Current Sink

Semi-analytical solution:

Consider the differential equation governing the current flow through the inductor:

$\frac{dI}{dt} = 10(I_S - I)$

This implies:

$\frac{dI}{dt} + 10I = 10I_S$

$\frac{d}{dt}\left(e^{10t} I\right) = 10e^{10t}I_S$

$d\left(e^{10t} I\right) = 10e^{10t}I_S dt$

When $t= 0$ then $I = 0$ . Then:

$e^{10t} I = \int_{0}^{t}10e^{10t}I_S dt$

Solving this monster on the right-hand side is very tedious. Here is my attempt.

It so happens that: $t = 0.15 = \frac{18\pi}{\omega}$

Let $p = \frac{\pi}{\omega}$

The integral on the right-hand side can be written as:

$e^{10t} I = \int_{0}^{p}f(s) ds + \int_{2p}^{3p}f(s) ds + + \int_{4p}^{5p}f(s) ds + + \int_{6p}^{7p}f(s) ds + \dots + \int_{16p}^{17p}f(s) ds + \int_{18p}^{t}f(s) ds$

Where $f(s) = 50e^{10s}\sin(\omega s)$ . In the integral above, $t>= 0.15$ .

Let:

$S = \int_{0}^{p}f(s) ds + \int_{2p}^{3p}f(s) ds + + \int_{4p}^{5p}f(s) ds + + \int_{6p}^{7p}f(s) ds + \dots + \int_{16p}^{17p}f(s) ds$

Then:

$e^{10t} I = S + \int_{18p}^{t}f(s) ds$

The current through the inductor is:

$I = e^{-10t}\left(S + \int_{18p}^{t}f(s) ds\right)$

Solving for $S$ was done using a Wolfram Alpha like tool. The integrals are very doable but too cumbersome.

Finally:

$I_R = I_S - e^{-10t}\left(S + \int_{18p}^{t}f(s) ds\right)$

$I_R$ is a function of $t$ such that $t\ge 0.15$ .

From here, one can take a more analytical approach of finding local maxima by differentiating with respect to $t$ , equating it to zero, and searching for the solution that lies between $t = 0.15$ and $t = 0.16$ . I have not done this. Instead, I have just plotted this time function within the specified interval and extracted the maxima.

So its not entirely analytical. I have made use of a computational tool at the end.

Numerical Approach:

Governing equations of the circuit are:

$\frac{dI_L}{dt} = 10(I_S - I_L)$ $I_S = I_R + I_L$ $I_L(0) = 0$

The following plot illustrates the current response of the circuit:

Simulation code is as follows:

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import math

#constants
w = 120*3.141592653589793

# Initialisations:
t = 0.0
dt = 10.0**(-6.0)
IL = 0.0
IRmax = 0.0

while t <= 0.2:

    # Source Current:
    IS = 5*math.sin(w*t)
    if IS <= 0:
        IS = 0

    # Euler integration:     
    IL = IL +  dt*(10*(IS -IL))

    # Current through resistor:
    IR = IS - IL

    # Computing max current through resistor in specified time interval:
    if t >= 0.15 and t <= 0.16:
        if IR >= IRmax:
            IRmax = IR

    t = t + dt

print(IRmax)
# Result: IRmax = 3.733878133739374 Amps