EBK PRECALCULUS W/LIMITS

4th Edition

ISBN: 9781337516853

Author: Larson

Publisher: CENGAGE CO

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Question

Chapter 7.3, Problem 65E

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 65E

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 64E

Chapter 7.3, Problem 66E

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

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If we apply Kirchhoff’s law to the electrical network in the figure, the currents I 1 , I 2 and I 3 , are the solutions of the system, what will be the currents? { I 1 − I 2 + I 3 = 0 3 I 1 + 2 I 2 = 7 2 I 2 + 4 I 3 = 8

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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

Answer to Problem 65E

Explanation of Solution

Chapter 7 Solutions

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