Chapter 11: Q. 74 (page 716)

An application of Kirchhoff’s Rules to the circuit shown results in the following system of equations:
$I_{3} = I_{1} + I_{2} 8 = 4 I_{3} + 6 I_{2} 8 I_{1} = 4 + 6 I_{2}$
Find the currents

Short Answer

Expert verified

The values of current are $\frac{11}{13}, \frac{6}{13}, \frac{17}{13}$ respectively.

Step by step solution

Given information

We are given a system of equation

$I_{3} = I_{1} + I_{2} 8 = 4 I_{3} + 6 I_{2} 8 I_{1} = 4 + 6 I_{2}$

Simplify the equations

Substitute equation 1 in equation 2 we get

Equation 2 becomes

$4 (I_{1} + I_{2}) + 6 I_{2} = 8 4 I_{1} + 10 I_{2} = 8 (4)$

Now we multiply equation 4 by -2 and add it to equation 3

We get,

$- 8 I_{1} - 20 I_{2} = - 16 + 8 I_{1} - 6 I_{2} = 4 - 26 I_{2} = - 12$

hence $I_{2} = \frac{12}{26}$

Now we find the remaining values

$4 I_{1} + 10 I_{2} = 8 4 I_{1} + 10 \cdot \frac{6}{13} = 8 I_{1} = \frac{11}{13}$

Now also,

$I_{3} = I_{1} + I_{2} I_{3} = \frac{11}{13} + \frac{6}{13} I_{3} = \frac{17}{13}$

Conclusion

The values of current are $\frac{11}{13}, \frac{6}{13}, \frac{17}{13}$ respectively

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An application of Kirchhoff’s Rules to the circuit shown results in the following system of equations:
$I_{3} = I_{1} + I_{2} 8 = 4 I_{3} + 6 I_{2} 8 I_{1} = 4 + 6 I_{2}$
Find the currents

Short Answer

Step by step solution

Given information

Simplify the equations

Conclusion

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An application of Kirchhoff’s Rules to the circuit shown results in the following system of equations:I3=I1+I28=4I3+6I28I1=4+6I2Find the currents

Short Answer

Step by step solution

Given information

Simplify the equations

Conclusion

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Geometry

Theoretical and Mathematical Physics

Discrete Mathematics

Logic and Functions

Probability and Statistics

Decision Maths

Study anywhere. Anytime. Across all devices.

An application of Kirchhoff’s Rules to the circuit shown results in the following system of equations:
$I_{3} = I_{1} + I_{2} 8 = 4 I_{3} + 6 I_{2} 8 I_{1} = 4 + 6 I_{2}$
Find the currents