198. A $\mathrm{K}$ source with an electromotive force (emf) equal to $e(t)$ is connected to a circuit consisting of a coil with inductance $L$, an ohmic resistance $R$, and a capacitance $C$ connected in series (Fig. 206). Find the current $i$ in the circuit as a function of time $t$, if at the initial moment the current in the circuit and the charge on the capacitor are zero.
Solution. $1^{0}$. According to Kirchhoff's law, the electromotive force in the circuit is equal to the sum of the voltage drops across the inductance, resistance, and capacitance:

$$
e(t)=U_{L}+U_{R}+U_{c}
$$

which are related to the current by the relations

$$
U_{L}=L \frac{d i}{d t} ; U_{R}=R i ; U_{c}=\frac{1}{C} \int_{0}^{1} i(t) d t
$$

Note that the last equality is derived from the relationship between the current and the charge on the capacitor: $i=\frac{d q}{d t}$. 

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-402.jpg?height=215&width=402&top_left_y=1598&top_left_x=737)

Fig. 206
where $q=\int_{0}^{t} i(t) d t+q_{0}$, and since $U_{C}=\frac{q}{C}$, we have $U_{C}=\frac{1}{C} \int_{0}^{t} i(t) d t+\frac{q_{0}}{C}$. In this problem, $q_{0}=0$ by condition.

Thus, we obtain the equation

$$
e(t)=L \frac{d i}{d t}+R_{i}+\frac{1}{C} \int_{0}^{t} i(t) d t
$$

$2^{0}$. This equation is an integro-differential equation. Differentiating it with respect to $t$, we obtain a linear second-order differential equation with constant coefficients:

$$
L \frac{d^{2} i}{d t^{2}}+R \frac{d i}{d t}+\frac{1}{C} i=\frac{d e}{d t}\left(\frac{d i}{d t}=i^{\prime}, \frac{d^{2} i}{d t^{2}}=i^{\prime \prime}\right)
$$

Here, two cases are possible: 1) $e(t)=$ const; 2) $e(t)=E \sin \omega t$.

Consider the first case. Here $\frac{d e}{d l}=0$. Then we obtain the homogeneous equation

$$
\frac{d^{2} i}{d t^{2}}+\frac{R}{L} \cdot \frac{d i}{d t}+\frac{1}{L C} \cdot i=0
$$

The characteristic equation

$$
r^{2}+\frac{R}{L} r+\frac{1}{L C}=0
$$

has roots

$$
r_{1.2}=-\frac{R}{2 L} \pm \sqrt{\frac{R^{2}}{4 L^{2}}-\frac{1}{L C}}=-\frac{R}{2 L} \pm \sqrt{\frac{R^{2} C-4 L}{4 L^{2} C}}
$$

If $R^{2} C-4 L \geqslant 0$, then the roots of the characteristic equation are real and the general solution is a non-periodic function. Accordingly, the current is aperiodic, and no electrical oscillations will occur in the circuit.

If $R^{2} C-4 L<0$, then the general solution

$$
i=e^{-\delta t} \cdot\left(C_{1} \cos \omega t+C_{2} \sin \omega t\right)
$$

where $\delta=\frac{R}{2 L}, \omega^{2}=\frac{1}{L C}-\frac{R^{2}}{4 L^{2}}$, defines electrical oscillations.

$3^{0}$. Since $\left.L \frac{d i}{d t}\right|_{t=0}=E$, then $\left.\frac{d i}{d t}\right|_{t=0}=\frac{E}{L}$. Therefore, the initial conditions can be written as

$$
\left.i\right|_{t=0}=0 ;\left.\frac{d i}{d t}\right|_{t=0}=\frac{E}{L}
$$

Differentiating $i$ with respect to $t$, we have

$$
\frac{d i}{d t}=e^{-\delta t} \cdot\left(-\delta\left(C_{1} \cos \omega_{1} t+C_{2} \sin \omega_{1} t\right)+\omega\left(-C_{1} \sin \omega_{1} t+C_{2} \cos \omega_{1} t\right)\right)
$$

Substituting $t=0$ into the expressions for $i$ and $\frac{d i}{d t}$, we get $0=C_{1}, \frac{E}{L}=$ $=-\delta C_{1}+\omega_{1} C_{2}$, from which $C_{1}=0$ and $C_{2}=\frac{E}{L \omega}$.

Thus, the particular solution of the equation takes the form

$$
i=\frac{E}{L \omega_{1}} \cdot e^{-\delta t} \cdot \sin \omega_{1} t
$$

Remark. In the second case, a linear non-homogeneous second-order differential equation with constant coefficients is obtained

$$
L \frac{d^{2} i}{d t^{2}}+R \frac{d i}{d t}+\frac{1}{C} i=E \omega \cos \omega t
$$

which is not considered in this manual.

## Questions and problems for note-taking