hw5_ext_sol

# hw5_ext_sol - EE 101 Professor Abbas El Gamal Handout 15-a...

This preview shows pages 1–3. Sign up to view the full content.

EE 101 Handout # 15-a Professor Abbas El Gamal Feb. 14, 2003 Homework #5 Extra Problem Solutions 1. Here we use superposition. First turn o± the 1 and 2V sources. The circuit reduces to current source in series with two 1Ω parallel resistors and v 1 = - 0 . 5V. Next we turn the 1V source on and the other sources o±. We get v 2 = - 1V. Next we turn on the 2V source only and get v 3 = 1V. Thus v = v 1 + v 2 + v 3 = - 0 . 5V. 2. This problem is best attacked via superposition, as implied by the presence of multiple sources. Keeping the leftmost current source and setting the other two to zero results in the following: 1A i 1 By application of the current divider rule, i 1 = 0 . 5A. Keeping the voltage source and setting the current sources to zero, we obtain the following: 1V i 2 Ohm’s law tells us that 1V = - i 2 (2Ω), implying that i 2 = - 0 . 5A. Finally, keeping the rightmost current source, we get the circuit: 1A i 3 1

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

View Full Document
So, i 3 = 1A. Adding the three current components, we get
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 3

hw5_ext_sol - EE 101 Professor Abbas El Gamal Handout 15-a...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online