# Transient processes of the second order, example of solution – Classical and operator methods

**Given**

R=5 ohms;

L=0,01 H;

C=2·10-4 F;

E=98 V;

**Find**

i

_{C}(t)—?(classical method)

i

_{C}(t)—?(operator method — transient components)

i

_{C}(t)—?(diagram)

## Solution

Circuit arrangement:## Classical method.

The calculation of the characteristic impedance of the circuit p.We find the input impedance of the circuit and equate it to zero. By finding the input impedance it is assumed that all switching occurred, the electromotive force is replaced with the short circuit, and current sources are replaced with the break. Then any point of the scheme is transformed into two terminals and we find concerning them the circuit resistance:

The type of the transient process at complex-conjugate roots is defined by the formula:

We find final values of the required parameters. We construct an equivalent diagram after long time. The key worked, all

**transient processes**are complete. At DC the inductor is equal to the short circuit, the capacitor is equal to the break:

Final value of the unknown current:

i

_{c уст}=0 А Analysis of the condition before switching for finding the conserved quantities: the voltage across the capacitor and current through the inductor.

At DC the inductor is equal to the short circuit; the capacitor is equal to the break. Circuit arrangement before switching:

Analysis of the initial conditions of the zero order. We consider the time right after the switching.

It is possible to write down according to the laws of switching:

I

_{Lafter switching0}=I

_{L before switching}=6,53 А;

U

_{Cafter switching0}=U

_{C before switching}=65,33 V;

Тогда:

I

_{R after switching0}=U

_{C after switching0}/R=65,33/5=13,07 А;

I

_{C after switching0}=I

_{L after switching0}-I

_{R after switching0}=6,53-13,07=-6,53 А;

U

_{L after switching0}=E-RI

_{L after switching0}-U

_{C after switching0}=98-5·6,53-65,33 =0 V.

Analysis of initial conditions of the first order (derivatives from initial values). We consider the time right after the switching. Since the electromotive force is constant and equivalent diagram is constructed for derived values, so the source will be absent.

Definition of the independent constants from initial conditions. We write down the formula for the current through the capacitor for initial time. We write down also the formula for the current derivative through the capacitor in initial time. We solve the system of equations and find unknown A and φ. i

_{C}(0)=-6,53 А; iC'(0)=6530 А⁄s; i

_{C уст}=0 А;

From the second equation follows:

## Operator method.

We construct the operator equivalent circuit:We make Laplace transformation for electromotive force sources:

E(t)=98;

E(p)=98/p;

We take the initial values for the current through the inductor and voltage across the capacitor from the previous calculation:

I

_{L}(0)=6,53 А;

U

_{C}(0)=65,33 V;

We find the potential of point a by the nodal pair method.

We mark the non-grounded node "a", then:

Y·U

_{a}=I;

U

_{a}=I/Y,where I — entering currents in a node, and Y — conductance of branches.

Using the formula of the supplementary angle:

According to the task it is necessary to find only free current components, but because its final value is equal to zero, then the total current is received. The characteristic of the capacitive currents coincides with the received characteristic by a classical method that confirms the correctness of calculations::