# Analysis of a circuit with transformer, R and C components 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 Method B This is not a homework problem. I was practicising problems when I came across this one.

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

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

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.

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

I calculated v using phasor analysis using y-parameter. Please help me in how to write time domain representation of v.  • 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 $2.656\cdot \cos(3t + 5.48^\circ)$. – Sven B Jul 10 '18 at 14:48