Steady State Analysis of 'Complex' Circuit - Part 2

Electricity and Magnetism Level pending

Consider the arrangement above.

V S = V o sin ⁡ ( ω t ) V_S = V_o \sin(\omega t)

The following function is defined: f ( ω ) = ∣ Z ( ω ) ∣ − 10 f(\omega) = \lvert Z(\omega)\rvert - 10

Here, Z ( ω ) Z(\omega) is the circuit impedance and ∣ Z ( ω ) ∣ \lvert Z(\omega)\rvert is its magnitude. Find the area A A bounded by this curve, the horizontal ' ω \omega ' axis and the vertical line ω = 0 \omega = 0 . Enter your answer as ⌊ 100 A ⌋ \lfloor 100A \rfloor .

Bonus: In the previous problem , the qualitative behaviour of the circuit was understood for large frequencies. This time, describe the circuit behaviour at very low frequencies.


The answer is 7.

Karan Chatrath by Karan Chatrath , 27, Netherlands

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1 solution

Steven Chase
Oct 13, 2020

At DC, the capacitor is effectively open, and the effective impedance is 40 Ω 40 \Omega . The attached code sweeps the angular frequency and plots the function f f . It also accumulates area. 

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import math

# Constants

R1 = 10.0
R2 = 30.0
C = 30.0

dw = 10.0**(-6.0)

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

A = 0.0

w = dw
count = 0

while w <= 10.0:

    ZC = complex(0.0,-1.0/(w*C)) # capacitor impedance

    ZP = R2*ZC/(R2+ZC)  # parallel of ZC and R2

    Z = R1 + ZP  # total impedance

    f = abs(Z) - 10.0  # function f

    dA = f*dw    # accumulate area
    A = A + dA

    w = w + dw
    count = count + 1

    #print w,f

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

print dw
print w
print A
print math.floor(100.0*A)

#>>> 
#1e-05
#1.00001
#0.0765975575078
#7.0
#>>> ================================ RESTART ================================
#>>> 
#1e-05
#10.0000099998
#0.076680890202
#7.0
#>>> ================================ RESTART ================================
#>>> 
#1e-06
#10.0000009993
#0.0768158901978
#7.0
#>>>

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