ps4_soln

Irwin, Basic Engineering Circuit Analysis, 10/E 1

Chapter 2: Resistive circuits Problem 2.90

SOLUTION:

2 Irwin, Basic Engineering Circuit Analysis, 10/E

Problem 2.90 Chapter 2: Resistive circuits



SOLUTION:

24V

6

16



SOLUTION:

6V



SOLUTION:



SOLUTION:

-



SOLUTION:

1

2FE-10

Strategy

1. Reduce circuit to a single voltage source plus equivalent resistance.

2. To do so, see 8+2 in series, then in parallel with 10, etc.... Or for practice, could convertbottom “∆” into “Y”. (Calling bottom right corner a; b and c going clockwise from a.

3. After finding Req, find source current (total current supplied by voltage source).

4. Then backtrack the simplifying steps in part 1 until we have Ix back in the circuit.

5. Find Ix using current division.

Req = [(8 + 2)//10 + 1]//3Ω + 1Ω = 3Ω

Is =12V

3Ω= 4A

Ix =1 + 4 + 1

3 + 1 + 4 + 1Is =

2

34A

=8

3A

Using “Y-∆” (for practice–but top way is easier).

Ra =2 · 8

2 + 8 + 10Ω =

4

5Ω

Rb =8 · 10

2 + 8 + 10Ω = 4Ω

Rc =10 · 2

2 + 8 + 10Ω = 1Ω

Req = [1 + (1 + 4 + 1)//3]Ω = 3Ω

Is =12V

3Ω= 4A

Ix =1 + 4 + 1

3 + 1 + 4 + 1Is =

2

34A

=8

3A

created by Deborah Won for EE204: Circuit Analysis

ps4_soln

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