hw3soln

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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
1 -2j 2j 2 4 p V1 0 3 0 V = I I 2 I 1 Transform the circuit: The total impedance seen by the voltage source is: 4( j 2 ) | | ( 2j 2 ) (j2)(2 j2) 44 2 j 2 j2 2 j2 6j 2 Z =+ =++ +− Using this we can calculate I: 10 30 1.58111.565 A 2 V I Z == = + Now we use current dividers to calculate I 1 and I 2 : For the source: For the 4 resistor: () ( ) 2 2 11 1.581 4 5 W 22 PI R = For the 2 resistor: ( ) 2 2 2 1.581 2 2.5 W 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 3 023 01 2 12 3 1 2 3 0.5 0.5 4 60 0.5 0.5 0.5 0 00 . 2 5 4V 45 0 0.5 0.5 0 4 60 0.5 0.5 0.5 0 1 0 . 2 5 1 0 451 0 0 jj V jV A V A VV VVV V V A ⎡⎤ ⎢⎥ −− = −+ ⎣⎦ =− = = 0.25 S 0.5 S 1 S j 3 V 2 V Solving for V 3 : 3 9.992 5.06 V V = 2 3 P 24.96 W V R For the 4 resistor: 0 4V 460 A -0.5 S j 0 V - +
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/19/2012 for the course EECE 253 taught by Professor Shuo,tang during the Spring '11 term at UBC.

Page1 / 3

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online