Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 11

MATH152 April 2016

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Question A 11
Solve for the loop currents $i_{1}$ and $i_{2}$ in the circuit below.

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
Consider Kirchoff's laws. Keep in mind that both $i_{1}$ and $i_{2}$ contribute to the current through the 1 $\Omega$ resistor.

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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. By Kirchoff's voltage law, the sum of the voltage drops across the left and right loop must be zero. This leads to the equations ${\begin{aligned}2i_{1}+1(i_{1}-i_{2})&=9\\1(i_{1}-i_{2})-3i_{2}&=14\end{aligned}}$ We solve this system using row reduction. This system is represented by the augmented matrix $\left[{\begin{array}{cc|c}3&-1&9\\-1&4&-14\end{array}}\right]$. We add $1/3$ of the first row to the second row: $\left[{\begin{array}{cc|c}3&-1&9\\0&11/3&-11\end{array}}\right]$. Now, we multiply the second row by $3/11$: $\left[{\begin{array}{cc|c}3&-1&9\\0&1&-3\end{array}}\right]$. We add the second row to the first row: $\left[{\begin{array}{cc|c}3&0&6\\0&1&-3\end{array}}\right]$. Finally, we divide the first row by 3: $\left[{\begin{array}{cc|c}1&0&2\\0&1&-3\end{array}}\right]$. Therrefore, we get that $i_{1}=2{\text{ Amperes}}$ and that $i_{2}=-3{\text{ Amperes}}$. We now verify that this answer makes sense. The direction of $i_{2}$ is opposite the direction of the current, because the current is running from high voltage to low voltage. Therefore, we expect that $i_{2}$ should be negative and $i_{1}$ should be positive. This is confirmed by our calculations. Answer: $\color {blue}i_{1}=2{\text{A}},i_{2}=-3{\text{A}}$

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