Chapter 1 : Introduction

Example 1.1 Page No : 1

In [1]:
# Given
m = 10.;        # Acceleration is 2.0m/s**2
a = 2.;         # mass

# Calculation and Results
#a)")
F = m*a
print "Force is %dN"%(F)
#b)")
t = 4;            # time
x = (a*t*t)/2
KE = (F*x)
P = KE/t
print "Position is %dm"%(x)
print "Kinetic energy  = %3.1fJ"%(KE)
print "Power  = %3.1fW"%(P)

Force is 20N
Position is 16m
Kinetic energy  = 320.0J
Power  = 80.0W

Example 1.2 Page No : 3

In [2]:
# Given
i = 5.;         # current flow

# Calculation
#As electroms/min is asked so we need to convert A(C/s) to C/min
i1 = 5*60;
#Let e be electronic charge
e = 1.602*10**-19
n = (i1/e)

# Results
print "Number of electrons  = %3.2f electrons/min"%(n)

Number of electrons  = 1872659176029962633216.00 electrons/min

Example 1.3 Page No : 8

In [3]:
# Given
E = 9.25*10**-6         ##Energy is 9.25uJ")
q = 0.5*10**-6;         # #Charge to be transferred is 0.5uC")

# Calculation
#1 volt is 1 joule per coulomb
V = E/q;

# Results
print "Potential difference between two points a and b is %3.1fV"%(V)

Potential difference between two points a and b is 18.5V

Example 1.4 Page No : 10

In [4]:
# Given
#Potential difference is 50V")
#Charge per minute is 120C/min")
V = 50.
x = 120;

# Calculation
#As Electrical energy is to be calculated charge per minute is to be converted in charge per second
#Charge per second is nothing but the current
i = x/60;
P = i*V;
#Since is 1W = 1J/s

# Results
print "Rate of energy conversion is %dJ/s"%(P)

Rate of energy conversion is 100J/s