Circuit Analysis Survival: Kirchhoff's Laws & KCL/KVL Equations, Study notes of Law

Download Circuit Analysis Survival: Kirchhoff's Laws & KCL/KVL Equations and more Study notes Law in PDF only on Docsity! Survival Skills for Circuit Analysis What you need to know from ECE 109 Phyllis R. Nelson prnelson@csupomona.edu Professor, Department of Electrical and Computer Engineering California State Polytechnic University, Pomona P. R. Nelson Fall 2010 WhatToKnow —- – p. 1/46 Basic Circuit Concepts All circuits can be analyzed by many methods. P. R. Nelson Fall 2010 WhatToKnow —- – p. 2/46 Basic Circuit Concepts All circuits can be analyzed by many methods. Some methods have specific advantages for calculating a particular result in a specific circuit. Some methods yield useful intuition about circuit behavior. No single analysis method will be the best for all possible circuits! P. R. Nelson Fall 2010 WhatToKnow —- – p. 2/46 Kirchhoff’s Laws These two methods are the most general tools of circuit analysis. Sign errors in writing Kirchhoff’s laws are the most common error I observe in student work in higher-level classes. Sign errors are not trivial errors! P. R. Nelson Fall 2010 WhatToKnow —- – p. 3/46 Kirchhoff’s current law is equivalent to stating that electrical charge cannot be stored, created, or destroyed locally in a conductor. The algebraic sum of all currents entering a node is zero. Alternate statements: The algebraic sum of the currents leaving a node is zero. The total current entering a node is equal to the total current leaving the node. P. R. Nelson Fall 2010 WhatToKnow —- – p. 4/46 Solving K*L . . . Solving KCL means finding the node voltages. Solving KVL means finding the branch currents. ⇒ need relationships between current and voltage for each circuit element. These relationships describe the operation of the circuit elements. P. R. Nelson Fall 2010 WhatToKnow —- – p. 7/46 Ohm’s law + V − I V = IR Ohm’s law is an example of a current-voltage (I-V ) relation. To use K*L, you must know the I-V relation for every circuit element. I-V relations include definition of the relationship between the direction of the current and the sign of the voltage. P. R. Nelson Fall 2010 WhatToKnow —- – p. 8/46 Series resistors, voltage divider R1 + V1 − R2 + V2 − Ri + Vi − Rn + Vn − −VT+ I VT = I Req Prove that these equations are true: Req = n ∑ j=1 Rj Vi = Ri ∑n j=1 Rj VT P. R. Nelson Fall 2010 WhatToKnow —- – p. 9/46 I = I1 + I2 = V R1 + V R2 = V ( 2 ∑ i=1 R−1 i ) = V Req ⇒ R−1 eq = 2 ∑ i=1 R−1 i Req = 1 R1 + 1 R2 = R1R2 R1 + R2 P. R. Nelson Fall 2010 WhatToKnow —- – p. 12/46 Let x = 1 and y = 2. Then I1 = V R1 = I Req R1 = ( R1R2 R1+R2 ) R1 I = ( R2 R1 + R2 ) I If x = 2 and y = 1 then I2 = ( R1 R1 + R2 ) I P. R. Nelson Fall 2010 WhatToKnow —- – p. 13/46 For n resistors in parallel, I = n ∑ j=1 Ij = n ∑ j=1 V Rj = V ( n ∑ j=1 R−1 j ) = V Req ⇒ R−1 eq = n ∑ j=1 R−1 j This is the only formula for