1. Sole circuits (i.e., find currents and oltages of interest) by combining resistances in series and parallel. 2. Apply the oltage-diision and current-diision principles. 3. Sole circuits by the node-oltage technique. ELECTICAL
4. Find Théenin and Norton equialents. 5. Apply the superposition principle. 6. Draw the circuit diagram and state the principles of operation for the Wheatstone bridge. ELECTICAL
Circuit Analysis using Series/Parallel Equialents 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. edraw the circuit with the equialent resistance for the combination found in step 1. ELECTICAL
3. epeat 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. Sole for the currents and oltages in the final equialent circuit. ELECTICAL
Voltage Diision 1 1 = 1i = total 1 + 2 + 3 2 2 = 2i = total 1 + 2 + 3 ELECTICAL
Application of the Voltage- Diision Principle 1 = 1 total 1 + 2 + 3 + 4 1000 = 1000 + 1000 + 2000 + 6000 = 1.5V 15 ELECTICAL
Current Diision i = = 2 i 1 total + 1 1 2 i = = 1 i 2 total + 2 1 2 ELECTICAL
Application of the Current- Diision Principle eq i = 23 30 60 = = Ω + 30 + 60 20 2 3 20 15 10 + 20 eq 1 = is = = 1 + eq 10A ELECTICAL
Although they are ery important concepts, series/parallel equialents and the current/oltage diision principles are not sufficient to sole all circuits. ELECTICAL
Node Voltage Analysis ELECTICAL
Writing KCL Equations in Terms of the Node Voltages for Figure 2.16 1 = s ELECTICAL 2 1 2 2 3 + + = 2 4 3 1 3 3 2 + + = 1 5 3 3 0 0
0 2 2 1 1 1 = + + s i 0 4 3 2 3 2 2 1 2 = + + i s = + 4 2 3 5 3
Circuits with Voltage Sources We obtain dependent equations if we use all of the nodes in a network to write KCL equations. ELECTICAL
( 15) ( 15) 1 1 2 2 + + + = 2 1 4 3 0 ELECTICAL
0 10 2 1 = + + 1 3 3 2 2 3 1 1 1 = + + 0 4 3 3 2 3 2 1 3 = + + 1 4 3 1 1 = +
Node-Voltage Analysis with a Dependent Source First, we write KCL equations at each node, including the current of the controlled source just as if it were an ordinary current ELECTICAL source.
i s i x 2 1 2 1 + = 0 3 3 2 2 2 1 1 2 = + + 0 2 4 3 3 2 3 = + + i x
Next, we find an expression for the controlling ariable i x in terms of the node oltages. i x = 3 3 2 ELECTICAL
Substitution yields 3 2 3 1 2 1 2 i s + = 0 3 3 2 2 2 1 1 2 = + + 0 2 3 2 3 4 3 3 2 3 = + +
ELECTICAL Node-Voltage Analysis 1. Select a reference node and assign ariables for the unknown node oltages. If the reference node is chosen at one end of an independent oltage source, one node oltage is known at the start, and fewer need to be computed.
2. Write network equations. First, use KCL to write current equations for nodes and supernodes. Write as many current equations as you can without using all of the nodes. Then if you do not hae enough equations because of oltage sources connected between nodes, use KVL to write additional equations. ELECTICAL
3. If the circuit contains dependent sources, find expressions for the controlling ariables in terms of the node oltages. Substitute into the network equations, and obtain equations haing only the node oltages as unknowns. ELECTICAL
4. Put the equations into standard form and sole for the node oltages. 5. Use the alues found for the node oltages to calculate any other currents or oltages of interest. ELECTICAL
Mesh Current Analysis ELECTICAL
Choosing the Mesh Currents When seeral mesh currents flow through one element, we consider the current in that element to be the algebraic sum of the mesh currents. Sometimes it is said that the mesh currents are defined by soaping the window panes. ELECTICAL
Writing Equations to Sole for Mesh Currents If a network contains only resistances and independent oltage sources, we can write the required equations by following each current around its mesh and applying KVL. ELECTICAL
Using this pattern for mesh 1 of Figure 2.32a, we hae For mesh 2, we obtain For mesh 3, we hae ( i i ) + ( i i ) 0 2 1 s 3 1 2 A = ( i i ) + i + 0 3 2 1 4 2 B = ELECTICAL ( i i ) + i 0 2 3 1 1 3 B =
In Figure 2.32b i ( i i ) + ( i i ) 0 1 1 + 2 1 4 4 1 2 A = ( i i ) + ( i ) 0 5 2 + 4 2 1 6 2 i3 = i ( i i ) + ( i ) 0 7 i3 + 6 3 2 8 3 i4 = i ( i i ) + ( i ) 0 + i 3 4 2 4 1 8 4 3 = ELECTICAL
Mesh Currents in Circuits Containing Current Sources A common mistake made by beginning students is to assume that the oltages across current sources are zero. In Figure 2.35, we hae: ELECTICAL i = 2A 1 ) 5 10 0 10( 1 2 i i + i + = 2
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: i ( i i ) + 4( i ) + 10 0 1 + 2 1 3 2 i3 = ( i i ) + 2( i ) 0 3i + 4 3 2 3 i1 3 = i 2 i1 = 5 ELECTICAL
20 + 4i + 6i2 + 2i2 1 = 0 x 4 = i 2 i 1 x = 2i 2 ELECTICAL
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. ELECTICAL
2. Write network equations, stopping after the number of equations is equal to the number of mesh currents. First, use KVL to write oltage 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. ELECTICAL
3. If the circuit contains dependent sources, find expressions for the controlling ariables in terms of the mesh currents. Substitute into the network equations, and obtain equations haing only the mesh currents as unknowns. ELECTICAL
4. Put the equations into standard form. Sole for the mesh currents by use of determinants or other means. 5. Use the alues found for the mesh currents to calculate any other currents or oltages of interest. ELECTICAL
Théenin Equialent Circuits ELECTICAL
Théenin Equialent Circuits V t = oc t = i oc sc ELECTICAL
Finding the Théenin esistance Directly When zeroing a oltage source, it becomes an open circuit. When zeroing a current source, it becomes a short circuit. We can find the Théenin resistance by zeroing the sources in the original network and then computing the resistance between the terminals. ELECTICAL
Step-by-step Théenin/Norton- Equialent-Circuit Analysis 1. Perform two of these: a. Determine the open-circuit oltage V t = oc. b. Determine the short-circuit current I n = i sc. c. Zero the sources and find the Théenin resistance t looking back into the terminals. ELECTICAL
2. Use the equation V t = t I n to compute the remaining alue. 3. The Théenin equialent consists of a oltage source V t in series with t. 4. The Norton equialent consists of a current source I n in parallel with t. ELECTICAL
Source Transformations ELECTICAL
Maximum Power Transfer The load resistance that absorbs the maximum power from a two-terminal circuit is equal to the Théenin resistance. ELECTICAL
SUPEPOSITION PINCIPLE The superposition principle states that the total response is the sum of the responses to each of the independent sources acting indiidually. In equation form, this is ELECTICAL r = r + r + L + r T 1 2 n
WHEATSTONE BIDGE The Wheatstone bridge is used by mechanical and ciil engineers to measure the resistances of strain gauges in experimental stress studies of machines and buildings. x = 2 3 1 ELECTICAL