LRC II

Just for fun, I will continue with the hill-climbing technique. This time, the program solves for four different unknown parameters $(E, X_L, R, X_C)$ . There are three checks: one for the $U_1$ magnitude, one for the $U_2$ magnitude, and a dot product orthogonality check between $U_1$ and $U_2$ . Note that there are multiple solutions for the parameters in general, but the source voltage always ends up being $13.892$ .

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import math
import cmath
import random

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

delta = 0.001  # mutation size

minres = 99999999.0 # initialize cumulative residual to large value 
                                     # eventually, this should become close to zero

E = 50.0*random.random()   # source voltage - random initialization
R = 5.0 * random.random()  # resistance - random initialization
XL = 5.0 * random.random()  # inductive reactance magnitude - random initialization
XC = 5.0 * random.random()  # capacitive reactance magnitude - random initialization

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

for j in range(0,5*10**6):

    Em = E + delta * (-1.0 + 2.0 * random.random())   # mutate parameters
    Rm = R + delta * (-1.0 + 2.0 * random.random())
    XLm = XL + delta * (-1.0 + 2.0 * random.random())  
    XCm = XC + delta * (-1.0 + 2.0 * random.random())

    ZL = complex(0.0,XLm)  # calculate complex impedances
    ZC = complex(0.0,-XCm)

    I = Em/(ZL + Rm + ZC)  # complex current

    U1 = I*(ZL+Rm)  # complex voltage U1
    U2 = I*(Rm+ZC)  # complex voltage U2

    U1R = U1.real   # real/imag parts of voltages
    U1I = U1.imag

    U2R = U2.real
    U2I = U2.imag

    res1 = math.fabs(abs(U1) - 20.0)  # difference between U1 magnitude and 20
    res2 = math.fabs(abs(U2) - 15.0)  # difference between U2 magnitude and 15

    dot = (U1R*U2R + U1I*U2I)/(abs(U1) * abs(U2)) # dot product orthogonality check
    res3 = math.fabs(dot)

    res = res1 + res2 + res3  # form residual (should be zero)

    if res < minres:   # accept mutated XL,XC if cumulative residual gets smaller

        minres = res

        E = Em  
        R = Rm
        XL = XLm 
        XC = XCm

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


print "minimum residual"
print minres
print ""
print "XL - ohms"
print XL
print ""
print "XC - ohms"
print XC
print ""
print "RMS Source Voltage (E)"
print E
print ""
print "Resistance"
print R


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


# Results

#minimum residual
#5.85605402858e-06

#XL - ohms
#4.14944926514

#XC - ohms
#2.33406251656

#RMS Source Voltage (E)
#13.8924514996

#Resistance
#3.11208840141

This problem can be solved using phasor diagrams . An approximate phasor diagram of this circuit is given in the picture above, where $U_R$ , $U_L$ and $U_C$ are voltages across the resistor, inductor and capacitor respectively and $I$ is the current flowing through the circuit. If we take $I$ as referential phasor, then:

• $U_R$ is collinear with $I$ , because the voltage across a resistor is in phase with the current through it

• $U_L$ is counter-clockwise perpendicular to $I$ and $U_R$ , because the voltage across an inductor leads the current through it by $90°$

• $U_C$ is clockwise perpendicular to $I$ and $U_R$ , because the voltage across a capacitor lags the current through it by $90°$

• $U_R$ and $U_L$ sum up to $U_1$

• $U_R$ and $U_C$ sum up to $U_2$

• $U_1$ leads $U_2$ by $90°$

Points marked as $A, B, C$ and $D$ on the phasor diagram form three similar right-angled triangles: $\triangle{ACB}, \triangle{ADC}$ and $\triangle{BDC}$ . Side lengths of each of these triangles correspond to the following RMS values of voltages:

• $AC = U_1$

• $BC = U_2$

• $AB = U_L + U_C$

• $AD = U_L$

• $BD = U_C$

• $CD = U_R$

Since $AB^2 = AC^2 + BC^2$ , then $U_L + U_C = \sqrt{{U_1}^2 + {U_2}^2} = 25 \si{V}$ .

$U_R$ can be determined using triangle similarity:

$\dfrac{U_R}{U_1} = \dfrac{U_2}{U_L + U_C} \Rightarrow U_R = \dfrac{U_1 * U_2}{U_L + U_C} = \dfrac{15 * 20}{25} \si{V} = 12 \si{V}$

$U_L = \sqrt{{U_1}^2 - {U_R}^2} = \sqrt{20^2 - 12^2} \si{V} = 16 \si{V}$

$U_C = 25 \si{V} - U_L = 9 \si{V}$

Finally, for a series RLC circuit , RMS voltage of the source is determined as:

$\boxed{ E = \sqrt{{U_R}^2 + (U_L - U_C)^2} = \sqrt{12^2 + (16 - 9)^2} = \sqrt{193} \si{V} \approx 13.892 \si{V}}$