# Chapter 7 Heating effect of electric curent¶

## Example 7.1 Page no 373¶

In [6]:
#Given
V=240                   #V
P=60
P1=100

#Calculation
R=V**2/P
R1=V**2/P1

#Result
print"Resistance of a bulb for 60 W is", R,"ohm and for 100 W is",R1,"ohm"
print"Hence The resistance of 60 W,240 V bulb is more than that of 100 W,240 V bulb"

Resistance of a bulb for 60 W is 960 ohm and for 100 W is 576 ohm
Hence The resistance of 60 W,240 V bulb is more than that of 100 W,240 V bulb


## Example 7.2 Page no 373¶

In [10]:
#Given
V=230                #v
P=100
t=20*60
V1=115                   #V

#Calculation
R=V**2/P
E=(V1**2*t)/R

#Result
print"Heat and light energy is", E,"J"

Heat and light energy is 30000 J


## Example 7.3 Page no 374¶

In [18]:
#Given
P=500             #W
V=200.0              #V
V1=240

#Calculation
I=P/V
R=V1-V
R1=R/I

#Result
print"The value of R=",R1,"ohm"
print"Current in a circuit is",I,"A"

The value of R= 16.0 ohm
Current in a circuit is 2.5 A


## Example 7.4 Page no 374¶

In [25]:
#Given
P1=100.0                     #W
P=1100.0                      #W
V=250

#Calculation
P2=P-P1
R=V**2/P2

#Result
print"The value of unknown resistance is", R,"ohm"

The value of unknown resistance is 62.5 ohm


## Example 7.6 Page no 374¶

In [29]:
#Given
V=220
P=200.0
P1=100

#Calculation
R1=V**2/P
R2=V**2/P1
H=R1/R2

#Result
print"The ratio of heats genetated in them is", H

The ratio of heats genetated in them is 0.5


## Example 7.7 Page no 375¶

In [36]:
#Given
m=1
c=1
a=100                  #W
b=15
t=7.5                   #second
P=1                     #KW
C=860                     #Kcal

#Calculation
A=m*c*(a-b)
B=P*t/60.0
D=B*C
n=A*a/D

#Result
print"Efficiency of the kettle is", round(n,1),"%"

Efficiency of the kettle is 79.1 %


## Example 7.8 Page no 375¶

In [39]:
#Given
P1=9                    #W
R1=8
R2=12.0

#Calculation
P2=(P1*R1)/R2

#Result
print"Power dissipated in 12 ohm resistor is", P2,"W"

Power dissipated in 12 ohm resistor is 6.0 W


## Example 7.9 Page no 375¶

In [44]:
#Given
H1=10
a=5.0
b=4.2

#Calculation
I1=(H1*b)/(a*4)
A=I1*4/b

#Result
print"Heat generated in 4 ohm resistor is", A,"cal/sec"

Heat generated in 4 ohm resistor is 2.0 cal/sec


## Example 7.10 Page no 375¶

In [55]:
#Given
E=12                   #V
I=1                     #A
r=0.5                   #ohm

#Calculation
P1=E*I
P2=I**2*r
P=P1-P2

#Result
print"(i) Rate of consumption of chemical energy is", P1,"W"
print"(ii) Rate Of energy dissipated inside the battery is",P2,"W"
print"(iv) Rate of energy dissipated in the resistor is", P,"W"
print"(v) Power output of the source is",P,"W"

(i) Rate of consumption of chemical energy is 12 W
(ii) Rate Of energy dissipated inside the battery is 0.5 W
(iv) Rate of energy dissipated in the resistor is 11.5 W
(v) Power output of the source is 11.5 W


## Example 7.11 Page no 376¶

In [68]:
#Given
P=110                      #W
P1=100                     #W
n=5
V=220                      #V
t=2                     #hours
n1=4
P2=1120                   #W
m=1.5                         #per KWh

#Calculation
W=n*P1
W1=V*t
W2=n1*P
W3=W+W1+W2+P2
E=(W3*t)*10**-3
E2=E*30
B=m*E2

#Result
print"Electricity bill for the month of september is", B,"Rs"

Electricity bill for the month of september is 225.0 Rs


## Example 7.12 Page no 376¶

In [75]:
#Given
V=220                         #V
P=60.0                            #W
P1=85                            #w

#Calculation
import math
R=V**2/P
V1=math.sqrt(P1*R)

#Result
print"Maximum voltage is", round(V1,1),"V"

Maximum voltage is 261.9 V


## Example 7.13 Page no 376¶

In [82]:
#Given
V=200              #V
P=500.0                 #W
V1=160               #v

#Calculation
R=V**2/P
H=V1**2/R
P1=P-H
H1=P1*100/P

#Result
print"Heat percentage is", H1,"%"

Heat percentage is 36.0 %


## Example 7.14 Page no 376¶

In [86]:
#Given
P1=500                           #W
P2=100

#Calculation
R=P1/P2

#Result
print"Since P1'=5P2', 100W bulb will glow brighter"

Since P1'=5P2', 100W bulb will glow brighter


## Example 7.15 Page no 377¶

In [95]:
#Given
m=900
w=100.0
c=1
a=80
b=4.2
V=210                   #V
x=12
y=60

#Calculation
Hout=(m+w)*c*a
Hin=(V*x*y)/b
Hin1=90/w*Hin
I=Hout/Hin1

#Result
print"Strength of the current is", round(I,3),"A"

Strength of the current is 2.469 A


## Example 7.16 Page no 377¶

In [99]:
#Given
a=0.8

#Calculation
H=a**2
H1=(1-H)*100

#Result
print"Decreased percentage is", H1,"%"

Decreased percentage is 36.0 %


## Example 7.17 Page no 377¶

In [108]:
#Given
a=14
b=60
c=24
d=7.0

#Calculation
t=a*b/60.0
t1=(c/d)

#Result
print"(i) Time in series is", t,"minute"
print"(ii) Time in parallel is",round(t1,2),"minute"

(i) Time in series is 14.0 minute
(ii) Time in parallel is 3.43 minute


## Example 7.19 Page no 378¶

In [113]:
#Given
I=0.5
R=100
t=30
a=4.2
m=200                      #g
w=10                      #g

#Calculation
H=I**2*R*t*60/a
A=H/(m+w)

#Result
print"The rise of temperature is", round(A,2),"degree C"

The rise of temperature is 51.02 degree C


## Example 7.20 Page no 379¶

In [120]:
#Given
c=4.2                           #KJ/Kg/C
m=0.2                           #Kg
a=90
b=20
t=30
V=230

#calculation
d=a-b
H=c*m*d
P=H/t
I=P/V

#Result
print"The value of current is", round(I*10**3,2),"A"

The value of current is 8.52 A