U210_sol01 - 17 . 1318 17 . 1918 ) 600 ( 376920 ) 600 ( 5 ....

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
ECEU210 Solutions to HW #1 P 1.6 P 1.14 P 1.20 Comment for P 1.20 (c) -> The current and voltage references for the 4-volt battery are a "load set", so the power output of the battery is p=-iv. The numerical result of the power output is negative, indicating that in this circuit the battery is actually receiving power (it is being charged).
Background image of page 1

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

View Full DocumentRight Arrow Icon
P 1.24 P 1.28 P 1.32
Background image of page 2
P 1.40 (a) P 1.40 (b)
Background image of page 3

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

View Full DocumentRight Arrow Icon
Background image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 17 . 1318 17 . 1918 ) 600 ( 376920 ) 600 ( 5 . 196 ) 600 ( 5 . 523 720 5 . 523 ) 600 ( 720 5 . 523 ) 600 ( 720 ) 600 ( 720 5 . 523 2 720 327 ) 600 ( || 720 ! ! " ! " # " # " # ! " # $ ! " " " # $ ! " ! " R R R R R R R R P 1.42 (c) Apply series/parallel reduction from right to left The voltage across the rightmost 4 ! resistor is v = 4 i 3 = 4 i 1 /4 = 4 * 1.75/4 = 1.75 V...
View Full Document

This note was uploaded on 04/07/2008 for the course ECE U210 taught by Professor Shekel during the Spring '08 term at Northeastern.

Page1 / 4

U210_sol01 - 17 . 1318 17 . 1918 ) 600 ( 376920 ) 600 ( 5 ....

This preview shows document pages 1 - 4. Sign up to view the full document.

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