Two Inductors - Part 2

My solution is similar to that of @Karan Chatrath, but with some stylistic differences. Recall from the previous problem that when the applied angular frequency is $\frac{1}{\sqrt{LC}}$ , the parallel $LC$ branch has infinite impedance in AC steady-state. Consequently, current only flows in the circuit during an initial transient phase, which eventually dies out. This is why it is reasonable to propose that the energy dissipated in the lamp over infinite time is finite.

The state variables are the inductor currents and the capacitor voltage.

Fundamental equations:

$V_{L1} = L_1 \dot{I}_{L1} \\ V_{L2} = L_2 \dot{I}_{L2} \\ I_C = C \dot{V}_C$

Write the left sides in terms of other state variables.

$V_{S} - R I_{L1} - V_C = L_1 \dot{I}_{L1} \\ V_{C} = L_2 \dot{I}_{L2} \\ I_{L1} - I_{L2} = C \dot{V}_C$

All that remains is to isolate the derivative terms and numerically integrate. The energy dissipated in the lamp is:

$E_{lamp} = \int_0^{\infty} I_{L1}^2 R \, dt$

In practice, all we have to do is integrate over a large finite time span.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61

import math

dt = 10.0**(-5.0)

R = 1.0
L1 = 1.0
L2 = 1.0
C = 1.0

###################################

t = 0.0
count = 0

IL1 = 0.0
IL2 = 0.0
VC = 0.0

ER = 0.0

VS = 10.0*math.cos(t)

IL1d = (VS - R*IL1 - VC)/L1
IL2d = VC/L2
VCd = (IL1 - IL2)/C

while t <= 100.0:

    IL1 = IL1 + IL1d*dt
    IL2 = IL2 + IL2d*dt
    VC = VC + VCd*dt

    VS = 10.0*math.cos(t)

    PR = (IL1**2.0)*R
    ER = ER + PR*dt

    IL1d = (VS - R*IL1 - VC)/L1
    IL2d = VC/L2
    VCd = (IL1 - IL2)/C

    t = t + dt
    count = count + 1

    #if count % 1000 == 0:
        #print t,IL1

###################################

print ""
print ""

print dt
print t
print ER

#>>> 
#1e-05
#100.000000021
#50.0015005846
#>>> 

Let the current through the resistor be $I$ , that through the parallel inductor be $I_2$ and that through the capacitor be (I_1)

Charge on the capacitor is $Q_1$

Circuit equations:

$-V_S + I + \dot{I} + \dot{I}_2=0$ $I_1 + I_2 = I$ $\dot{Q}_1 = I_1$ $Q_1 = \dot{I}_2$

$I(0) = I_1(0) = Q_1(0) = I_2(0) = 0$

Manipulating and rearrangine the differential equations to obtain an ODE governing $I$ yields:

$\frac{d^3I}{dt^3} + \frac{d^2I}{dt^2} + 2\frac{dI}{dt} + I=0$

$I(0) = 0$ $\dot{I}(0) = 10$ $\ddot{I}(0) = -10$

This final ODE is solved numerically. The resulting heat dissipated can be computed as such (also done numerically):

$H = \int_{0}^{\infty} I^2R \ dt$ $H \approx 50$