327_pdfsam_math 54 differential equation solutions odd

327_pdfsam_math 54 differential equation solutions odd -...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Exercises 5.6 voltage source. We apply Kirchhof’s voltage law, E L + E R + E C = E ( t ), to Loop 1 and Loop 2 to get two equations connecting currents in the network. (Similarly to Example 2 and Problem 11, there is no need to apply Kirchhof’s voltage law to Loop 3 because the resulting equation is just a linear combination oF those For other two loops.) Loop 1: E L + E R =0 0 . 5 dI 1 dt +1 · I 2 dI 1 dt +2 I 2 . (5.37) Loop 2: E L + E C =cos3 t 0 . 5 dI 1 dt + q 3 0 . 5 t dI 1 dt +4 q 3 =2cos3 t. (5.38) Additionally, at joint points, by Kirchhof’s current law, I 1 + I 2 + I 3 ⇒− I 1 + I 2 + dq 3 dt . (5.39) Putting (5.37)–(5.39) together yields the Following system:
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.
Ask a homework question - tutors are online