project final.docx

Let’s consider the following circuit:

R = R1 = R2 = R3 = 25 Ω

C = 190 µF

L = 125 mH

E = 100 V

where switches S1 and S2 are closed at t = 0 and t = τ (where τ is the time constant of the LR circuit), according the following behavior:

S1=OffS2=Offdla t<0S1=OnS2=Offdla 0 ≤t<τS1=OnS2=Ondla t ≥ τ

The goal is to find the solution of the above circuit, basically the current iL(t) through the inductor and the voltage v(t)C on the capacitor. Knowing v(t)C we can easily find the current i1(t). We will obtain the solution with 2 different methods:

-         classic method

-         Laplace method

Classic method

·        For t < 0 there is no closed net. As a consequence there is no current flowing in the circuit:

iL(t) = iC(t) = 0

·        For 0 ≤ t < τ switch S1 is On. There is a closed net representing a RL circuit. Applying Kirchoff equations we have:

E-R3i3-Ldi1dt-R1i1=0   i2=0  ===>  i1=i2=iL  

We obtain a First Order Linear Differential equation with initial conditions:

LdiLdt+2RiL=EiL0=0

To find the solution, first we consider the associated homogenous equation and the characteristic equation:

LdiLdt+2RiL=0      è     Lλ+2R=0                  è    λ=-2RL

so

iLt=Ceλt=Ce-2RLt

To find a particular solution we assume that the derivative of the current iL is zero:

2RiL=E   è     iL=E2R

So:

iLt=Ce-2RLt+E2R

To calculate the value of constant C, we use the initial condition:

0=C+E2R è    C=-E2R

Finally the solution of the differential equation is:

iLt=E2R 1-e-2RLt      for    0 ≤ t ≤ τ

where τ is:

τ=1λ=L2R

Replacing the values for R, L and E we have:

iLt=2 1-e-400t      for   0 ≤ t ≤ 2,5 msec

·        For  t > τ switch S1 and S2 are On. Applying Kirchoff equations we have:

E-R3i3-LdiLdt-R1iL=0R3i3-vC-R2CdvCdt=0iL=CdvCdt+i3

Considering the current iL(t), we obtain a Second Order Linear Differential equation

2RLCd2iLdt2+5R2C+LdiLdt+2RiL=E

The initial conditions can be obtained by the solution:

iLt=E2R 1-e-2RLt      dla   0 ≤ t ≤ τ

when t = τ. So we obtain:

iLτ=E2R 1-e-2RLτ=E2R 1-1e=1,26 A

and:

iL'τ=diL(t)dtt=τ =EL e-2RLtt=τ=EL 1e=295 A/sec

Finally we obtain a Second Order Linear Differential equation with initial conditions

2RLCd2iLdt2+3R2C+LdiLdt+2RiL=EiLτ=E2R 1-1e    iL'τ=EL 1e

To find the solution, first we consider the associated homogenous equation and the characteristic equation:

2RLCd2iLdt2+5R2C+LdiLdt+2RiL=0   è     2RLCλ2+5R2C+Lλ+2R=0     

Substituting values for R, L and C we have:

1,1875*10-3λ2+481,25*10-3λ+50=0

and solving the second order equation, we obtain two complex solutions:

λ1,2=α±jβ=-202,6±j32,3

so:

iLt=eαtc1cosβt+c2sinβt=e-202,6tc1cos32,3t+c2sin32,3t

To find a particular solution we assume that the derivative of the current iL is zero:

2RiL=E   è     iL=E2R

So:

iLt=e-202,6tc1cos32,3t+c2sin32,3t+E2R

To calculate the value of constant c1 and c2, we use the initial condition on iLτ:

1,26 = e-202,6*2,5*10-3c1cos32,3*2,5*10-3+c2sin32,3*2,5*10-3+2

-1,226 = c1cos0,08075+c2sin0,08075

0,81565*c1+0,00115*c2=-1

and the initial condition on iL'τ:

295=-202,6*e-202,6*2,5*10-3c1cos32,3*2,5*10-3+c2sin32,3*2,5*10-3+e-202,6*2,5*10-3- 32,3*c1sin32,3*2,5*10-3+32,3*c2cos32,3*2,5*10-3

122,2474*c1+19,6623*c2...