2
\$\begingroup\$

Original question from the text book with answer provided: enter image description here

My attempt at solving this problem using mesh current loops: enter image description here

For loop 1: $$V_H + V_x + 1(i_1 - i_2) + 2V_x = 0$$

For loop 2: $$-2V_x + 1(i_2 - i_1) + 4 i_2 = 0$$

where

\begin{align} i &= -i_1\\ V_H &= L\frac{d_{i1}}{d_t} = 2\frac{di_1}{dt}\\ V_x &= i_1 4\\ i(0) &= 5A \implies i_1(0) = -5A\\ \end{align}

Simplify loop 2: \begin{aligned} -2(i_14)+i_2-i_1+4i_2&=0\\ -8i_1+i_2-i_1+4i_2&=0\\ -9i_1+5i_2&=0\\ 5i_2&=9i_1\\ i_2&=\frac{9}{5}i_1\\ \end{aligned}

Simplify loop 1: \begin{aligned} L\frac{di_1}{dt}+4i_1+i_1-i_2+24i_1&=0\\ 2\frac{di_1}{dt}+4i_1+i_1-i_2+8i_1&=0\\ 2\frac{di_1}{dt}+12i_1+i_1-i_2&=0\\ 2\frac{di_1}{dt}+13i_1&=i_2\\ 2\frac{di_1}{dt}+13i_1&=\frac{9}{5}i_1\\ 2\frac{di_1}{dt}&=\frac{9}{5}i_1-13i_1\\ 2\frac{di_1}{dt}&=\frac{9}{5}i_1-\frac{65}{5}i_1\\ 2\frac{di_1}{dt}&=-\frac{56}{5}i_1\\ \frac{di_1}{dt}&=-\frac{56}{10}i_1\\ \frac{di_1}{dt}&=-5.6i_1\\ \frac{di_1}{i_1}&=-5.6dt\\ \int\frac{di_1}{i_1}&=\int-5.6dt\\ \ln{i_1}&=-5.6t+A\\ e^{\ln{i_1}}&=Ae^{-5.6t}\\ i_1(t)&=Ae^{-5.6t} \end{aligned}

Final answer: \begin{align} i(t)&=-5e^{-5.6t}\\ V_x(t)&=20e^{-5.6t} \end{align}

I have tried to solver it for a few times, but I don't get how the book has \$\tau\$ to be \$1/4\$.

\$\endgroup\$
3
  • 1
    \$\begingroup\$ +1 for all the MathJAX effort. On EE.SE use \$ for inline MathJAX. You have only used the $ in the last sentence. \$\endgroup\$ – Transistor Mar 30 '19 at 22:03
  • \$\begingroup\$ As stated on the problem, \$i(0) = 5\$ A; so your solution has wrong sign. \$\endgroup\$ – Dirceu Rodrigues Jr Mar 31 '19 at 3:47
  • \$\begingroup\$ i realized the sign was wrong, but my main concern is the tau, the book shows a 1/\$\tau\$ = 4, where i got 5.6 \$\endgroup\$ – user97662 Mar 31 '19 at 6:57
1
\$\begingroup\$

My solution: Consider \$v_L\$ as the voltage on inductor (in accordance with the direction of \$i\$) and \$v_A\$ as the voltage on upper node (called here node \$A\$). Applying the KCL on this node:

$$ \frac{v_A-v_L}{4}+ \frac{v_A-2v_x}{1} + \frac{v_A}{4}=0 $$

If $$ \left\{\begin{matrix}v_L=2 \frac{\mathrm{d}i }{\mathrm{d} t} \\v_x=-4i \end{matrix}\right. $$

Then

$$ \frac{\mathrm{d}i }{\mathrm{d} t}-16i-3v_a=0 $$

Replacing \$ v_a=2\frac{\mathrm{d}i }{\mathrm{d} t}+4i \$ leads to:

$$ 5\frac{\mathrm{d}i }{\mathrm{d} t}+28i=0 $$

So, the current \$i(t)\$ is given by (in Ampere): $$i(t) = 5e^{-5.6t} A$$

Remebering that \$ v_x=-4i \$, this voltage is given by (in Volt):

$$v_x(t) =-20e^{-5.6t} $$

\$\endgroup\$
2
  • \$\begingroup\$ looks like the answer from the book is wrong then, it gave a tau of -0.25. hmmm, guess I am not crazy after all. \$\endgroup\$ – user97662 Mar 31 '19 at 6:56
  • \$\begingroup\$ ... it also gave the unit of current as [V] in the answer. \$\endgroup\$ – Chu Mar 31 '19 at 8:43

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.