parallel resistors that extends easily to more than two resistors. P. R. Nelson Fall 2010 WhatToKnow —- – p. 14/46 KCL Example − +Vx R1 R2 R3 R4 Iy Find VR2 and VR4 . Hint: Since KCL is applied at nodes, how many KCL equations will be required? P. R. Nelson Fall 2010 WhatToKnow —- – p. 16/46 KCL Example - 1 − +Vx R1 R2 R3 R4 Iy Find VR2 and VR4 . P. R. Nelson Fall 2010 WhatToKnow —- – p. 17/46 KCL Example - 1 − +Vx R1 R2 R3 R4 Iy Find VR2 and VR4 . 1. Since KCL is applied at nodes, count the nodes. P. R. Nelson Fall 2010 WhatToKnow —- – p. 17/46 KCL Example - 2, 3 − +Vx R1 R2 R3 R4 Iy Find VR2 and VR4 . 2. Choose one node as ground. Choose for convenience! 3. Label node voltages. P. R. Nelson Fall 2010 WhatToKnow —- – p. 18/46 − +Vx R1 R2 R3 R4 Iy V2 V4 P. R. Nelson Fall 2010 WhatToKnow —- – p. 19/46 − +Vx R1 R2 R3 R4 Iy V2 V4 Both VR2 and VR4 are connected to ground. The voltage source is connected to ground ⇒ one less KCL equation. P. R. Nelson Fall 2010 WhatToKnow —- – p. 19/46 KCL Example - Eqn’s − +Vx R1 I1 R2 I2 R3 I3 R4 I4 Iy V2 V4 KCL at 2: I1 − I2 − I3 = 0 P. R. Nelson Fall 2010 WhatToKnow —- – p. 21/46 KCL Example - Eqn’s − +Vx R1 I1 R2 I2 R3 I3 R4 I4 Iy V2 V4 KCL at 2: I1 − I2 − I3 = 0 KCL at 4: P. R. Nelson Fall 2010 WhatToKnow —- – p. 21/46 KCL Example - Eqn’s − +Vx R1 I1 R2 I2 R3 I3 R4 I4 Iy V2 V4 KCL at 2: I1 − I2 − I3 = 0 KCL at 4: I3 − I4 − Iy = 0 P. R. Nelson Fall 2010 WhatToKnow —- – p. 21/46 KCL Example - Substitution Vx − V2 R1 − V2 R2 − V2 − V4 R3 = 0 V2 − V4 R3 − V4 R4 − Iy = 0 Check! 2 equations in 2 unknowns This pair of equations can be solved, but it’s messy unless resistor values are given. P. R. Nelson Fall 2010 WhatToKnow —- – p. 23/46 KVL Example − +VA R1 R2 R3 I3 IB R4 Find I3. P. R. Nelson Fall 2010 WhatToKnow —- – p. 24/46 KVL Example − +VA R1 R2 R3 I3 IB R4 Find I3. Hint: Since KVL is applied around closed loops, how many KVL equations will be required? P. R. Nelson Fall 2010 WhatToKnow —- – p. 24/46 KVL Example - 1 − +VA R1 R2 R3 I3 IB R4 Find I3. 1. Count the number of “windowpanes” in the circuit. (3) P. R. Nelson Fall 2010 WhatToKnow —- – p. 25/46 KVL Example - 2, 3 − +VA R1 R2 R3 I3 IB R4 Find I3. P. R. Nelson Fall 2010 WhatToKnow —- – p. 26/46 KVL Example - 2, 3 − +VA R1 R2 R3 I3 IB R4 Find I3. 2. Choose one closed loop for each windowpane. Choose for convenience! 3. Choose a direction for each loop. P. R. Nelson Fall 2010 WhatToKnow —- – p. 26/46 KVL Example - I & V − +VA R1 R2 R3 IB R4 I1 I3 I4 − +VA R1 R2 R3 IB R4 + V1 − + V3 − V2 − + V4 − P. R. Nelson Fall 2010 WhatToKnow —- – p. 29/46 KVL Example - Eqn’s KVL for loop 1: P. R. Nelson Fall 2010 WhatToKnow —- – p. 30/46 KVL Example - Eqn’s KVL for loop 1: −VA + V1 + V2 = 0 P. R. Nelson Fall 2010 WhatToKnow —- – p. 30/46 KVL Example - Eqn’s KVL for loop 1: −VA + V1 + V2 = 0 KVL for loop 3: −V2 + V3 + V4 = 0 KVL for loop 4: ??!? P. R. Nelson Fall 2010 WhatToKnow —- – p. 30/46 KVL Example - Eqn’s KVL for loop 