# Example: Determine the power supplied by each of the sources, independent and dependent, in this circuit:

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1 Example: Determine the power supplied by each of the sources, independent and dependent, in this circuit: Solution: We ll begin by choosing the bottom node to be the reference node. Next we ll label the other nodes and some element oltages: Notice that the 8 Ω resistor, the 0 Ω resistor and the two independent current sources are all connected in parallel. Consequently, the element oltages of theses elements can be labeled so that they are equal. Similarly, the 4 Ω resistor and the dependent current source are connected in parallel so their oltages can be labeled so as to be equal. Using Ohm s Law we see that the current directed downward in the 8 Ω resistor is directed downward in the 0 Ω resistor is Ω resistor is, current 8, and the current directed from left to right in the 0. Applying Kirchhoff s Current Law (KCL) at node a gies 5 = =.5 () 8 0 Using Ohm s Law we see that the current directed downward in the 4 Ω resistor is Kirchhoff s Current Law (KCL) at node a gies. Applying 4

2 +.5 = = 0 () 4 Applying Kirchhoff s Voltage Law (KVL) to the mesh consisting of the 0 Ω resistor, the Ω resistor and the dependent source to get + = 0 () Equations, and comprise a set of three simultaneous equations in the three unknown oltages, and. We can write these equations in matrix form as = 0 0 We can sole this matrix equation using MATLAB: Hence = V, = V and = V

3 The power supplied by the 5 A current source is ( ) supplied by the.5 A current source is ( ) 5 = = W. The power.5 = = 0.47 W. The power.5 = = W. supplied by the dependent current source is ( ) ( )( ) Obseration: Changing the order of the 8 Ω resistor, the 0 Ω resistor and the two independent current sources only changes the order of the terms in the KCL equation at node a. We know that addition is commutatie, so change the order of the terms will not affect the alues of the oltages, and. For example, if the positions of the.5 A current source and 8 Ω resistor are swithched: The KCL equation at node a is Similarly, when the circuit is drawn as 5 = = The KCL equation at node a is 5 = = The changes do not affect the alues of the oltages, and.

4 Example: Determine the power supplied by each of the sources, independent and dependent, in this circuit: Solution: We ll begin by choosing the bottom node to be the reference node. Next we ll label the other nodes and some element currents: Notice that two 4 Ω resistors, an 8 Ω resistor and the two independent oltage sources are all connected in series. Consequently, the element currents of theses elements can be labeled so that they are equal. Similarly, a 4 Ω resistor and the dependent oltage source are connected in series so their currents can be labeled so as to be equal. The current in each resistor has been labeled so we can use Ohm s Law to calculate resistor oltages from the resistor currents and the resistances. Apply Kirchhoff s Voltage Law (KVL) to the left mesh to get 6 + 8i + 8i + 4i 5 + 4i = 0 6i + 8i = 9 () Apply Kirchhoff s Voltage Law (KVL) to the right mesh to get 4i + 5i 8i = 0 () Applying Kirchhoff s Current Law (KCL) at node a to get i = i + i i+ i + i = 0 () 4

5 Equations, and comprise a set of three simultaneous equations in the three unknown oltages, and. We can write these equations in matrix form as i i = 0 0 i We can sole this matrix equation using MATLAB: Hence i = A, i = A and i = 0.0 A The power supplied by the 5 V oltage source is ( ) supplied by the 6 V oltage source is i ( ) the dependent oltage source is ( i ) i ( )( ) 5i = = 6.65 W. The power 6 = =.4546 W. The power supplied by 5 = = W. 5

6 Obseration: Changing the order of the two 4 Ω resistors, an 8 Ω resistor and the two independent oltage sources in the left mesh changes the order of the terms in the KVL equation for that mesh. We know that addition is commutatie, so change the order of the terms will not affect the alues of the currents, and i. For example, when the circuit is drawn as i i The KVL equation for the left mesh is i + 4i + 8i + 4i = 0 6i + 8i = 9 When the circuit is drawn as The KVL equation for the left mesh is 4i + 4i + 8i i = 0 6i + 8i = 9 These changes do not affect the alues of the currents, and i. i i 6

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