EBK PRECALCULUS W/LIMITS

3rd Edition

ISBN: 9781285607160

Author: Larson

Publisher: CENGAGE CO

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Question

Chapter 7.3, Problem 67E

To determine

To calculate:

If we apply Kirchhoff’s law to the electrical network in the figure, the currents I1 , I2 and I3 , are the solutions of the system, what will be the currents?

{I1−I2+I3=03I1+2I2=72I2+4I3=8

Expert Solution & Answer

Answer to Problem 67E

The values of the currents are I1=17,I2=−22,I3=13

Explanation of Solution

Given information:

The given equations are {I1−I2+I3=03I1+2I2=72I2+4I3=8

Calculation:

We can solve these equations by elimination method.

First, let us add equation 2 and 3 and then we get a equation with all three variables. Then we can eliminate any one of the variable.

3I1+2I2+2I2+4I3=7+83I1+4I2+4I3=15

Now we can solve equation 1 and the above formed equation to eliminate the variables.

Let us multiply equation 1 by 4 so that we can eliminate.

4I1−4I2+4I3=0

Let us add these both equations now,

4I1−4I2+4I3+3I1+4I2+4I3=15+07I1+8I3=15

Now we have a equation in two variable we can make into a equation which can be substituted into other.

I1=15−8I37

Now we can substitute this in equation 2.

3I1+2I2=73(15−8I37)+2I2=745−24I3+14I2=4914I2−24I3=47I2−12I3=2

Now solving the above equation and equation 3 we get,

Multiply equation 3 by 7 and the above equation by 2 then we subtract them we get

7(2I2+4I3=8)→32(7I2−12I3=2)→414I2+28I3=56→314I2−24I3=4→4

By subtracting them we get,

14I2+28I3−14I2+24I3=56−44I3=52I3=13

Substitute this value in equation 3 we get

2I2+4I3=82I2+4(13)=82I2+52=8I2=−44/2I2=−22

Substitute this in equation 2 we get the other variable.

3I1+2I2=73I1+2(−22)=73I1−44=73I1=7+44I1=51/3I1=17

Conclusion:

The values of the currents are I1=17,I2=−22,I3=13

Chapter 7.3, Problem 66E

Chapter 7.3, Problem 68E

Chapter 7 Solutions

EBK PRECALCULUS W/LIMITS

Ch. 7.1 - Prob. 1E Ch. 7.1 - Prob. 2E Ch. 7.1 - Prob. 3E Ch. 7.1 - Prob. 4E Ch. 7.1 - Prob. 5E Ch. 7.1 - Prob. 6E Ch. 7.1 - Prob. 7E Ch. 7.1 - Prob. 8E Ch. 7.1 - Prob. 9E Ch. 7.1 - Prob. 10E

Ch. 7.1 - Prob. 11E Ch. 7.1 - Prob. 12E Ch. 7.1 - Prob. 13E Ch. 7.1 - Prob. 14E Ch. 7.1 - Prob. 15E Ch. 7.1 - Prob. 16E Ch. 7.1 - Prob. 17E Ch. 7.1 - Prob. 18E Ch. 7.1 - Prob. 19E Ch. 7.1 - Prob. 20E Ch. 7.1 - Prob. 21E Ch. 7.1 - Prob. 22E Ch. 7.1 - Prob. 23E Ch. 7.1 - Prob. 24E Ch. 7.1 - Prob. 25E Ch. 7.1 - Prob. 26E Ch. 7.1 - Prob. 27E Ch. 7.1 - Prob. 28E Ch. 7.1 - Prob. 29E Ch. 7.1 - Prob. 30E Ch. 7.1 - Prob. 31E Ch. 7.1 - Prob. 32E Ch. 7.1 - Prob. 33E Ch. 7.1 - Prob. 34E Ch. 7.1 - Prob. 35E Ch. 7.1 - Prob. 36E Ch. 7.1 - Prob. 37E Ch. 7.1 - Prob. 38E Ch. 7.1 - Prob. 39E Ch. 7.1 - Prob. 40E Ch. 7.1 - Prob. 41E Ch. 7.1 - Prob. 42E Ch. 7.1 - Prob. 43E Ch. 7.1 - Prob. 44E Ch. 7.1 - Prob. 45E Ch. 7.1 - Prob. 46E Ch. 7.1 - Prob. 47E Ch. 7.1 - Prob. 48E Ch. 7.1 - Prob. 49E Ch. 7.1 - Prob. 50E Ch. 7.1 - Prob. 51E Ch. 7.1 - Prob. 52E Ch. 7.1 - Prob. 53E Ch. 7.1 - Prob. 54E Ch. 7.1 - Prob. 55E Ch. 7.1 - Prob. 56E Ch. 7.1 - Prob. 57E Ch. 7.1 - Prob. 58E Ch. 7.1 - Prob. 59E Ch. 7.1 - Prob. 60E Ch. 7.1 - 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