Circuit Analysis using Series/Parallel Equivalents 1.Begin by locating a combination of resistances that are in series or parallel. Often the place to start is farthest from the source. 2.Redraw the circuit with the equivalent resistance for the combination found in step 1. 3.Repeat steps 1 and 2 until the circuit is reduced as far as possible. Often (but not always) we end up with a single source and a single resistance. 4.Solve for the currents and voltages in the final equivalent circuit.
Writing Equations to Solve for Mesh Currents If a network contains only resistors and independent voltage sources, we can write the required equations by following each current around its mesh and applying KVL.
For mesh 1, we have For mesh 2, we obtain For mesh 3, we have
Determine the two mesh currents, i 1 and i 2, in the circuit below. For the left-hand mesh, -42 + 6 i 1 + 3 ( i 1 - i 2 ) = 0 For the right-hand mesh, 3 ( i 2 - i 1 ) + 4 i 2 - 10 = 0 Solving, we find that i 1 = 6 A and i 2 = 4 A. (The current flowing downward through the 3- resistor is therefore i 1 - i 2 = 2 A. )
Mesh Currents in Circuits Containing Current Sources *A common mistake is to assume the voltages across current sources are zero. Therefore, loop equation cannot be set up at mesh one due to the voltage across the current source is unknown Anyway, the problem is still solvable.
As the current source common to two mesh, combine meshes 1 and 2 into a supermesh. In other words, we write a KVL equation around the periphery of meshes 1 and 2 combined. Mesh 3: It is the supermesh. Three linear equations and three unknown
Find the three mesh currents in the circuit below. Creating a “supermesh” from meshes 1 and 3: -7 + 1 ( i 1 - i 2 ) + 3 ( i 3 - i 2 ) + 1 i 3 = 0  Around mesh 2: 1 ( i 2 - i 1 ) + 2 i 2 + 3 ( i 2 - i 3 ) = 0 Rearranging, i 1 - 4 i 2 + 4 i 3 = 7 -i 1 + 6 i 2 - 3 i 3 = 0 i 1 - i 3 = 7 Solving, i 1 = 9 A, i 2 = 2.5 A, and i 3 = 2 A. Finally, we relate the currents in meshes 1 and 3: i 1 - i 3 = 7
supermesh of mesh1 and mesh2 current source branch current
Mesh-Current Analysis 1. If necessary, redraw the network without crossing conductors or elements. Then define the mesh currents flowing around each of the open areas defined by the network. For consistency, we usually select a clockwise direction for each of the mesh currents, but this is not a requirement. 2. Write network equations, stopping after the number of equations is equal to the number of mesh currents. First, use KVL to write voltage equations for meshes that do not contain current sources. Next, if any current sources are present, write expressions for their currents in terms of the mesh currents. Finally, if a current source is common to two meshes, write a KVL equation for the supermesh. 3. If the circuit contains dependent sources, find expressions for the controlling variables in terms of the mesh currents. Substitute into the network equations, and obtain equations having only the mesh currents as unknowns. 4. Put the equations into standard form. Solve for the mesh currents by use of determinants or other means. 5. Use the values found for the mesh currents to calculate any other currents or voltages of interest.
Superposition Superposition Theorem – the response of a circuit to more than one source can be determined by analyzing the circuit’s response to each source (alone) and then combining the results Insert Figure 7.2
Finding the Thévenin Resistance Directly When zeroing a voltage source, it becomes a short circuit. When zeroing a current source, it becomes an open circuit. We can find the Thévenin resistance by zeroing the sources in the original network and then computing the resistance between the terminals.
Applications of Thevenin’s Theorem Load Voltage Ranges – Thevenin’s theorem is most commonly used to predict the change in load voltage that will result from a change in load resistance
Applications of Thevenin’s Theorem Maximum Power Transfer –Maximum power transfer from a circuit to a variable load occurs when the load resistance equals the source resistance –For a series-parallel circuit, maximum power occurs when R L = R TH
Applications of Thevenin’s Theorem Multiload Circuits Insert Figure 7.30
Norton’s Theorem Norton’s Theorem – any resistive circuit or network, no matter how complex, can be represented as a current source in parallel with a source resistance
Norton’s Theorem Norton Current (I N ) – the current through the shorted load terminals Insert Figure 7.35
Norton’s Theorem Norton Resistance (R N ) – the resistance measured across the open load terminals (measured and calculated exactly like R TH )
Norton’s Theorem Norton-to-Thevenin and Thevenin-to-Norton Conversions Insert Figure 7.39
Step-by-step Thévenin/Norton- Equivalent-Circuit Analysis 1. Perform two of these: a. Determine the open-circuit voltage V t = v oc. b. Determine the short-circuit current I n = i sc. c. Zero the sources and find the Thévenin resistance R t looking back into the terminals.
2. Use the equation V t = R t I n to compute the remaining value. 3. The Thévenin equivalent consists of a voltage source V t in series with R t. 4. The Norton equivalent consists of a current source I n in parallel with R t.