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Electrical Circuits_Lecture4 Circuit Theorems Lecture 4 Objectives: To study the Superposition theorem To study the Source Transformation techniques Voltage to current transformation Current to voltage transformation INTRODUCTION A large complex circuits Simplify circuit analysis Circuit Theorems ‧Thevenin’s theorem ‧Circuit linearity ‧source transformation ‧ Norton theorem ‧ Superposition ‧ max. power transfer Linearity Property A linear circuit is one whose output is linearly related (or directly proportional) ito its input. I0 Suppose vs = 10 V gives i = 2 A. According to the linearity principle, vs = 5 V will give i = 1 A. V0 v Superposition The superposition principle states that the voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltages across (or currents through) that element independent source acting alone. due to each Steps to apply superposition principle 1. Turn off all independent sources except one source. Find the output (voltage or current) due to that active source using nodal or mesh analysis. 2. 3. Turn off voltages sources = short voltage sources; make it equal to zero voltage Turn off current sources = open current sources; make it equal to zero current Repeat step 1 for each of the other independent sources. Find the total contribution by adding algebraically all the contributions due to the independent sources. Dependent sources are left intact. Example 1: use Superposition principle to find Io 2mA Source Contribution 2kW 2mA 1kW I’0 I’0 = -4/3 mA 2kW 4mA Source Contribution 2kW 4mA 1kW I’’0 I’’0 = 0 2kW 12V Source Contribution 12V 2kW – + 1kW I’’’0 I’’’0 = -4 mA 2kW Final Result I’0 = -4/3 mA I’’0 = 0 I’’’0 = -4 mA I0 = I’0+ I’’0+ I’’’0 = -16/3 mA Example 2: find v using superposition one independent source at a time, dependent source remains KCL: i = i1 + i2 Ohm's law: i = v1 / 1 = v1 KVL: 5 = i (1 + 1) + i2(2) KVL: 5 = i(1 + 1) + i1(2) + 2v1 10 = i(4) + (i1+i2)(2) + 2v1 10 = v1(4) + v1(2) + 2v1 v1 = 10/8 V Consider the other independent source KCL: i = i1 + i2 KVL: i(1 + 1) + i2(2) + 5 = 0 i2(2) + 5 = i1(2) + 2v2 Ohm's law: i(1) = v2 v2(2) + i2(2) +5 = 0 => i2 = (5+2v2)/2 i2(2) + 5 = i1(2) + 2v2 -2v2 = (i - i2)(2) + 2v2 -2v2 = [v2 + (5+2v2)/2](2) + 2v2 -4v2 = 2v2 + 5 +2v2 -8v2 = 5 => v2 = - 5/8 V from superposition: v = -5/8 + 10/8 v = 5/8 V Source Transformation A source transformation is the process of replacing a voltage source vs in series with a resistor R by a current source is in parallel with a resistor R, or vice versa Source Transformation vs vs is R or is R Source Transformation Vs Rs I s Vs Is Rs Source Transformation Equivalent sources can be used to simplify the analysis of some circuits. A voltage source in series with a resistor is transformed into a current source in parallel with a resistor. A current source in parallel with a resistor is transformed into a voltage source in series with a resistor. Example 3: Use source transformation to find vo in the circuit in the following figure. we use current division to get 2 i (2) 0.4A 28 and vo 8i 8(0.4) 3.2V Example5: Find vx in the circuit using source transformation 3 5i vx 18 0 Applying KVL around the loop igives 3 5i vx 18 0 (1) Appling KVL to the loop containing only the 3V voltage source, the resistor, and vx yields (2) 3 1i vx 0 vx 3 i Substituting this into Eq.(1), we obtain 15 5i 3 0 i 4.5A Alternatively vx 4i vx 18 0 i 4.5A thus vx 3 i 7.5V