Chapter 18 : Capacitive Circuits

Example No. 18_1 Page No. 545

In [2]:
from math import sqrt,pi,atan
# If a R=30ohms and Xc=40ohms are in series with 100V applied, find the following: Zt, I, Vr, Vc and Theta z. What is the phase angle between Vc and Vr with respect to I? Prove that the sum of the series voltage drop equals the applied voltage Vt

# Given data

R = 30.#     # Resistance=30 Ohms
Xc = 40.#    # Capacitive Reactance=40 Ohms
Vt = 100.#   # Applied Voltage=100 Volts

R1 = R*R#
Xc1 = Xc*Xc#

Zt = sqrt(R1+Xc1)#
print 'Zt = %0.2f Ohms'%Zt

I = (Vt/Zt)#
print  'I = %0.2f Ampers'%I

Vr = I*R#
print  'Voltage Across Resistor = %02.f Volts'%Vr

Vc = I*Xc#
print  'Voltage Across Capacitive Reactance = %0.2f Volts'%Vc

Oz = atan(-(Xc/R))*180/pi
print  'Theta z =%0.2f degree'%Oz

#Prove that the sum of the series voltage drop equals the applied voltage Vt

Vt = sqrt((Vr*Vr)+(Vc*Vc))#
print  'Sum of Voltage Drop is Equal to Applied Voltage of 100V = %0.2f Volts'%Vt

Zt = 50.00 Ohms
I = 2.00 Ampers
Voltage Across Resistor = 60 Volts
Voltage Across Capacitive Reactance = 80.00 Volts
Theta z =-53.13 degree
Sum of Voltage Drop is Equal to Applied Voltage of 100V = 100.00 Volts

Example No. 18_2 Page No. 548

In [6]:
from math import sqrt,pi,atan
# A 30-mA Ir is in parallel with another branch current of 40 mA for Ic. The applied voltage Va is 72 V. Calculate It, Zeq and Theta I.

# Given data

Ir = 30.*10**-3#  # Current Ir=30 mA
Ic = 40.*10**-3#  # Current Ic=40 mA
Va = 72.#        # Applied Voltage=72 Volts

A = Ir*Ir#
B = Ic*Ic#

It = sqrt(A+B)#
print 'The Total Current = %0.2f Amps'%It
print 'i.e 50 mAmps'

Zeq = Va/It#
print 'The Equivqlent Impedence = %0.2f Ohms'%Zeq
print 'i.e 1.44 kohms'

Oi = atan(Ic/Ir)*180/pi
print 'The Value of Theta I = %0.2f degrees'%Oi

The Total Current = 0.05 Amps
i.e 50 mAmps
The Equivqlent Impedence = 1440.00 Ohms
i.e 1.44 kohms
The Value of Theta I = 53.13 degrees