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\$.
Best Answer
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):
$$v_x(t) =-20e^{-5.6t} $$