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I'm doing a practice problem and I'm unsure how my approach did not receive the same result as the solution. Any help would be appreciated. Below is the Figure along with the solution as well as my hand written approach. I hope it is readable. Thank you..

Figure

My attempted solution

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  • \$\begingroup\$ one error I spotted is that \$I_3=-10\$ \$\endgroup\$
    – obataku
    Commented Dec 8, 2014 at 2:33

2 Answers 2

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You wrote your \$I_3\$ equation for the third mesh incorrectly; it should read \$I_3=\color{red}-10\angle0^\circ\,\text{A}\$. Solving this system of equations yields \$I_1\approx-2.4805-5.4297j\approx5.9695\angle-114.553^\circ\$ and so \$I_0=-I_1\approx5.9695\angle65.447^\circ\$. Note that \$I_0\$ is a phasor; for the time-varying instantaneous current we write \$i_0(t)=\Re\{I_0e^{i\omega t}\}\approx5.9695\cos(\omega t+65.447^\circ)\$.

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enter image description here

Sir first I used source transformation to make it easy to find the current in loop two .. I don't know if you can read my solution clear but I got it right..

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    \$\begingroup\$ I have marked your answer up. However first you need to go to the help section and take the tour to understand how to use the site. Next you need to understand how to use the maths markup language here to give clear answers of this type. In the meantime it would be useful if you could do some improvement to the photographs of your notes to make them easier to read. \$\endgroup\$
    – RoyC
    Commented Mar 8, 2018 at 17:36

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