Nodal and Loop Analysis, Summaries of Law

Download Nodal and Loop Analysis and more Summaries Law in PDF only on Docsity! Nodal and Loop Analysis The process of analyzing circuits can sometimes be a difficult task to do. Examining a circuit with the node or loop methods can reduce the amount of time required to get important information on a circuit. These analyses apply Kirchhoff's laws in a step-by-step approach to solve for unknown voltages and current. Once these values are obtained, further information such as electrical power can be analyzed. This module introduces Kirchhoff's laws and their relation to circuit analysis. Different methods for analysis are explained, covering both node and mesh analysis. These processes are explained in detail to provide an understanding of how to analyze circuits. Kirchhoff's Laws In 1847, Gustav Kirchhoff formulated his voltage law and current law. These laws were derived from the conservation of charge and conservation of energy laws and applied to circuits. These laws are used to develop equations for circuit analysis. Kirchhoff's Current Law (KCL) Kirchhoff current law states that the algebraic sum of all currents entering a node of a circuit is always zero. A node in a circuit is the place where circuit elements are connected together. The direction of the current must be considered during the analysis. If the current is entering the node, the current should be subtracted from the algebraic sum. Likewise, if the current is directed away or out from a node then the current is said to be positive. Another simpler way to analyze the current is to set the sum of the currents directed away from the node equal to the sum of the currents directed towards the node. The following circuit can be analyzed with KCL. Figure 1: KCL Analysis of a Circuit The node in the center of this circuit involves four currents. The sum of the two currents entering the node (the current from elements A and D) are equal to the two currents leaving the node (towards elements B and C). Analyzing the center node of this circuit shows Kirchhoff's Voltage Law (KVL) Kirchhoff's voltage law states that the algebraic sum of the voltages around any loops in Figure 4: DC Circuit with Voltage and Current Sources Here, node c is an example of a supernode which is a connection between two nodes via an independant or dependant voltage source. Because supernodes are connected to a voltage source we can find their voltage immediately. In this case, with the ground at , the voltage across the source will be , therefore . Similarily, node a is related to node b as a supernode, . We can substitute and into our KCL equations to solve for . Calculate the KCL equations at node a in figure 3. The current source is directing current into node a, and we will assume the current flows away from a towards node b and c. The KCL equation for node a is The current across the resistor represented by the node voltages is found through Ohm's law as the potential across the element divided by the resistance. Note the assumed direction of the current to ensure the correct polarity of the difference in potential. The current through node c will be equivalent to the potential across the resistor divided by the resistance. Substituting in the currents and and using the equations for and , the KCL equation at node A is found to be Solving for shows the voltage is 8 volts. Finally, the voltage at a is , so volts. Loop/Mesh Analysis Loop analysis is a special application of KVL on a circuit. We use a special kind of loop called a 'mesh' which is a loop that does not have any other loops inside of it. A mesh starts at a node and traces a path around a circuit, returning to the original node without hitting any nodes more than once. We can only apply mesh analysis to planar circuits, that is circuits without crossover connections. If a circuit cannot be redrawn without the intersecting disconnected lines then we cannot use mesh analysis. Similar to nodal analysis, we want to obtain the mesh equations to be able to interpret the circuit. The mesh equations are obtained by 1. Applying Kirchhoff's voltage law (KVL) to each mesh in the circuit. 2. Express the voltages of elements in terms of the mesh currents. The current along a mesh may not be uniform but this is accounted for by considering the current imposed by other meshes. Observe the circuit shown below. Figure 5: Mesh circuit This circuit has 2 meshes in it, outlined in green and blue dotted lines. The currents and are the currents around a mesh. In places where the currents affect each other, such as in figure 5, take the sum of the currents but consider the directions. Here, if we examine the green mesh, the current through will be because is going in the opposite direction as . The method behind mesh analysis is to examine the mesh in terms of the voltages of each element and express that with the mesh currents. Using Ohm's law , we can express the voltage across as . Use this approach across a circuit applying KVL to find the mesh equations, then you can use these equations to solve for unknown current. In the case where a dependent source is present in a circuit, the controlled current or voltage of that source must also be expressed by the mesh currents. Examine the circuit and find which part of the circuit controls the dependent source and express that voltage or current with the mesh currents that affect it. , and represent the voltages at nodes a, b and c respectively. We will begin with the KCL equation for node a. The currents and are entering the node, and assume the currents through the resistors leading to the ground are leaving the node. Rearranging this by extracting the node voltages leaves Use the same method to solve for the node equation at b. Solving for the node voltages, For node c, the KCL equation will be Again, rearranging this to isolate the coefficients of , and will complete our matix. This can be solved easily as a matrix with Maple using the solve command. Otherwise, use substitution and elimination with the KCL equations to solve for the values of the node voltages. = Thus the potential of the node voltages are , and . MapleSim Solution Step 1: Insert Components Drag the following components into a new workspace. Component Location Electrical > Analog > Common (6 required) Electrical > Analog > Common (3 required) Electrical > Analog > Sources > Current Step 2: Connect the components. Connect the components as shown in the diagram below. Figure 8: MapleSim model diagram Note the direction of the current sources. Green arrows have been added to the image to emphasize this. Step 3: Set up the resistors. For each resistor, 1. Select the Resistor block. On the 'Inspector' tab, set the R parameter to the following values: , , , , , and . Step 4: Set up the current sources. For all three of the current sources, 1. Select the Constant Current block and on the 'Inspector' tab, set the I parameter to the following values: , , . 2. Connect probes to the proper locations and ensure the 'v' option is checked on the (3.2.1.1) On the second mesh with the dependent source, the mesh equation will be, Of course we know the voltage of the dependent source so we can substitute this into the second mesh equation. Using the values for and from the mesh equations, the dependent source gain can finally be found. Therefore the voltage of the dependent source is 4 times greater than the current through . Alternatively we could use Maple's solve function to determine the currents and gain in one step. MapleSim Solution Step 1: Insert Components Drag the following components into a new workspace. Component Location Electrical > Analog > Common (2 required) Electrical > Analog > Common (2 required) Electrical > Analog > Sources > Voltage Step 2: Connect the components. Connect the components as shown in the diagram below. Figure 12: MapleSim model The current source on the right is replacing the dependent voltage because the question supplied the voltage across the element. It will be easier to use the known value. Note that the direction of the probe arrow is dependent on the direction you moved to connect the nodes. For example, here the top node was connected to the bottom node instead of the bottom connecting to the top, so the arrow points downwards. It is an assumption made by the program. Step 3: Set Up the Circuit 1. Set the resistance to and for and respectively. Select the Resistor block and change the parameter on the 'Inspector' tab. 2. Set the left voltage source at 36 V on the 'Inspector' tab. Also, set the right voltage source to -7.2 volts and confirm the positive terminal is at the top of the circuit. This emulates the controlled voltage source. 3. Connect a probe along the middle connection between the resistors. On the 'Inspector' tab, deselect the voltage ' v ' and select the current ' i '. An arrow will show up at the probe designating which direction the program assumes the current will be travelling. If the current is negative, then it is travelling in the opposite direction of the arrow. Step 3: Run Simulation Run the simulation. The probe will return the value of . Divide the voltage across the dependent source by the current to get the magnitude of the gain. Step 4: Check the Solution To check your solution, 1. Connect a current-controlled voltage source (Electrical > Analog > Sources > Controlled > CCV) in place of the right voltage source. On the 'Inspector' tab, set the gain to 4. 2. Connect a voltmeter (Electrical > Analog > Sensors > Voltage Sensor)