#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"
#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"
#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"
#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"
#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
#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),"%"
#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"
#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"
#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"
#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"
#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"
#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,"%"
#Given
P1=500 #W
P2=100
#Calculation
R=P1/P2
#Result
print"Since P1'=5P2', 100W bulb will glow brighter"
#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"
#Given
a=0.8
#Calculation
H=a**2
H1=(1-H)*100
#Result
print"Decreased percentage is", H1,"%"
#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"
#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"
#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"