Chapter 16 : Capacitance

Example No. 16_1 Page No. 492

In [4]:
# How much charge is stored in a 2 uF capacitor connected across a 50-V supply?

# Given data

V = 50#         # Voltage=50 Volts
C = 2*10**-6#    # Capacitor=2 uFarad

Q = C*V#
print 'The Charge Stored = %0.2e Coulomb'%Q
print 'i.e 100*10**-6 Coulombs'
The Charge Stored = 1.00e-04 Columb
i.e 100*10**-6 Columbs

Example No. 16_2 Page No. 492

In [5]:
# How much charge is stored in a 40 uF capacitor connected across a 50-V supply?

# Given data

V = 50#         # Voltage=50 Volts
C = 40*10**-6#   # Capacitor=2 uFarad

Q = C*V#
print 'The Charge Stored = %0.2e Coulomb'%Q
print 'i.e 2000*10**-6 Coulombs'
The Charge Stored = 2.00e-03 Columb
i.e 2000*10**-6 Columbs

Example No. 16_3 Page No. 493

In [6]:
# A constant current of 2 uA charges a capacitor for 20 s. How much charge is stored? Remember I=Q/t or Q=I*t.

# Given data

I = 2*10**-6#        # Current=2 uAmps
t = 20#             # Time=20 Sec

Q = I*t
print 'The Charge Stored = %0.2e Coulomb'%Q
print 'i.e 40*10**-6 Coulombs OR 40 uCoulomb'
The Charge Stored = 4.00e-05 Columb
i.e 40*10**-6 Columbs OR 40 uColumb

Example No. 16_4 Page No. 494

In [7]:
# The voltage across the charged capacitor is 20 V. Calculate C.

#Given data

V = 20#         # Voltage=20 Volts
Q = 40*10**-6#   # Charge=40 uCoulomb

C = Q/V
print 'The Capacitance = %0.2e Farad'%C
print 'i.e 2 uF'
The Capacitance = 2.00e-06 Farad
i.e 2 uF

Example No. 16_5 Page No. 495

In [8]:
# A constant current of 5 mA charges a 10 uF capacitor for 1 s. How much is the voltage across the capacitor?

# Given data

I = 5*10**-3#       # Current=5 mAmps
t = 1#             # Time=1 Sec
C = 10*10**-6#      # Cap=10 uFarad

Q = I*t#

V = Q/C#
print 'The Voltage across Capacitor = %0.2f Volts'%V
The Voltage across Capacitor = 500.00 Volts

Example No. 16_6 Page No. 496

In [9]:
# Calculate C for two plates, each with an area 2 sqm, separated by 1 cm with a dielectric of air.

# Given data

c = 8.85*10**-12#    # Constant=8.85 p
A = 2#              # Area=2 sqm
d = 1*10**-2#        # Distance=1 cm
K = 1               # Permeability=1

C = K*c*(A/d)#
print 'The Capacitance = %0.2e Farad'%C
print 'i.e 1700*10**-12 F OR 1770 pF'
The Capacitance = 1.77e-09 Farad
i.e 1700*10**-12 F OR 1770 pF

Example No. 16_11 Page No. 514

In [10]:
# The high-voltage circuit for a color picture tube can have 30 kV across 500 pF of C . Calculate the stored energy.

# Given data

V = 30*10**3#         # Voltage=30 kVolts
C = 500*10**-12#      # Cap=500 pFarad

E = 0.5*C*V*V
print 'The Energy Stored = %0.2f Joules'%E
The Energy Stored = 0.23 Joules