Capacitor discharging in RLC series circuit. Equation (8) general solution for the discharging current,
i(t)=e^(-αt) (A cos⁡(ωt)+B sin⁡(ωt)) (8)
Now we will find the constant A and B using the initial condition assuming the capacitor voltage VDC and current IDC. The initial conditions are taken just before the fault occurs.
At t=0
i(0)=I_DC=A
Then the nominal voltage at the initial condition is equal to to the nominal voltage.
Vc(0)=V_DC
To find B take the derivative of the equation (8) we get
(di(t))/dt=e^(-αt) (-α(A cos⁡(ωt)+B sin⁡(ωt) )+ω(B cos⁡(ωt)-A sin⁡(ωt)) (9)
At t=0
(di(0))/dt=-αA+ωB
The initial rate of change of current is equal to
(di(0))/dt=(V_L (0))/L=-αA+ωB
Using
A=I_DC
B=((V_L (0))/L+αI_DC)/ω
Since the RLC is a series circuit and initial conditions are the same thus the current is also the same across R, L, and C.
(V_L (0))/L=I_L=I_DC
B=(I_DC+αA)/ω (10)
Putting the values of A and B in the equation (8) we get
i(t)=e^(-αt) (I_DC cos⁡(ωt)+(I_DC+αA)/ω sin⁡(ωt)) (11)
For the capacitor
1/C i(0)=V_c (0) and for initial condition i(0)=I_DC so 1/C I_DC=V_DC
“A” becomes equal to
A=I_DC=CV_DC
Expressing the equation in terms of the A and B from equation (10)
I_DC=Bω-Aα
So putting in equation (10)
i(t)=[(Bω-αA) cos⁡(ωt)-(Aω+Bα) sin⁡(ωt)]e^(-αt) (12)Capacitor discharging in RLC series circuit. Equation (8) general solution for the discharging current,
i(t)=e^(-αt) (A cos⁡(ωt)+B sin⁡(ωt)) (8)
Now we will find the constant A and B using the initial condition assuming the capacitor voltage VDC and current IDC. The initial conditions are taken just before the fault occurs.
At t=0
i(0)=I_DC=A
Then the nominal voltage at the initial condition is equal to to the nominal voltage.
Vc(0)=V_DC
To find B take the derivative of the equation (8) we get
(di(t))/dt=e^(-αt) (-α(A cos⁡(ωt)+B sin⁡(ωt) )+ω(B cos⁡(ωt)-A sin⁡(ωt)) (9)
At t=0
(di(0))/dt=-αA+ωB
The initial rate of change of current is equal to
(di(0))/dt=(V_L (0))/L=-αA+ωB
Using
A=I_DC
B=((V_L (0))/L+αI_DC)/ω
Since the RLC is a series circuit and initial conditions are the same thus the current is also the same across R, L, and C.
(V_L (0))/L=I_L=I_DC
B=(I_DC+αA)/ω (10)
Putting the values of A and B in the equation (8) we get
i(t)=e^(-αt) (I_DC cos⁡(ωt)+(I_DC+αA)/ω sin⁡(ωt)) (11)
For the capacitor
1/C i(0)=V_c (0) and for initial condition i(0)=I_DC so 1/C I_DC=V_DC
“A” becomes equal to
A=I_DC=CV_DC
Expressing the equation in terms of the A and B from equation (10)
I_DC=Bω-Aα
So putting in equation (10)
i(t)=[(Bω-αA) cos⁡(ωt)-(Aω+Bα) sin⁡(ωt)]e^(-αt) (12) Capacitor discharging in RLC series circuit. Equation (8) general solution for the discharging current,
i(t)=e^(-αt) (A cos⁡(ωt)+B sin⁡(ωt)) (8)
Now we will find the constant A and B using the initial condition assuming the capacitor voltage VDC and current IDC. The initial conditions are taken just before the fault occurs.
At t=0
i(0)=I_DC=A
Then the nominal voltage at the initial condition is equal to to the nominal voltage.
Vc(0)=V_DC
To find B take the derivative of the equation (8) we get
(di(t))/dt=e^(-αt) (-α(A cos⁡(ωt)+B sin⁡(ωt) )+ω(B cos⁡(ωt)-A sin⁡(ωt)) (9)
At t=0
(di(0))/dt=-αA+ωB
The initial rate of change of current is equal to
(di(0))/dt=(V_L (0))/L=-αA+ωB
Using
A=I_DC
B=((V_L (0))/L+αI_DC)/ω
Since the RLC is a series circuit and initial conditions are the same thus the current is also the same across R, L, and C.
(V_L (0))/L=I_L=I_DC
B=(I_DC+αA)/ω (10)
Putting the values of A and B in the equation (8) we get
i(t)=e^(-αt) (I_DC cos⁡(ωt)+(I_DC+αA)/ω sin⁡(ωt)) (11)
For the capacitor
1/C i(0)=V_c (0) and for initial condition i(0)=I_DC so 1/C I_DC=V_DC
“A” becomes equal to
A=I_DC=CV_DC
Expressing the equation in terms of the A and B from equation (10)
I_DC=Bω-Aα
So putting in equation (10)
i(t)=[(Bω-αA) cos⁡(ωt)-(Aω+Bα) sin⁡(ωt)]e^(-αt) (12) matrix, mathematics, electrical, circuit design MATLAB Answers — New Questions

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