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15 V 6.0 0 20 V 3.5 2l 4.0 Ω 5.0 Ω 2.0 n 2.5 0. 4.0 0 7.0 0 4.5 0 12 V 2.5 0 10.0 2 (a) (b)

15 V 6.0 0 20 V 3.5 2l 4.0 Ω 5.0 Ω 2.0 n 2.5 0. 4.0 0 7.0 0 4.5 0 12 V 2.5 0 10.0 2 (a) (b)

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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Use Kirchhoff’s Laws to solve for the currents in the 4.0 Ω resistors in the circuits in (a) and (b)

15 V 6.0 0 20 V 3.5 2l 4.0 Ω 5.0 Ω 2.0 n 2.5 0. 4.0 0 7.0 0 4.5 0 12 V 2.5 0 10.0 2 (a) (b)
Transcribed Image Text:15 V 6.0 0 20 V 3.5 2l 4.0 Ω 5.0 Ω 2.0 n 2.5 0. 4.0 0 7.0 0 4.5 0 12 V 2.5 0 10.0 2 (a) (b)
Expert Solution
Step 1

(a).

The given circuit diagram is:

Electrical Engineering homework question answer, step 1, image 1

 

Step 2

Applying Kirchhoff’s voltage law (KVL):

 

Kirchhoff’s voltage law: It states that the algebraic sum of voltage in a closed circuit is equal to zero.

                  Electrical Engineering homework question answer, step 2, image 1

The modified circuit diagram is:

Electrical Engineering homework question answer, step 2, image 2 

                                 

Step 3

The loop equations are:

Electrical Engineering homework question answer, step 3, image 1

Solving equation 1,2 and 3:

Electrical Engineering homework question answer, step 3, image 2

Step 4

The current in 4Ω resistor is equal to the I3. Hence,

Electrical Engineering homework question answer, step 4, image 1

The current in 4Ω resistor is 0.112 A.

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