0 State response of First order circuit

The 0 state response is the response of the circuit in the initial state of 0 (the initial energy storage of the dynamic element is 0), which is caused by the external excitation.

0 State response of RC circuit

At t=0 time, the switch S is closed and the circuit is connected to the DC voltage source us. According to KVL, there are

Ur+uc=us

(kvl∑u=0

The direction of the specified loop is clockwise, the voltage reference direction of R and C is consistent with the direction of the bypass, the voltage front to take "+" number, the US voltage reference direction and the direction of the detour inconsistent, the front to take "-" number.

According to KVL, there are

Ur+uc-us=0

Have

US=UR+UC)

The differential equation of the circuit is obtained by substituting the ur=ri,i=cduc/dt into the

RC (DUC/DT) +uc=us

This equation is a first order linear non-homogeneous equation. Solution of the equation by the special solution of the non-homogeneous UC '
and the corresponding homogeneous equation of the general solution UC ' ' two components, i.e.

Uc=uc ' +uc '

It is not difficult to find a special solution for

UC ' =us

(for first-order linear differential equations Dy/dx+p (x) y=q (x)

General solution of non-homogeneous linear equations

Y=e-∫p (x) dx (∫q (x) e∫p (x) dxdx+c)

Get the c=0 and give a special solution

　　　　Y=e-∫p (x) dx∫q (x) e∫p (x) DXDX

For RC (DUC/DT) +uc=us,uc=e-(T/RC) * * * But there is no need to behave like this for solving this practical problem

Finally, the circuit is stable, C is equivalent to open circuit, the C voltage is equal to the voltage source voltage at both ends, so the special solution is ready, that is, UC ' =us.

And the general solution of rcduc/dt+uc=0 homogeneous equation is

UC ' =ae-t/Tau

Which Τ=RC. So

Uc=us+ae-t/τ

Substituting the initial value, you can obtain

A=-us

and

Uc=us-use-t/τ=us (1-e-t/τ)

I=c (DUC/DT) = (US/R) e-t/τ

Wr= (CUS2) (only 50% charging efficiency)

0 State response of First order circuit