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Lecture #3 OUTLINE KCL; KVL examples Dependent sources.

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Presentation on theme: "Lecture #3 OUTLINE KCL; KVL examples Dependent sources."— Presentation transcript:

1 Lecture #3 OUTLINE KCL; KVL examples Dependent sources

2 Use reference polarities to determine whether a voltage is dropped – with no concern about actual voltage polarities Using Kirchhoff’s Voltage Law (KVL) Consider a branch which forms part of a loop: +v1_+v1_ loop voltage “drop” –v2+–v2+ loop voltage “rise” (negative drop)

3 Formulations of Kirchhoff’s Voltage Law Formulation 1: Sum of voltage drops around loop = sum of voltage rises around loop Formulation 2: Algebraic sum of voltage drops around loop = 0 Voltage rises are included with a minus sign. Formulation 3: Algebraic sum of voltage rises around loop = 0 Voltage drops are included with a minus sign. (Conservation of energy)

4 A Major Implication of KVL KVL tells us that any set of elements which are connected at both ends carry the same voltage. We say these elements are connected in parallel. Applying KVL in the clockwise direction, starting at the top: v b – v a = 0  v b = v a + v a _ + v b _

5 Path 1: Path 2: Path 3: vcvc vava ++ ++ 3 2 1 +  vbvb v3v3 v2v2 + - Three closed paths: a b c KVL Example

6 Example 1

7 Example 2

8 Find v using KVL and KCL

9 Simplify a circuit before applying KCL and/or KVL:  + 7 V R 1 = R 2 = 3 k  R 3 = 6 k  R 4 = R 5 = 5 k  R 6 = 10 k  I R1R1 R2R2 R4R4 R5R5 R3R3 R6R6 Find I

10 Find Vx, Vy and Vz

11 Dependent Sources A linear dependent source is a voltage or current source that depends linearly on some other circuit current or voltage. We can have voltage or current sources depending on voltages or currents elsewhere in the circuit.

12 Here, the voltage V provided by the dependent source (right) is proportional to the voltage drop over Element X. The dependent source does not need to be attached to the Element X in any way.

13 The 4 Basic Dependent Sources

14 Circuit with Dependent Source Example 1

15 Example 2

16 Find i 2, i 1 and i o

17 –+–+ –+–+ 80 V 5i5i 20  10  200  2.4 A ii Exercise


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