3
\$\begingroup\$

enter image description here

I am not able to solve the above problem. I don't think my equations are wrong but I am not able to solve them. Please help me with this.

The answer given in my workbook is 3.93cost(3t+59.9)

Method A

enter image description here

Method B

enter image description here

This is not a homework problem. I was practicising problems when I came across this one.

\$\endgroup\$
  • \$\begingroup\$ Can you try KCL and KVL with a current source? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 10 '18 at 1:56
  • 1
    \$\begingroup\$ Can you please show some steps. I am not able to understand. \$\endgroup\$ – Nikhil Kashyap Jul 10 '18 at 2:09
  • \$\begingroup\$ Start with Norton transform. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 10 '18 at 2:16
  • 1
    \$\begingroup\$ Again derivative term will cause problems. Please show the equations. \$\endgroup\$ – Nikhil Kashyap Jul 10 '18 at 2:58
2
\$\begingroup\$

While I agree with Chu that you can save yourself some pain by using phasors/Fourier transform, it is possible to solve these equations.

$$\begin{align} &v_x)\quad &\frac{v_s-v_x}{1} + \frac{v_1-v_x}{2} + \frac{1}{3}\frac{d(v-v_x)}{dt} &= 0 \\ &v_1)\quad &\frac{v_x - v_1}{2} - i_1 &= 0 \\ &v)\quad &\frac{1}{3}\frac{d(v_x-v)}{dt} + i_2 - \frac{v}{5} &= 0\\ &tf)\quad &v &= 4v_1\\ & & i_1 &= 4i_2 \end{align}$$

You can eliminate \$\frac{1}{3}\frac{d(v-v_x)}{dt}\$ using the third equation by substituting it in the first equation.

$$\frac{1}{3}\frac{d(v-v_x)}{dt} = i_2 - \frac{v}{5}$$

You can then solve all equations but the third to all unknowns except \$v\$ to find that (I used a CAS like Maxima to solve it)

$$\begin{align} i_1 &= \frac{20v_s-9v}{55} \\ i_2 &= \frac{20v_s-9v}{220} \\ v_1 &= \frac{v}{4} \\ v_x &= \frac{160v_s - 17v}{220} \end{align}$$

which allows us to plug \$i_2\$ and \$v_x\$ back in:

$$\begin{align} \frac{1}{3}\frac{d(v-v_x)}{dt} &= i_2 - \frac{v}{5} \\ &\Downarrow \\ 79\frac{dv}{dt} + 53v &= 80\cos(3t) - 640\sin(3t) \end{align}$$

The homogeneous solution is not important, as we're only interested in the steady-state solution. The particular solution is given by

$$\begin{align} v(t) &= \frac{77960}{29489}\cos(3t) -\frac{7480}{29489}\sin(3t) \\ &= 2.656\cos(3t + 5.48^\circ) \end{align}$$

Now, I'm not sure how you would get the reference answer of your book. When I verified using LTSpice, it seemed to support this answer rather than the one from your book.

Verification using LTSpice

\$\endgroup\$
  • \$\begingroup\$ please check my answer and help me with the last step. \$\endgroup\$ – Nikhil Kashyap Jul 10 '18 at 12:59
  • \$\begingroup\$ Also is it necessary to use that CAS Maxima thing. Can't it be done by hand ? \$\endgroup\$ – Nikhil Kashyap Jul 10 '18 at 13:03
  • \$\begingroup\$ A CAS is not necessary. It's just faster. To avoid people asking what my intermediary steps are, I include the mention. \$\endgroup\$ – Sven B Jul 10 '18 at 13:57
2
\$\begingroup\$

You appear to be doing a transient analysis; this problem calls for complex notation. Represent the capacitor's reactance as: \$\small -j\:X_C=-\large\frac{j}{\omega C}\small =-j\$, let the source voltage be: \$\small V_S=4+j0=4\$, and then use nodal analysis.

\$\endgroup\$
  • \$\begingroup\$ please check my answer and help me with the last step. \$\endgroup\$ – Nikhil Kashyap Jul 10 '18 at 13:03
1
\$\begingroup\$

I calculated v using phasor analysis using y-parameter. Please help me in how to write time domain representation of v.

enter image description here

enter image description here

\$\endgroup\$
  • 1
    \$\begingroup\$ That is correct too. The final cosine would have an amplitude of \$0.664\cdot 4 = 2.656\$, and the phase shift is \$5.48^\circ\$, so you get \$2.656\cdot \cos(3t + 5.48^\circ)\$. \$\endgroup\$ – Sven B Jul 10 '18 at 14:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.