage method to determine the (i) voltages across the 1 Ω and 12 Ω resistors, (ii) v as shown, and (iii) subsequently, we ll use the results to find the current through the 12 Ω resistor. Observation: Note that since there are 3 essential nodes, we may describe the circuit with 2 (i.e. one less than the number of essential nodes) node-voltage equations. Step 1: Choose a reference node (indicated by the solid arrow) notice it s the node with the most branches. Step 2: Label the remaining essential nodes with their voltages with respect to the reference. Here we have used v 1 and v 2. Step 3: Use KCL at each of the nodes in Step 3. In doing this the text s convention is to sum the currents leaving each of these nodes to zero. Here, we get 2
Step 4: Use the node-voltages to determine any required branch currents and powers. (NOTE: An example involving dependent sources and supernodes i.e. nodes between which a voltage source exists will be completed in a Tutorial.) It may be noted that if a voltage source, v s, is connected directly between two essential nodes, then it constrains the voltage difference between these nodes to have a value v s. For example, in the diagram below, where the lower node is taken to be the reference, we know that v 1 = v s without having to apply KCL at node 1 this reduces the number of unknown voltages and simplifies the analysis. 3
2.2 Mesh-Current Method Recall that a mesh is a circuit loop that does not enclose any other loops. In the circuit shown below there are 3 meshes. A mesh current may be thought of as a current that exists in the perimeter of a mesh. This need not be an actual current, but as we will see, actual currents may be determined from the mesh currents as we shall see in the following example. The mesh-current method of circuit analysis actually makes use of KVL for individual meshes. Step 1: Assign a mesh current to each of the meshes as indicated by the arrowed curves labelled i 1, i 2, and i 3. Step 2: Apply KVL to each mesh to obtain a system of equations involving the mesh currents. Solve these for the mesh currents. 4
Step 3: Obtain any necessary branch currents from the mesh currents. This allows any voltages and powers of interest to be calculated. Here we illustrate by finding (a) the power delivered by the 80 V source and (b) the power dissipated in the 8 Ω and 26 Ω resistors. (It is also easy to check that power delivered = power absorbed.) 5
Extra notes on mesh currents: (1) If there is a dependent source in the circuit, the mesh equations will have to be supplemented by an equation involving that source. Illustration: (2) When a branch contains a current source, the voltage across it is not known. Then we may ignore that branch and create a supermesh which avoids it. The supermesh must use the same mesh currents as originally chosen and a separate equation relating the previously ignored source to these currents will be required also. Illustration: 6
2.3 Source Transformations and Equivalent Circuits Source Transformations It is possible to replace a voltage source in series with a resistor with a current source in parallel with the same resistor (or vice versa) so that the voltage v ab across a load, say R L, remains the same. Consider the following in which we require v ab to be the same in each case: In the first circuit, v ab =... (1) while in the second, v ab =... (2) From (1) and (2), it is thus required that This means that i s = v s R... (3). As far as the behaviour at the terminal a and b is concerned, both circuits are equivalent if (3) holds. 7
Thévenin Equivalent Circuits Consider a resistive network possibly containing both dependent and independent sources. We wish to reduce the circuit to a single equivalent independent voltage and resistance such that the current and voltage associated with a load connected across a and b will be the same in the simplified circuit as it was in the original network. Illustration: When this is the case, the voltage and resistance are referred to as the Thévenin voltage and resistance, symbolized as V Th and R Th, respectively. This equivalence must hold for all possible values of load resistance R L. It must therefore hold in the two extreme cases: (i) R L (i.e. an open circuit) and (ii) R L 0 (i.e. a short circuit). In the first of these cases, v ab = V Th = v oc... (4) where v oc is the open-circuit voltage. Since, by definition, the Thévenin voltage is the same for any load, equation (4) indicates we may get it by simply open-circuiting the load. In the second case, the short-circuit current i sc must be given by i sc = V Th R Th... (5) again since both V Th and R Th are required to be unchanged for any load. From (5) R Th = V Th i sc... (6) and equations (4) and (6) thus define the Thévenin equivalent circuit. 8
Norton Equivalent Circuit Before elaborating on Thévenin s equivalent circuits we note the following source transformation: Illustration: From our discussion on source transformations at the beginning of this subsection, we know that the above circuits are equivalent as long as i s = In this case the circuit on the right is referred to as the Norton equivalent circuit. 9
Preliminary Example on Thévenin/Norton Equivalent Circuits Find the Thévenin and Norton equivalent circuits for Thévenin Equivalent Circuit: Using source transformation, the circuit may be represented as The Thévenin voltage and resistance may calculated simply as follows: Norton Equivalent Circuit: 10
More on Thévenin Equivalent Circuits Depending on the type of circuit, reduction to its Thévenin equivalent may involve a variety of techniques: (1) In general, v oc, i sc, and/or R Th may be determined using the results of nodevoltage and mesh-current analysis. (2) As above, sometimes use of equivalent sources is helpful. (3) If the circuit contains only independent sources, R Th may be obtained easily by (i) short-circuiting the voltage supplies and (ii) open-circuiting the current supplies. This works only because R Th does not depend on the values of the independent voltage and current supplies in the original circuit. Consider this for the above example: (4) If the circuit contains dependent sources, R Th may be determined by (i) shortcircuiting the independent voltage supplies, (ii) open-circuiting the independent current supplies and (iii) connecting a test source across the terminals and determining the equivalent resistance of the revised circuit from its response to the test source. Consider Drill Exercise 4.20, p. 119 of the text: 11
2.4 Preliminary Look at Maximum Power Transfer Later we shall be concerned with maximum power being delivered from a circuit which has elements other than resistors to a load which is not purely resistive. Here, however, we review the idea of maximum power transfer from a purely resistive network to a purely resistive load. The purely resistive network may be represented by its Thévenin equivalent and is depicted below as being attached to a load resistance R L. We wish to establish the value of R L that permits maximum power transfer from the network to the R L. Illustration: We begin by noting that power delivered to the load is a function of R L as given by p = i 2 R L = =... (1) Since we wish to determine the value of R L which maximizes p, we may take the first derivative of p in (1) with respect to R L, set the result to 0 and solve for R L : dp dr L = =... (2) 12
Clearly, R L = R TH produces an extreme value. The second derivative of p with respect to R L is negative (SHOW THIS) for this value. Thus, R L = R TH... (3) is the value of R L which maximizes the power delivered to the load. From (1), this power is given by p max =... (4) Example: Similar to Problem 4.79, page 139 of text. 13
2.5 Superposition The principle of superposition states that whenever a linear system is driven by more than one independent source of energy, the total response of the system equals the sum of the responses from each of the sources acting individually. The application of superposition to the problems encountered thus far, in which all independent sources are dc, does not offer an improvement over the techniques already used. However, for illustrative purposes we ll consider superposition briefly. In applying superposition: (i) to find the effects of independent voltage sources (one at a time) the independent current sources are deactivated by open-circuiting them; (ii) to find the effects of independent current sources (one at a time) the independent voltage sources are deactivated by short-circuiting them; (iii)dependent sources are never deactivated. Example: Consider the following circuit. We wish to find the voltage v as shown. 14