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3 k2 ww 4 k2 2 k2 5 kΩ 1 k2 V 3 V 10 k2 4 V + 1)

3 k2 ww 4 k2 2 k2 5 kΩ 1 k2 V 3 V 10 k2 4 V + 1)

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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In the circuit given below, V = 10 V. Determine the current through the 10-kΩ resistor. 

3 ΚΩ 4 ΚΩ 2 ΚΩ 5 k2 ww 1 k2 V 3 V 10 k2 4 V (+|
Transcribed Image Text:3 ΚΩ 4 ΚΩ 2 ΚΩ 5 k2 ww 1 k2 V 3 V 10 k2 4 V (+|
Expert Solution
Step 1

Apply KCL at node 1, we will get

         7V1-2V2=26 ...........................(1)

 

Apply KCL at node 2, we will get

         -5V1+8V2=-6 ...........................(2)

By solving equation 1 and 2 we will get

V1=4.26 VV2=1.913 V

 

Current through 10Kohm resistor is

                      I=1.91310=0.1913 mAmps

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