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2.7 Source Transformations A method called Source Transformations will allow the transformations of a voltage source in series with a resistor to a current source in parallel with resistor. R a a vs is + - R b b The double arrow indicate that the transformation is bilateral , that we can start with either configuration and drive the other R a a vs + - is RL iL R RL iL b b Voltage source with a series resistor Current source with a parallel resistor R iL is R + RL vs iL R + RL Equating we have , vs R is R + RL R + RL is vs R OR v s Ri s Polarity of Vs in one form must be such that it tends to push in the direction of Is in the other form + - Ex. 2.29: Determine i and ix using source transformation Single loop KVL 15V - 10V i 1A 2 + 3 ix should be found from the original circuit i x - (i - 5) = 5 - i 4A Ex 2.30: Determine I and V Transform the 3V in series with the 2 Ohm resistor We get a single-node-pair circuit 14 3 3 4 V A + 2A * 2 || 4 A + 2A * V 3 2 2 3 V 7 A Using Ohm’s law in the original circuit: I 4 6 Example (a) find the power associated with the 6 V source (b) State whether the 6 V source is absorbing or delivering power We are going to use source transformation to reduce the circuit, however note that we will not alter or transfer the 6 V source because it is the objective. 40 8A 5 (20 || 5) 4Ω (8A )(4Ω) 32 V (6 + 4) 10Ω (10 + 10) 20Ω 32 1.6A 20 (30 || 20) 12Ω (4 + 12) 16Ω (1.6A )(12Ω) 19.2 V i i= - 0.825 A + P6V (0.825)(6) 4.95 W It should be clear if we transfer the 6V during these steps you will not be able to find the power associated with it