Given:
Series RL circuit for example 7
V
s
(
t
)
=
120
2
c
o
s
(
377
t
+
120
∘
)
{\displaystyle V_{s}(t)=120{\sqrt {2}}cos(377t+120^{\circ })}
Prior Work calculating Steady State/Particular Solution
edit
Have already found the steady state/Particular solution:
i
(
t
)
=
I
m
c
o
s
(
ω
t
+
α
)
{\displaystyle i(t)=I_{m}cos(\omega t+\alpha )}
Where:
I
m
=
V
m
R
2
+
(
L
ω
)
2
{\displaystyle I_{m}={\frac {V_{m}}{\sqrt {R^{2}+(L\omega )^{2}}}}}
α
=
ϕ
−
∠
a
r
c
t
a
n
(
L
ω
R
)
{\displaystyle \alpha =\phi -\angle arctan({\frac {L\omega }{R}})}
Or numerically:
i
(
t
)
s
P
=
15.9
c
o
s
(
377
t
+
1.73
)
{\displaystyle i(t)_{s_{P}}=15.9cos(377t+1.73)}
Calculating the transient/Homogeneous Solution
edit
Need to find the transient/Homogeneous Solution to:
R
∗
i
(
t
)
+
L
∗
d
d
t
i
(
t
)
=
0
{\displaystyle R*i(t)+L*{d \over dt}i(t)=0}
There is no VS ... this makes the homogeneous solution easy!
Guess:
i
(
t
)
s
H
=
A
∗
e
−
t
τ
{\displaystyle i(t)_{s_{H}}=A*e^{\frac {-t}{\tau }}}
Finding the time constant:
R
∗
A
∗
e
−
t
τ
+
L
A
−
τ
e
−
t
τ
=
0
{\displaystyle R*A*e^{\frac {-t}{\tau }}+{\frac {LA}{-\tau }}e^{\frac {-t}{\tau }}=0}
R
−
L
τ
=
0
{\displaystyle R-{\frac {L}{\tau }}=0}
τ
=
L
/
R
=
.001
{\displaystyle \tau =L/R=.001}
Now find see if it works:
R
A
e
−
t
L
/
R
+
L
A
(
−
R
L
)
∗
e
−
t
L
/
R
=
?
?
?
0
{\displaystyle RAe^{\frac {-t}{L/R}}+LA(-{\frac {R}{L}})*e^{\frac {-t}{L/R}}{\overset {\underset {\mathrm {???} }{}}{=}}0}
divide through by A, cancel L's
R
e
−
t
L
/
R
−
R
e
−
t
L
/
R
=
?
?
?
0
{\displaystyle Re^{\frac {-t}{L/R}}-Re^{\frac {-t}{L/R}}{\overset {\underset {\mathrm {???} }{}}{=}}0}
It works, therefore it must be the solution:
i
(
t
)
=
i
(
t
)
s
P
+
i
(
t
)
s
H
=
599
cos
(
377
t
+
3.30
)
+
A
e
−
t
0.0001
{\displaystyle i(t)=i(t)_{s_{P}}+i(t)_{s_{H}}=599\cos(377t+3.30)+Ae^{\frac {-t}{0.0001}}}
Now must find the initial conditions.
Determining the Constants
edit
Finding two initial conditions
edit
mupad and matlab code to find constants A and C .. code
Two equations are necessary to find A and C.
Initially the current through the inductor and the entire circuit is going to be zero:
i
(
0
−
)
=
0
{\displaystyle i(0_{-})=0}
, thus
i
(
0
+
)
=
0
{\displaystyle i(0_{+})=0}
.
This means that setting t=0, have one equation:
i
(
t
)
s
=
15.9
c
o
s
(
377
t
+
1.73
)
+
A
∗
e
−
t
.001
+
C
=
0
{\displaystyle i(t)_{s}=15.9cos(377t+1.73)+A*e^{\frac {-t}{.001}}+C=0}
Evaluating this at t=0:
A
+
C
−
2.58
=
0
{\displaystyle A+C-2.58=0}
The second equation comes from the loop:
v
r
(
t
)
+
v
L
(
t
)
−
V
s
=
0
{\displaystyle v_{r}(t)+v_{L}(t)-V_{s}=0}
R
∗
i
(
t
)
s
+
L
∗
d
i
(
t
)
s
d
t
−
V
s
=
0
{\displaystyle R*i(t)_{s}+L*{di(t)_{s} \over dt}-V_{s}=0}
Substituting for i(t)S and V(t)S , taking the differential and then evaluating at t=0, get:
1.86
∗
10
−
9
−
10.0
∗
C
=
0
{\displaystyle 1.86*10^{-9}-10.0*C=0}
So solving get:
C
=
0
{\displaystyle C=0}
A
=
2.5781
{\displaystyle A=2.5781}
i
(
t
)
=
15.9
c
o
s
(
377
t
+
1.73
)
+
2.58
∗
e
−
t
.001
{\displaystyle i(t)=15.9cos(377t+1.73)+2.58*e^{\frac {-t}{.001}}}
This agrees with the Laplace solution and simulation.