1: −VA + V1 + V2 = 0 KVL for loop 3: −V2 + V3 + V4 = 0 KVL for loop 4: ??!? −V4 + V4 = 0 is a tautology . . . P. R. Nelson Fall 2010 WhatToKnow —- – p. 30/46 KVL Example - Eqn’s KVL for loop 1: −VA + V1 + V2 = 0 KVL for loop 3: −V2 + V3 + V4 = 0 KVL for loop 4: ??!? −V4 + V4 = 0 is a tautology . . . . . . but the current source current is IB = I4 − I3 P. R. Nelson Fall 2010 WhatToKnow —- – p. 30/46 KVL Example - Substitution −VA + R1 I1 + (I1 − I3) R2 = 0 − (I1 − I3) R2 + I3 R3 + I4 R4 = 0 I4 − I3 = IB Check! 3 equations in 3 unknowns Collect terms, then solve this set of equations by Kramer’s rule. P. R. Nelson Fall 2010 WhatToKnow —- – p. 32/46 KVL Example - Matrix    R1 + R2 −R2 0 −R2 R2 + R3 R4 0 −1 1       I1 I3 I4    =    VA 0 IB    ∆ =    R1 + R2 −R2 0 −R2 R2 + R3 R4 0 −1 1    = (R1 + R2) (R2 + R3) + (R1 + R2) R4 − R2 2 = R1R2 + R1R3 + R1R4 + R2R3 + R2R4 P. R. Nelson Fall 2010 WhatToKnow —- – p. 33/46 ∆ I3 =    R1 + R2 VA 0 −R2 0 R4 0 IB 1    = R2VA − (R1 + R2) R4VA I3 = R2 VA − (R1 + R2) R4 IB R1R2 + R1R3 + R1R4 + R2R3 + R2R4 P. R. Nelson Fall 2010 WhatToKnow —- – p. 34/46 Norton equivalent Isc Req − V + I I = Isc − V Req V = Req Isc − Req I If the terminals are shorted, V = 0 and I = Isc. If the terminals are connected to an open circuit, I = 0 and V = Req Isc. P. R. Nelson Fall 2010 WhatToKnow —- – p. 37/46 Graphical representation All possible combinations of pairs of voltage and current values for either equivalent circuit model can be represented as the line through the points (V = 0, ISC) and (VOC , 0). I V ISC VOC P. R. Nelson Fall 2010 WhatToKnow —- – p. 38/46 Superposition A voltage or current in a linear circuit is the superposition (sum) of the results for each source alone. Superposition can be used along with parallel and series resistors, voltage and current dividers, and Thèvenin and Norton equivalent circuits to simplify the analysis of a circuit. Superposition often gives solutions in a mathematical form that enhances intuition. P. R. Nelson Fall 2010 WhatToKnow —- – p. 39/46 Superposition Example - V Set Iy = 0 and solve for the current I2V . − +Vx R1 R2 I2V R3 I2V = ( R3 R1R2 + R1R3 + R2R3 ) Vx P. R. Nelson Fall 2010 WhatToKnow —- – p. 41/46 Superposition Example - I, Result Set Vx = 0 and solve for the current I2I . R1 R2 I2I R3 Iy P. R. Nelson Fall 2010 WhatToKnow —- – p. 42/46 Superposition Example - I, Result Set Vx = 0 and solve for the current I2I . R1 R2 I2I R3 Iy I2I = − ( R1R3 R1R2 + R1R3 + R2R3 ) Iy P. R. Nelson Fall 2010 WhatToKnow —- – p. 42/46 Dependent Source Example − +Vx R1 R2 + V2 − +− BV2 R3 + V3 − Find V3. P. R. Nelson Fall 2010 WhatToKnow —- – p. 44/46 Dependent Source Example − +Vx R1 R2 + V2 − +− BV2 R3 + V3 − Find V3. Hint: The answer may only contain Vx, R1, R2, R3, and B. P. R. Nelson Fall 2010 WhatToKnow —- – p. 44/46 Dependent Source Example - Eqn’s − +Vx R1 R2 + V2 − +− BV2 R3 + V3 − Find V3. P. R. Nelson Fall 2010 WhatToKnow —- – p. 45/46