EE 105 Discussion #1: Fundamentals of Circuit Analysis 1.1 Ohm s Law V = ir i = V/R 1.2 KCL & KVL Kirchoff s Current Law (KCL) Kirchoff s Voltage Law (KVL) The algebraic sum of all currents entering a node equals zero. Currents leaving a node are included with a minus sign. i1+ i2 i3 i4 = 0 1.3 Node Voltage and Mesh Current Methods The Node Voltage Method (allows us to describe a circuit in n-1 equations) 1. Select one of the essential nodes (e.g. any node where three or more circuit elements join) to be the reference node => Look for the most connections 2. Define the unknown node voltages on the circuit diagram =>A node voltage is defined as the voltage rise from the reference node to a non-reference node 3. Write KCL at each unknown essential node by expressing current in terms of node voltages using the I-V relationships of the branch elements The algebraic sum of all voltages around any closed path in a circuit equals zero. Va + V 2 V 3+ Vc = 0 Va + V 2 + Vb = 0 Vb V 3 + Vc = 0 The Mesh Current Method (allows us to describe a circuit in b (n-1) equations) 1. Select M independent mesh (e.g. a loop with no other loops inside of it) currents such that at least one mesh passes through each branch. => M = #branches - #nodes + 1 2. Apply KVL to each meah, expressing voltages in terms of mesh currents => M equations for M unknown mesh currents. 3. Solve for mesh currents and determine node voltages. Additional Considerations:
4. Solve N node-voltage equations for the N unknown node voltages. Additional Considerations 4. Supermeshes 5. Dependent Sources 5. Supernodes 6. Dependent Sources Example 1: (node-voltage method) What to do in the case of a supernode Equ 1: I1+ Va R2 + Vb R4 I2 = 0 Equ 2: Vb Va = V LL Thus, given I1, I2, R2, R4, and V LL we can solve for Va and Vb and any branch currents in the circuit. Example 2: Solve for Vx using node voltage method Equ 1: 1A + V1 V1 2Vx + + V1 V 2 = 0 5 15 10 Equ 2: 2A + V 2 5 + V 2 2Vx 10 Equ 3: Vx = V2 V1 + V 2 V1 = 0 10 Example 3: Solve v using mesh current method Equ #1: i1 = 2A Equ #2: 6(i2 - i1) + 2*i2 + 6(i2 i4) + 4(i2 i3) = 0-6*i1 + 18*i2 4*i3 6*i4 = 0 Equ #3: i3 = -3A Equ #4: 12V + 3*(i4 i3) + 6(i4 i2) = 0-6*i2 3*i3 + 9*i4 = -12V Then after solving i1, i2, i3 and i4 We find that v = 4*(i2 i3)
Question: If you where to solve the following circuit for each of the branch currents would you want to use the node voltage method, or the mesh current method. Anwser: Clearly, the node voltage method would require you to solve two simultaneous equations while the mesh current method would require you to solve five. You should always look to keep your analysis as simple as possible. 1.4 Thevenin and Norton Equivalent Circuits Concept: Any linear two-terminal network of independent and dependent voltage sources, independent and dependent current sources, and resistors can be replaced by either an independent voltage source in series with a resistor (Thevenin equivalent), or an independent current source in parallel with a resistor (Norton Equivalent) without affecting the operation of the rest of the circuit.
Procedure 1. Find the open circuit voltage between the terminals of interest. Define this voltage as V th 2. Find the Thevenin Resistance by one of the following methods i. Short the output of the terminals and solve for the short-circuit current (e.g. the current flowing in the direction of the open circuit voltage drop across the terminals). R th = Vth isc ii. If the network contains only independent sources and resistors, deactivate all independent sources and calculate the resistance looking into the network at the designated terminal pair. Independent Voltage Source! Short Circuit Independent Current Source! Open Circuit R th = Resistance seen looking into the network iii. If the network contains dependent sources, deactivate all independent sources as in ii. Above. Apply a 1V test voltage source and solve for the current going through the test voltage source. R th = 1V i where i = the current delivered by the test source Example 4: Assume Va = Vb = Vc=5V. Find the Thevenin Equivalent circuit. Equ 1: V1 Va + V1 V2 =0 3V1 2V2 =Va R Equ 2: V2 Vb +V2 V1 R +V2 V3 R =0 2V1+5V2 2V2=Vb Equ 3: V3 Vc + V3 + V3 V2 =0 2V2+ 4V3=Vc R Solving for V1, V2, V3 yields Vth = V1 = 4.735V Rth = R by inspection (you shouldn t have to due calculations here) 1.5 Additional Topics Phasors, Source transformations, Capacitors, and Transfer Functions Example 5: For the circuit below find the transfer function Vout Vs
Step 1: Notice that we can do a source transformation on the Thevenin Equivalent circuit changing it into a Norton Equivalent circuit thus simplifying the analysis. Is = Vs Rs Is*(Rs Rin) Step 2: Solve for I! I'= 1 (Rs Rin) + jωcin Step 3: Solve for Vout!Vout = I'*A'*RL = ( Vs Rs )* Rs Rin Rs Rin + 1 * A'*RL jωc Step 4: Solve for the transfer function! Vout Vs RL = A'* Rs * Rs Rin Rs Rin + 1 jωc