Capacitor Charge Problem

Here is a question shared by Neeraj.

I have included in the image the analysis necessary to determine the steady state capacitor charge magnitude. The Python code below uses numerical integration to compute the evolution of the capacitor charge over time. Within about 20 microseconds, the code solution converges to the known steady state solution.

import math

dt = 10.0**(-9.0)

u = 10.0**(-6.0)

C = 3.0*u

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

t = 0.0
count = 0

VC = 0.0

I1 = (0.0 - (10.0+VC))/2.0
I2 = (0.0 - (2.0+VC))/4.0
I3 = (10.0 - (2.0+VC))/2.0

IC = I1 + I2 + I3
# IC = C*VCd

VCd = IC/C

while t <= 30.0*u:

    VC = VC + VCd*dt

    I1 = (0.0 - (10.0+VC))/2.0
    I2 = (0.0 - (2.0+VC))/4.0
    I3 = (10.0 - (2.0+VC))/2.0

    IC = I1 + I2 + I3
    # IC = C*VCd

    VCd = IC/C

    Q = C*VC

    t = t + dt
    count = count + 1

    if count%10 == 0:
        print t,(abs(Q)/u)

#ElectricityAndMagnetism

Easy Math Editor

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Comments

@Lil Doug

Steven Chase - 10 months, 1 week ago

Yeah Very nice, I wanted this method only. Maybe this is long but its crazy.
Thank you so much.

Talulah Riley - 10 months, 1 week ago

You're welcome. What is long?

Steven Chase - 10 months, 1 week ago

@Steven Chase The code is long for me. Maybe for you not because you are a experienced LPE.
Thanks again .

Talulah Riley - 10 months, 1 week ago

@Talulah Riley – Yeah, I think it took me about 3 minutes

Steven Chase - 10 months, 1 week ago

@Steven Chase – @Steven Chase it will take me minimum 10 min to type this code. You are a legend ,smart and stud.

Talulah Riley - 10 months, 1 week ago

@Steven Chase try the 45th problem

Talulah Riley - 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$