Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10EE201 Lecture 11 P. 1Source TransformationsA technique for analyzing circuits that does not require solving simultaneous equations.Terminology:A two-terminal network is an interconnection of circuit elements which has only two accessible terminals for connection to other networks.+__+ivSource TransformationsEE201 Lecture 11 P. 2Equivalent two-terminal networks are two-terminal networks with the same terminal current-voltage relationship, but different internal details.+_ 6 V_+i 3 A 2 2 v_+ivi + 3 = v / 2 v = 6 + 2i v = 6 + 2i RL RLThis is a source transformation.EE201 Lecture 11 P. 3Source transformations are possible for two terminal networks with dependent sources as |shown below. _+iv_+iv_+ R R IS VSIS VS / R RL RLEE201 Lecture 11 P. 4+_vs R_+vLLook at equivalence from I-V plots (compare with P. 5)vLiL RLiLRL = (open)RL = 0 (short)vs - R iL = vLiL = (vS/R) - (vL/R)vSvS/REE201 Lecture 11 P. 5_+ iS RConsider equivalence from I-V plots (compare with P. 4)vL RLiLvLiLRL = (open)RL = 0 (short)iSRiSis = iL + (vL/R)iL = iS - (vL/R)Equivalent if, is = vs/REE201 Lecture 11 P. 6 1 A 2 5 Example:_+ 3 + 2 18 V 15 VUse source transformations to find power absorbed by the 5 resistor._EE201 Lecture 11 P. 7 2 V 5 Solution:_+ 3 _+ 2 18 V 15 VStep 1: Replace current source-parallel resistor combination with voltage source-series resistor combination. Vs = Is R._+ 2 EE201 Lecture 11 P. 8 16 V 5 _+ 3 _+ 15 VStep 2: Sum series resistors and voltage source outputs to simplify circuit. 4 EE201 Lecture 11 P. 9Step 3: Replace each voltage source-resistor pair with parallel current source-resistor pair. 4 A 4 5 3 5 AStep 4: Find equivalent resistance of parallel resistors. 4 A (12/7) 5 5 AEE201 Lecture 11 P. 10Step 5: Combine effects of current sources in accordance with principle of superposition.Step 6: Using current division, find current through 5 resistor.I5 = [(2.4/12) / ((2.4/12)+(7/12))] 1 = (12/47) AStep 7: Calculate power through 5 resistor.P5 = I2 R = (12/47)2 5 = 0.326 W 1 A (12/7) 5
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