Chapter 3 : Ohm's Law

Example No. 3_1 Page No. 88

In [1]:
# A heater with the resistance of 8 Ohms is connected across the 120-V power line. How much is current I?

# Given data

V = 120#        # Voltage of Power line=120 Volts
R = 8#          # Heater Resistance=8 Ohms

I = V/R#
print 'The Current I = %0.2f Amps'%I

The Current I = 15.00 Amps

Example No. 3_2 Page No. 88

In [3]:
# A small lightbulb with a resistance of 2400 Ohms is connected across the 120-V power line. How much is current I?

# Given data

V = 120.#        # Voltage of Power line=120 Volts
R = 2400#       # Lightbulb Resistance=2400 Ohms

I = V/R#
print 'The Current I = %0.2f Amps'%I

The Current I = 0.05 Amps

Example No. 3_3 Page No. 88

In [4]:
# If a 12-Ohms resistor is carrying a current of 2.5 A, how much is its voltage?

# Given data

I = 2.5#        # Current=2.5 Amps
R = 12#         # Resistance=12 Ohms

V = I*R#
print 'The Voltage = %0.2f Volts'%V

The Voltage = 30.00 Volts

Example No. 3_4 Page No. 89

In [5]:
# How much is the resistance of a lightbulb if it draws 0.16 A from a 12-V battery?

# Given data

V = 12#          # Voltage of Battery=12 Volts
I = 0.16#        # Current drawn form Battery=0.16 Amps

R = V/I
print 'The Resistance  = %0.2f ohms'%R

The Resistance  = 75.00 ohms

Example No. 3_5 Page No. 89

In [6]:
# The I of 8 mA flows through a 5-kOhms Resistor. How much is the IR voltage?

# Given data

I = 8*10**-3#        # Current flowing through Resistor=8m Amps
R = 5*10**3#         # Resistance=5k Ohms

V = I*R#
print 'The Voltage = %0.2f Volts'%V

The Voltage = 40.00 Volts

Example No. 3_6 Page No. 90

In [8]:
# How much current is produced by 60 V across 12 kOhms?

# Given data

V = 60.0#        # Voltage=60 Volts
R = 12*10**3#   # Resistance=12k Ohms

I = V/R#
print 'The Current I = %0.3f Amps'%I
print 'i.e 5 mAmps'

The Current I = 0.005 Amps
i.e 5 mAmps

Example No. 3_7 Page No. 91

In [9]:
# A toaster takes 10 A from the 120-V power line. How much power is used?

# Given data

V = 120#        # Voltage of Power line=120 Volts
I = 10#         # Current drawn from Powerline=10 Amps

P = V*I#
print 'The Power used = %0.2f Watts'%P
print 'OR 1.2 kW'

The Power used = 1200.00 Watts
OR 1.2 kW

Example No. 3_8 Page No. 91

In [11]:
# How much current flows in the filament of a 300-W bulb connected to the 120-V power line?

# Given Data

V = 120.0#        # Voltage of Power line=120 Volts
P = 300#        # Power of Bulb=300 Watts

I = P/V#
print 'The Current I = %0.2f Amps'%I

The Current I = 2.50 Amps

Example No. 3_9 Page No. 91

In [13]:
# How much current flows in the filament of a 60-W bulb connected to the 120-V power line?

# Given Data

V = 120#        # Voltage of Power line=120 Volts
P = 60.0#         # Power of Bulb=60 Watts

I = P/V#
print 'The Current I = %0.2f Amps'%I
print 'OR 500 mA'

The Current I = 0.50 Amps
OR 500 mA

Example No. 3_10 Page No. 93

In [14]:
# Asuming that the cost of electricity is 6 cent per kWh, how much will it cost to light a 100-W lightbulb for 30 days?

h = 24*30#    # Total hours = 24 hrs * 30 days

kWh = 0.1*h#    # 100W=0.1kW

Cost = kWh*0.06#    # 6 cent = $0.06

print 'Cost = %0.2f $'%Cost

Cost = 4.32 $

Example No. 3_11 Page No. 95

In [15]:
# Calculate the power in a circuit where the source of 100 V produces 2 A in a 50 Ohms Resistor.

# Given data

I = 2#          # Current=2 Amps
R = 50#         # Resistance=50 Ohms
V = 100#        # Voltage Source=100 Volts

P = I*I*R#
print 'The Power = %0.2f Watts'%P

The Power = 200.00 Watts

Example No. 3_12 Page No. 95

In [17]:
 
# Calculate the power in a circuit where the source of 100 V produces 4 A in a 25 Ohms Resistor.

# Given data

I = 4#          # Current=4 Amps
R = 25#         # Resistance=25 Ohms
V = 100#        # Voltage Source=100 Volts

P = I*I*R#
print 'The Power = %0.2f Watts'%P

The Power = 400.00 Watts

Example No. 3_13 Page No. 96

In [18]:
# How much current is needed for a 600-W, 120-V toaster?

# Given data

V = 120.0#        # Applied Voltage=120 Volts
P = 600#        # Power of toaster=600 Watts

I = P/V#
print 'The Current I = %0.2f Amps'%I

The Current I = 5.00 Amps

Example No. 3_14 Page No. 97

In [20]:
# How much is the resistance of a 600-W, 120-V toaster?

# Given data

V = 120.0#        # Applied Voltage=120 Volts
P = 600#        # Power of toaster=600 Watts

R = (V*V)/P#
print 'The Resistance = %0.2f Ohms'%R

The Resistance = 24.00 Ohms

Example No. 3_15 Page No. 97

In [21]:
from math import sqrt
# How much current is needed for a 24 Ohms Resistor that dissipates 600 W?

# Given data

R = 24.0#         # Resistance=24 Ohms
P = 600.0#        # Power=600 Watts

I = sqrt(P/R)#
print 'The Current I = %0.2f Amps'%I

The Current I = 5.00 Amps

Example No. 3_16 Page No. 98

In [23]:
# Determine the required resistance and appropriate wattage rating of a resistor to meet the following requirements: The resistor must have a 30-V IR drop when its current is 20 mA. The resistors available have the following wattage ratings: 1⁄8, 1⁄4, 1⁄2, 1, and 2 W.

# Given data

I = 20.0*10**-3#   # Current=20m Amps
V = 30.0#         # Voltage Drop=30 Volts

R = V/I#
print 'The Resistor value = %0.2f Ohms'%R
print 'i.e 1.5 kohms'

P = I*I*R#
print 'The Power = %0.2f Watts'%P
print 'OR 600 mW'

The Resistor value = 1500.00 Ohms
i.e 1.5 kohms
The Power = 0.60 Watts
OR 600 mW

Example No. 3_17 Page No. 99

In [28]:
#Determine the required resistance and appropriate wattage rating of a carbonfilm resistor to meet the following requirements: The resistor must have a 225-V IR drop when its current is 150 uA. The resistors available have the following wattage ratings: 1⁄8, 1⁄4, 1⁄2, 1, and 2 W.

# Given data

I = 150.0*10**-6#   # Current=150 uAmps
V = 225#         # Voltage Drop=225 Volts

R = V/I#
print 'The Resistor value = %0.f Ohms'%R
print 'i.e 1.5 Mohms'

P = I*I*R#
print 'The Power = %0.5f Watts'%P
print 'i.e 33.75 mW'

The Resistor value = 1500000 Ohms
i.e 1.5 Mohms
The Power = 0.03375 Watts
i.e 33.75 mW