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Chapter 4 - Additional Analysis Techniques(for Linear Circuits): Linearity and Superposition

Chapter 4 - Additional Analysis Techniques(for Linear Circuits): Linearity and Superposition. Read Chapter 4, pages 113 - 120.

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Chapter 4 - Additional Analysis Techniques(for Linear Circuits): Linearity and Superposition

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  1. Chapter 4 - Additional Analysis Techniques(for Linear Circuits): Linearity and Superposition Read Chapter 4, pages 113 - 120 Definition: A linear circuit is a circuit constructed using elements that have linear voltage-current relationships. That is, V = f(I) and I = f(V) are linear relationships. ENGR201 Circuits I - Chapter 4

  2. Examples of linear functions: Examples of nonlinear functions: • Q: What is a linear function? • A: A linear function is a function which satisfies both of the following properties: • additivity • if y = f(x) and y1 = f(x1) and y2 = f(x2) then • y = f(x1 + x2) = y1 + y2 • homogeneity: • if y = f(x) then y1 = f(Kx) =Ky • additivity and homgeneity combined: • if y = f(x) and y1 = a1f(x1) and y2 = a21f(x2) then • y = f(a1x1 +a2x2) = a1y1 + a2y2 Note that Ohm’s law is a linear relationship: V = IR ENGR201 Circuits I - Chapter 4

  3. + Vo - Using Linearity to Analyze Circuits • What is the value of Vo in the circuit shown? • by the VDR, Vo = (1/3)  12V = 4v • What is the value of Vo if the 12-volt source is doubled to 24v? • Vo = (1/3)  24V = 8v  Vo is doubled • What is the value of Vo if the 12-volt source cut in half (6v)? • Vo = (1/3)  6V = 2v  Vo is halved Because the circuit is linear, Vo is linear function of the input (Vo = Vin/3) multiplying the input by a constant results in the output being multiplied by the same constant. ENGR201 Circuits I - Chapter 4

  4. Is + VA - I1 Io + Vo -  Io = 1v/2k = 0.5ma By KVL VA = 4k  Io + 1v =4k 0.5ma + 1v = 3v  I1 = 3v/3k = 1ma By KCL, Is = I1 + Io = 1.5ma Example 4.1 Use linearity to find Vo Solution: Assume that Vo is known and that Vs is unknown and work the problem “backwards”. Assume a convenient value for Vo: assume Vo = 1v Vs = 2k  Is + VA = 2k  1.5ma + 3v = 6v If we know the value of Vo for one value of Vs , we know Vo for any value of Vs. Since the actual value of Vs is 12v (2  6v), then the actual value of Vo is 2  1v = 2v ENGR201 Circuits I - Chapter 4

  5. Superposition Superposition is based upon the additivity property of linear circuits and applies if the circuit has more than one independent source. • Superposition: • For a source with multiple independent sources, determine the desired voltage or current when each source is applied with all other sources zeroed out. • a voltage source is zeroed by replacing it with a short circuit • a current source is zeroed by replacing it with an open circuit • The response when all independent sources are applied simultaneously is the sum of the responses to the individual sources. ENGR201 Circuits I - Chapter 4

  6. + Vo1 - I + Vo2 - Example 4.3 Find Vo using superposition Apply 3V source: + Vo - Apply 2mA source: ENGR201 Circuits I - Chapter 4

  7. Voltage Sources • Observations and definitions: • RL, IL, and VL are the load resistance, current, and voltage • Ri is the internal resistance of the voltage source • no load is defined as the condition IL = 0 A (open circuit  PL = 0 W) • by KVL, VL = VS – ILRi • the difference between the available voltage, VS, and the load voltage, VL, is the drop across the internal resistance • for an ideal voltage source, Ri = 0 ENGR201 Circuits I - Chapter 4

  8. VL Ideal Voltage Source VS IL no load VL Practical Voltage Source VS (slope = -Ri) IL full load no load Ideal Versus Practical Voltage Sources VL = VS – ILRi ENGR201 Circuits I - Chapter 4

  9. Current Sources • Observations and definitions: • RL, IL, and VL are the load resistance, current, and voltage • Ri is the internal resistance of the current source • no load is defined as the condition VL = 0 V (short circuit  PL = 0 W) • by KCL, IL = IS – IL • the difference between the available current, IS, and the load current, IL, is the amount of current flowing through the internal resistance • for an ideal current source, Ri =  ENGR201 Circuits I - Chapter 4

  10. IL Ideal Current Source IS VL no load IL Practical Current Source IS (slope = -1/Ri) VL full load no load Ideal Versus Practical Current Sources IL = IS – Ii ENGR201 Circuits I - Chapter 4

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