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Circuit Analysis 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. Working Backward Find current flowing each resistor Voltage Division R1 v1 R1i v total R1 R2 R3 R2 v2 R2 i v total R1 R2 R3 Application of the VoltageDivision Principle R1 v1 vtotal R1 R2 R3 R4 1000 15 1000 1000 2000 6000 1.5V Current Division R2 v i1 itotal R1 R1 R2 R1 v i2 itotal R2 R1 R2 Application of the Current-Division Principle R2 R3 30 60 Req 20 R2 R3 30 60 Req 20 i1 is 15 10A R1 Req 10 20 •Voltage division •Voltage division and •current division Current division Although they are very important concepts, series/parallel equivalents and the current/voltage division principles are not sufficient to solve all circuits. Mesh Current Analysis Sources Definition of a loop Definition of a mesh Choosing the Mesh Currents When several mesh currents flow through one element, we consider the current in that element to be the algebraic sum of the mesh currents. 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 R2 i1 i3 R3 i1 i2 v A 0 For mesh 2, we obtain R3 i2 i1 R4i2 v B 0 For mesh 3, we have R2 i3 i1 R1i3 vB 0 Determine the two mesh currents, i1 and i2, in the circuit below. For the left-hand mesh, -42 + 6 i1 + 3 ( i1 - i2 ) = 0 For the right-hand mesh, 3 ( i2 - i1 ) + 4 i2 - 10 = 0 Solving, we find that i1 = 6 A and i2 = 4 A. (The current flowing downward through the 3- resistor is therefore i1 - i2 = 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. i1 2 10(i2 i1 ) 5i2 10 0 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. It is the supermesh. i1 2i1i3 4i2 i3 10 0 Mesh 3: 3i3 4i3 i2 2i3 i1 0 i2 i1 5 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 ( i1 - i2 ) + 3 ( i3 - i2 ) + 1 i3 = 0 [1] Around mesh 2: 1 ( i2 - i1 ) + 2 i2 + 3 ( i2 - i3 ) = 0 [2] Finally, we relate the currents in meshes 1 and 3: i1 - i3 = 7 [3] Rearranging, i1 - 4 i2 + 4 i3 = 7 [1] -i1 + 6 i2 - 3 i3 = 0 [2] i1 [3] - i3 = 7 Solving, i1 = 9 A, i2 = 2.5 A, and i3 = 2 A. supermesh of mesh1 and mesh2 20 4i1 6i2 2i2 0 branch current v x 2i2 current source vx i2 i1 4 20 4i1 6i2 2i2 0 v x 2i2 vx i2 i1 4 Three equations and three unknown. 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 Superposition Insert Figure 7.3 Superposition • Analyze Separately, then Combine Results Use superposition to find the current ix. Current source is zero – open circuit as I = 0 and solve iXv Voltage source is zero – short circuit as V= 0 and solve iXv i X i Xv i Xc Use superposition to find the current ix. The controlled voltage source is included in all cases as it is controlled by the current ix. Voltage and Current Sources Insert Figure 7.7 Voltage and Current Sources Insert Figure 7.8 Voltage and Current Sources Insert Figure 7.9 Source Transformation iL RS VS + + _ VL RL _ iL + IS RP VL _ RL Under what condition, the voltage and current of the load is the same when operating at the two practical sources? For voltage source VS VL RL R S RL For current source i S RP , VL RL R P RL We have, VS iS RP R S RL RP RL VS RP RS , iS RS Voltage and Current Sources • Equivalent Voltage and Current Sources – for every voltage source, there exists an equivalent current source, and vice versa Thevenin’s Theorem • Thevenin’s Theorem – any resistive circuit or network, no matter how complex, can be represented as a voltage source in series with a source resistance Thevenin’s Theorem • Thevenin Voltage (VTH) – the voltage present at the output terminals of the circuit when the load is removed Insert Figure 7.18 Thevenin’s Theorem • Thevenin Resistance (RTH) – the resistance measured across the output terminals with the load removed Thévenin Equivalent Circuits Thévenin Equivalent Circuits Vt voc voc Rt isc Thévenin Equivalent Circuits 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. Computation of Thévenin resistance Equivalence of open-circuit and Thévenin voltage A circuit and its Thévenin equivalent Superposition As the voltage source does not contribute any output voltage, Only the current source has the effect. Determine the Thévenin and Norton Equivalents of Network A in (a). Source transformation Find the Thévenin equivalent of the circuit shown in (a). v As i = -1, therefore, the controlled voltage source is -1.5V. Use nodal analysis at node v, v (1.5) v 1, v 0.6 3 2 Thus, Rth =v/I = 0.6/1 = 0.6 ohms 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 RL = RTH 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 (IN) – the current through the shorted load terminals Insert Figure 7.35 Computation of Norton current Norton’s Theorem • Norton Resistance (RN) – the resistance measured across the open load terminals (measured and calculated exactly like RTH) Norton’s Theorem • Norton-to-Thevenin and Thevenin-to-Norton Conversions Insert Figure 7.39 Step-by-step Thévenin/NortonEquivalent-Circuit Analysis 1. Perform two of these: a. Determine the open-circuit voltage Vt = voc. b. Determine the short-circuit current In = isc. c. Zero the sources and find the Thévenin resistance Rt looking back into the terminals. 2. Use the equation Vt = Rt In to compute the remaining value. 3. The Thévenin equivalent consists of a voltage source Vt in series with Rt . 4. The Norton equivalent consists of a current source In in parallel with Rt . Maximum Power Transfer The load resistance that absorbs the maximum power from a two-terminal circuit is equal to the Thévenin resistance. Power transfer between source and load Graphical representation of maximum power transfer