Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 25

MATH152 April 2016

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Question A 25
Questions A25 and A26 concern the circuits above. In the left diagram, $I$ is the current through the current source and $E$ is the voltage across it, to be determined. A25: For the left circuit above with two voltage sources and one current source write the one linear equation that matches the loop currents to the current source.

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
Recall from Kirchhoff’s Laws that for any node, current in equals current out for any node, current in equals current out.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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. We follow the hint and apply Kirchoff's current law to the top node. The current coming into the node from the left is is $i_{1}$, the current entering the node from the bottom is $I$, and the current leaving the node from the right is $i_{2}$. We therefore have $i_{1}+I=i_{2}$. Solving for $I$ gives $I=i_{2}-i_{1}$. Answer: $\color {blue}I=i_{2}-i_{1}$