# Proper way to solve first order circuit with voltage dependent voltage source

Original question from the text book with answer provided:

My attempt at solving this problem using mesh current loops:

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\$$.

• +1 for all the MathJAX effort. On EE.SE use $ for inline MathJAX. You have only used the  in the last sentence. – Transistor Mar 30 '19 at 22:03 • As stated on the problem,$i(0) = 5$A; so your solution has wrong sign. – Dirceu Rodrigues Jr Mar 31 '19 at 3:47 • 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 – user97662 Mar 31 '19 at 6:57

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):

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

• 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. – user97662 Mar 31 '19 at 6:56
• ... it also gave the unit of current as [V] in the answer. – Chu Mar 31 '19 at 8:43