Science:Math Exam Resources/Courses/MATH152/April 2017/Question B 03 (a)

MATH152 April 2017

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Question B 03 (a)
Consider the resistor network below (a) Using the loop current technique as described in the online notes and computer labs, write three linear equations that can be solved for $i_{1}$, $i_{2}$, and $E$. Let the sources $I$ and $V$ be parameters in your equations.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

To clarify the diagram a bit, $i_{1}=J$ is the current through the first loop, $i_{2}$ is the current through the second loop, the strength of the current source is I, and the resistance of the resistors are 1 and 2 ohms. E is the voltage across the current source. To set up the equations, use Kirchoff's voltage and current laws. Write down the voltage drop across each component and ensure that the sum is zero. Make sure that the current at the top node is also equal to zero.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. To clarify the diagram a bit, $i_{1}=J$ is the current through the first loop, $i_{2}$ is the current through the second loop, the strength of the current source is I, and the resistance of the resistors are 1 and 2 ohms. We will use Kirchoff's voltage law on each loop. We will start with the left loop. The voltage drop across the voltage source is -V, the voltage drop accross the current source is E, and the voltage drop across the resistor is $i_{1}(1\Omega )$. We proceed to the right loop. The voltage drop across the current source is -E, and the voltage drop across the resistor is $i_{2}(2\Omega )$. Finally, we look at the node at the top of the circuit. The current at this node must sum to zero. This current is $-i_{1}+i_{2}+I.$ We then get the system of 3 equations: ${\begin{aligned}(1\Omega )i_{1}+E&=V\\-i_{1}+i_{2}&=I\\(2\Omega )i_{2}-E&=0\end{aligned}}$ Writing this in matrix notation (and ignoring the units) we get $\left({\begin{array}{ccc}1&0&1\\-1&1&0\\0&2&-1\\\end{array}}\right)\left({\begin{array}{c}i_{1}\\i_{2}\\E\end{array}}\right)=\left({\begin{array}{c}V\\I\\0\end{array}}\right)$ Answer: $\color {blue}\left({\begin{array}{ccc}1&0&1\\-1&1&0\\0&2&-1\\\end{array}}\right)\left({\begin{array}{c}i_{1}\\i_{2}\\E\end{array}}\right)=\left({\begin{array}{c}V\\I\\0\end{array}}\right)$