Question

Show that the linear system given in (18) describes the currents ${\mathbf{i}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$

and ${\mathbf{i}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$in the network shown in Figure 3.3.4. [Hint: $\frac{\mathbf{dq}}{\mathbf{dt}}\mathbf{=}{\mathbf{i}}_{\mathbf{3}}$]

Short answer

The given result is proved.

Step-by-step solution

Given Information

Draw the given diagram as:

Determining the current

We have an electrical network as illustrated below, and we must show that the differential equations that describe the currents$i\left(t\right)$and $i_3\left(t\right)$are equal.

$\begin{array}{c}L\frac{d{i}_1}{dt}+R{i}_2=E\left(t\right)\\ RC{i}_2+i_1-i_2=0\end{array}$

as in the following method:

First, because the current $i_1$of the network splits into $i_2$and $i_3$at point c, we can use Kirchhoff's first law to get at

${\mathrm{i}}_1={\mathrm{i}}_2+{\mathrm{i}}_3$ (1)

Second, for each loop of the network, we can use Kirchhoff's second law as follows: We can use abcd for the loop.

$\mathrm{E}\left(\mathrm{t}\right)=\mathrm{L}\frac{{\mathrm{di}}_1}{\mathrm{dt}}+{\mathrm{Ri}}_2$ (Verified) (2)

We can use abfe for the loop.

$E\left(t\right)=L\frac{d{i}_1}{dt}+\frac{1}{C}\frac{dq}{dt}$

that is,

$E\left(t\right)=L\frac{d{i}_1}{dt}+\frac{1}{C}{i}_3$ (3)

There is, $\frac{dq}{dt}=i_3$

We may now have $\mathrm{E}\left(\mathrm{t}\right)=\mathrm{E}\left(\mathrm{t}\right)$for equations (2) and (3) because we have

$\mathrm{E}\left(\mathrm{t}\right)=\mathrm{E}\left(\mathrm{t}\right)$

$\begin{array}{c}L\frac{d{i}_1}{dt}+R{i}_2=L\frac{d{i}_1}{dt}+\frac{1}{C}{i}_3\\ R{i}_2=\frac{1}{C}{i}_3\\ R{i}_2-\frac{1}{C}{i}_3=0\dots \left(4\right)\end{array}$

After that, we may solve the two equations (1) and (4) by substituting the value ${\mathrm{i}}_3$for each other.

$R{i}_2-\frac{1}{C}\left(i_1-i_2\right)=0\times C$

$RC{i}_2-i_1+i_2=0$ ... (Verified)

Determining the Result

To achieve the desired results, apply Kirchhoff's first and second laws, as well as some simplification.