Science:Math Exam Resources/Courses/MATH152/April 2022/Question B5 (a)

MATH152 April 2022

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Question B5 (a)
This problem concerns the two circuits below. The voltage drops are $V_{1},V_{2},E,$ the currents are $I,J_{1},J_{2},i_{1},i_{2}$. The capacitances are $C_{1}=2F$ and $C_{2}=3F,$ the inductance is $L=4H.$ Two circuits. (a) In the left circuit assume that $V_{1},V_{2}$ and $I$ are some given numbers. Use the loop current method to solve for $i_{1},i_{2},E$ in terms of $V_{1},V_{2},I.$

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
The loop currents for this problem, ${\textstyle i_{1}}$ and ${\textstyle i_{2}}$, are already assigned to loops. Recall that if a current source exists in a loop, the loop current must equal the current source.

Hint 2
You will have to use Kirchoff’s voltage law, which says that the sum of voltage changes in a loop must equal zero. Note that that in order to use this with the given circuits, you will also need Ohm’s law: if a current $I$ flows through a resistance $R$, then the voltage drop across the resistance is ${\textstyle V=IR}$.

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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. We are given the two loops to which the currents ${\textstyle i_{1}}$ and ${\textstyle i_{2}}$ correspond, and we can use Kirchoff’s voltage law on these loops to determine equations for ${\textstyle i_{1}}$, ${\textstyle i_{2}}$, and ${\textstyle E}$. In the left loop: ${\textstyle -V_{1}+i_{1}\cdot 1+-V_{2}=0}$ In the right loop: ${\textstyle V_{2}+i_{2}\cdot 2-E=0}$ And, we also know that if there is a current source in a loop, the loop current must equal the current source, so we have ${\textstyle i_{2}=I}$. Rearranging gives us $${\begin{aligned}i_{1}&=V_{1}+V_{2}\\i_{2}&=I\\E&=2I+V_{2}\end{aligned}}.$$