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


\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\$.

  • 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, 2019 at 22:03
  • \$\begingroup\$ As stated on the problem, \$i(0) = 5\$ A; so your solution has wrong sign. \$\endgroup\$ Mar 31, 2019 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, 2019 at 6:57

1 Answer 1


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


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

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):

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

  • \$\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, 2019 at 6:56
  • \$\begingroup\$ ... it also gave the unit of current as [V] in the answer. \$\endgroup\$
    – Chu
    Mar 31, 2019 at 8:43

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