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

MATH152 April 2022

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Question B5 (b)
In the right circuit the differential equations are $${\frac {dV_{1}}{dt}}=-{\frac {J_{1}}{C_{1}}},\ {\frac {dV_{2}}{dt}}=-{\frac {J_{2}}{C_{2}}},\ {\frac {dI}{dt}}=-{\frac {E}{L}}.$$ Find the second differential equation ${\textstyle {\frac {dV_{2}}{dt}}=...}$. The equation should only involve ${\textstyle I,V_{1},V_{2}}$.

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
Kirchoff’s current law states that the sum of the currents entering a junction must equal the sum of the currents exiting a junction. Can you use this and Kirchoff’s voltage law to write ${\textstyle J_{2}}$ in terms of ${\textstyle V_{1}}$, ${\textstyle V_{2}}$, and ${\textstyle I}$?

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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 ${\textstyle C_{2}=3F}$ in the question statement, so what remains is to write ${\textstyle J_{2}}$ in terms of ${\textstyle V_{1}}$, ${\textstyle V_{2}}$, and ${\textstyle I}$. First use Kirchoff’s current law on the upper middle junction: ${\textstyle J_{1}=J_{2}+I}$. Rearranging gives ${\textstyle J_{2}=J_{1}-I}$. Now, to determine ${\textstyle J_{1}}$, we use Kirchoff’s voltage law on the left loop: ${\textstyle -V_{1}+J_{1}\cdot 1-V_{2}=0}$. Rearranging this and substituting it back into the equation for ${\textstyle J_{2}}$ gives ${\textstyle J_{2}=V_{1}+V_{2}-I}$. Thus, the second differential equation is $${\frac {dV_{2}}{dt}}=-{\frac {1}{3}}(V_{1}+V_{2}-I).$$