Chapter17 : Capacitive Reactance

Example No. 17_1 Page No. 530

In [2]:
from math import pi
# How much is Xc for (a) 0.1 uF of C at 1400 Hz? (b) 1 uF of C at the same frequency?

# Given data

f = 1400#           # Frequency=1400 Hz
C1 = 0.1*10**-6#     # Cap1=0.1 uF
C2 = 1*10**-6#       # Cap2=1 uF

Xc1 = 1./(2.*pi*f*C1)#
print 'The Capacitive Reactance = %0.2f Ohms'%Xc1
print 'approx 1140 Ohms'

Xc2 = 1./(2.*pi*f*C2)#
print 'The Capacitive Reactance = %0.2f Ohms'%Xc2
print 'approx 114 Ohms'

The Capacitive Reactance = 1136.82 Ohms
approx 1140 Ohms
The Capacitive Reactance = 113.68 Ohms
approx 114 Ohms

Example No. 17_2 Page No. 530

In [3]:
from math import pi
#How much is the Xc of a 47-pF value of C at (a) 1 MHz? (b) 10 MHz?

# Given data

f1 = 1*10**6#   # Frequency1=1 MHz
f2 = 10*10**6#  # Frequency2=10 MHz
C = 47*10**-12# # Cap=47 pF

# For 1 MHz

Xc1 = 1./(2.*pi*f1*C)#
print 'The Capacitive Reactance = %0.2f Ohms'%Xc1
print 'approx 3388 Ohms'

# For 10 MHz

Xc2 = 1./(2.*pi*f2*C)#
print 'The Capacitive Reactance = %0.2f Ohms'%Xc2

The Capacitive Reactance = 3386.28 Ohms
approx 3388 Ohms
The Capacitive Reactance = 338.63 Ohms

Example No. 17_3 Page No. 532

In [4]:
from math import pi
# What C is needed for Xc of 100 Ohms at 3.4 MHz?

# Given data

f = 3.4*10**6#   # Frequency=3.4 MHz
Xc = 100# # Capacitive Reactance=100 Ohms

C = 1./(2.*pi*f*Xc)#
print 'The Capacitance = %0.2e Farads'%C
print 'approx 468 pF'

The Capacitance = 4.68e-10 Farads
approx 468 pF

Example No. 17_4 Page No. 533

In [5]:
from math import pi
# At what frequency will a 10 uF capacitor have Xc equal to 100 Ohms?

# Given data

Xc = 100# # Capacitive Reactance=100 Ohms
C = 10*10**-6# # Cap=10 uF

f = 1./(2.*pi*C*Xc)#
print 'The Frequency = %0.2f Hertz'%f
print 'approx 159 Hz'

The Frequency = 159.15 Hertz
approx 159 Hz

Example No. 17_5 Page No. 534

In [7]:
# Calculate the instantaneous value of charging current ic produced by a 6 uF C when its potential difference is increased by 50 V in 1 s.

# Given data

C = 6*10**-6#    # Cap=6 uF
dv = 50.#        # differential voltage increased by 50 Volts
dt = 1.#         # differectial time is 1 sec

ic = C*(dv/dt)#
print 'The Instantaneous Value of Charging Current ic produced = %0.2e Amps'%ic
print 'i.e 300 uAmps'

The Instantaneous Value of Charging Current ic produced = 3.00e-04 Amps
i.e 300 uAmps

Example No. 17_6 Page No. 535

In [8]:
# Calculate the instantaneous value of charging current ic produced by a 6 uF C when its potential difference is decreased by 50 V in 1 s.

# Given data

C = 6*10**-6#    # Cap=6 uF
dv = -50.#        # differential voltage decreased by 50 Volts
dt = 1.#         # differectial time is 1 sec

ic = C*(dv/dt)#
print 'The Instantaneous Value of Discharging Current ic produced = %0.2e Amps'%ic
print 'i.e -300 uAmps'

The Instantaneous Value of Discharging Current ic produced = -3.00e-04 Amps
i.e -300 uAmps

Example No. 17_7 Page No. 536

In [9]:
# Calculate ic produced by a 250-pF capacitor for a change of 50 V in 1 us.

# Given data

C = 250*10**-12#    # Cap=250 pF
dv = 50.#           # differential voltage increased by 50 Volts
dt = 1.*10**-6#      # differectial time is 1 usec

ic = C*(dv/dt)#
print 'The Instantaneous Value of ic produced = %0.2e Amps'%ic
print '12500 uAmps or 12.5 mAmps'

The Instantaneous Value of ic produced = 1.25e-02 Amps
12500 uAmps or 12.5 mAmps