TEXTBOOK OF ELECTRICITY W/MINDTAP >BI<
TEXTBOOK OF ELECTRICITY W/MINDTAP >BI<
6th Edition
ISBN: 9781305812505
Author: Herman
Publisher: Cengage Learning
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Chapter 9, Problem 3PP
To determine

The unknown parameters using Kirchhoff’s Laws.

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Answer to Problem 3PP

    ES1=6 V E1= 0.89 V E2= 20.89 V E3= 11.1 V
    ES2=60 V I1= 1.31 mA I2= 20.89 mA I3= 22.2 mA
    R1=1.6kΩ R2= 1.2 kΩ R3= 2.4 kΩ

Explanation of Solution

We will use Kirchhoff’s Voltage law to solve for the unknown parameters.

Consider two loops LOOP 1 and LOOP 2 with their respective loop currents I1 and I2

  TEXTBOOK OF ELECTRICITY W/MINDTAP >BI<, Chapter 9, Problem 3PP , additional homework tip  1

Using KVL in loop 1,

  +62400(I1+I2)1600I1=0        4000I12400I2+6=0  ... (1)

Using KVL in loop 2,

  +602400(I2+I1)1200I2=0         2400I13600I2+60=0    ... (2)

From equation (1), we can write I2as ,

  I2=64000I12400   ... (3)

Substitute this expression for I2in equation (2),

                   2400I13600I2+60=0 2400I13600(64000I12400)+60=0           2400I19+6000I1+60=0                                   3600I1+51=0                                                    I1=513600                                                    I1=14.16 mA    

The negative sign indicates that the direction of current is different from that assumed at the start.

Substitute the value of I1 in equation (3).

  I2=64000I12400     =64000(14.16×103)2400     =26.11 mA 

CurrentI3 is the sum of currents I1 and I2

  I3=I1+I2   =(14.16+26.11) mA   =11.95 mA

The voltage drops are obtained by using Ohm’s Law,

  E1=I1R1   =(14.16×103)×(1600)   =22.66 V

  E2=I2R2   =(26.11×103)×(1200)   =31.33 V

  E3=I3R3    =(11.95×103)×(2400)    =28.68 V

The results are summarized in the figure below,

  TEXTBOOK OF ELECTRICITY W/MINDTAP >BI<, Chapter 9, Problem 3PP , additional homework tip  2

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