EE110_W08_HW 1 Solution - Q2 For this circuit problem nodal...

Info icon This preview shows pages 1–3. Sign up to view the full content.

Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

Q2: For this circuit problem, nodal analysis will require 3 simultaneous nodal equations, then subtraction/ division steps to obtain the desired currents. Mesh analysis requires 1 mesh equation, 1 supermesh equation, 2 simple KCL equations and one subtraction step to determine the currents. If either technique has an edge in this situation, it’s probably mesh analysis. Thus, define four clockwise mesh equations: i a in the bottom left mesh, i b in the top left mesh, i c in the top right mesh, and i d in the bottom right mesh. At the a , b , c supermesh: -100 + 6 i a + 20 i b + 4 i c + 10 i c – 10 i d = 0 [1] Mesh d: 100 + 10 id – 10 i c + 24 i d = 0 [2] KCL: - i a + i b = 2 [3] and - i b + i c = 3 i 3 = 3 i a [4] Collecting terms & simplifying, 6 i a + 20 i b + 14 i c – 10 i d = 100 [1] -10 i c + 34 i d = -100 [2] - i a + i b = 2 [3] -3 i a i b + i c = 0 [4] Solving, i a = 0.1206 A, i b = 2.121 A, i c = 2.482 A, and i d = -2.211 A. Thus, i 3 = i a = 120.6 mA and i 10 = i c i d = 4.693 A. Q3: We need to find the Thévenin equivalent resistance of the circuit connected to R L , so
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.
  • Winter '08
  • Gupta
  • Mesh Analysis, Harshad number

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern