hw3soln

# hw3soln - Note all solutions assume rms values unless...

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1 -2j 2j 2 4 p V 10 30V = I I 2 I 1 Transform the circuit: The total impedance seen by the voltage source is: 4 (j2) || (2 j2) (j2)(2 j2) 4 4 2 j2 j2 2 j2 6 j2 Z = + = + = + + + = + Using this we can calculate I: 10 30 1.58111.565 A 6 j2 V I Z = = = + Now we use current dividers to calculate I 1 and I 2 : For the source: For the 4 resistor: ( ) ( ) 2 2 1 1 1.581 4 5 W 2 2 P I R = = = For the 2 resistor: ( ) ( ) 2 2 2 1 1 1.581 2 2.5 W 2 2 P I R = = = For the inductor and the capacitor: 0 W P = 1) Note all solutions assume rms values unless otherwise specified. A 435 . 33 236 . 2 2 2 2 A 565 . 101 581 . 1 2 2 2 1 ° = = ° = = I j I I j I ( )( ) W 7.5 VA 5 . 2 5 . 7 565 . 11 581 . 1 30 10 2 1 2 1 * = + = ° ° = = P j I V S 2) 1 2 3 0 2 3 0 1 2 1 2 3 1 2 3 0.5 0.5 4 60 0.5 0.5 0.5 0 0 0.25 4 V 4 5 0 0.5 0.5 0 4 60 0.5 0.5 0.5 0 1 0 0 0.25 1 0 4 5 1 0 0 j j V j V A j j V A V V V V V V V V j j V j V j j V A + = + = = + = +

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