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1
SOURCES
ENERGY OR POWER
Arun Umrao
4
0.1 Energy
Energy is the quantitative property that must be transferred to a body or physical
system to perform work on the body. Its SI unit is Joule and also referred as Calorie.
0.1.1 Energy Source
An energy source has capacity to supply the desired quantity of energy at a steady
rate over a long period of time. Ideally, energy of source declines with time and in
long and finite duration, its energy decreases to zero. In reality, no energy source can
have infinite energy. Total energy (E) of an energy source in infinite time duration is
finite.
0 < E < ∞
Power of an energy source is non zero initially but after sufficient long time, power
becomes zero as energy exhausted. For infinite duration, average total power is zero.
Pavg = lim
t=∞
P
Et
t
=
E
∞
= 0
X
Et = E is finite energy in infinite time.
0.1. ENERGY 5
If Energy Decreases to Zero at Infinity : In this case there is no zero energy at
t = ∞ and average power (Pavg) shall never be zero (Pavg > 0 and is very very small
quantity). In other words, “decrease to zero” means the amount of energy received
from a source has no utilization.
6
12
2 4 6 8
t(h)
V
T Von T Voff
For example, assume that a 12 volt alkaline battery is operating an old black and
white TV at 12 volts. When potential of battery drops to 10 volts in three hours, TV
goes switched off. Now, we assume that energy of source “dropped to zero in three
6
hours” for that TV even if we can operate a device that works at 10 volts.
Solved Problem 0.1 See the energy time graph given below. A genetically modified
tree (having life of 120 days) is kept in laboratory for its survival and further repro-
duction under controlled environment. Tree needs 1.5 Joule energy per second for
surviving (threshold energy for surviving). According to this graph find when tree
will star dying? Explain your answer.
Solution
1
2
2 4 6 8 10 12
−2
−4
−6
t(d)
E(J)
0.1. ENERGY 7
From this graph it is clear that after six days from now (t = 0), energy supplied
to tree would decrease below 1.5 Joules. The energy below 1.5 Joules is useless for
tree. Hence it will start dying. In other words, we can say that energy of source is
“zero” for the “life of tree” at and after t = 6 days from now.
Solved Problem 0.2 14 kilogram coal is burned to generate energy. It gives constant
energy 12J (i.e. per second release of energy) for six hours and after that energy
generation decayed linearly to zero in next 24 hours. Find whether it is energy source
or power source.
Solution First we find energy generated by the source in infinite time duration.
From the given problem, energy generation lost in 24 hours. So, total energy generated
by this source is
E = 12 × 6 × 3600 +
1
2
× 12 × 24 × 3600
8
6
12
6 12 18 24 30 36
−6
−12
−18
t(h)
E(J)
It gives total energy released by source is 7.78 × 105
J. After thirty hours, source
does not release any energy. Hence we can say in infinite time being, energy released
by this source is
E = 7.78 × 105
J
This is finite value greater than zero but less than infinity. Again, average power in
infinite time being is
Pavg = lim
t→∞
E
t
= lim
t→∞
7.78 × 105
J
t
=
7.78 × 105
J
∞
= 0
0.1. ENERGY 9
So, average power of this source for infinite time being is zero. Hence it satisfy the
conditions
0 < 7.78 × 105
J < ∞; Pavg = 0
Hence this source is an energy source.
0.1.2 Power Source
A power source has capacity to supply a fixed amount of energy for a infinite period of
time. Power source provides non zero finite power for time infinite, hence its average
power for time being is non zero finite value.
Pavg = lim
t=∞
P
Et
t
=
∞
∞
= 1 > 0
Here,
X
Et = ∞ is energy of the power source. In other form
0 < Pavg < ∞
A power source can not emit infinite energy, if do so, it exhausted energy of energy
source. For continuous sustained power supply, a power source must have infinite
10
energy for infinite duration.
E = ∞
As concept of energy and power source are reversely related, hence a source can be
either energy source or can be power source but not both at a same time.
Solved Problem 0.3 Assume that an energy source is emitting energy in packets (as
a continuous train of energy packets) as shown below. Find whether source is energy
source or power source.
Solution
t
A
This is train of energy packet being emitted from the source. There is no drop in
energy packets at any instant of time. So, energy is not zero at any point. Assume
0.1. ENERGY 11
each packet has energy E′
then total energy of all packets emitted by the source in
infinite time is
E = lim
n→∞
E′
× n = E × ∞ = ∞
Now, average power is given by
Pavg = lim
t→∞
E
t
=
∞
∞
= 1 > 0
Mathematically we can also induce that average power of energy being emitted from
the source will be equal to one packet of this train of energy packets (P′
> 0).
Therefore, Pavg is finite.
0 < Pavg < ∞
Hence the source from which this train of energy packets being emitted is a power
source.
Neither Power Nor Energy Source If a source has infinite energy and infinite
power then it is neither power source nor energy source. Mathematically
E = ∞; Pavg = ∞

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Energy and power source

  • 2. 4 0.1 Energy Energy is the quantitative property that must be transferred to a body or physical system to perform work on the body. Its SI unit is Joule and also referred as Calorie. 0.1.1 Energy Source An energy source has capacity to supply the desired quantity of energy at a steady rate over a long period of time. Ideally, energy of source declines with time and in long and finite duration, its energy decreases to zero. In reality, no energy source can have infinite energy. Total energy (E) of an energy source in infinite time duration is finite. 0 < E < ∞ Power of an energy source is non zero initially but after sufficient long time, power becomes zero as energy exhausted. For infinite duration, average total power is zero. Pavg = lim t=∞ P Et t = E ∞ = 0 X Et = E is finite energy in infinite time.
  • 3. 0.1. ENERGY 5 If Energy Decreases to Zero at Infinity : In this case there is no zero energy at t = ∞ and average power (Pavg) shall never be zero (Pavg > 0 and is very very small quantity). In other words, “decrease to zero” means the amount of energy received from a source has no utilization. 6 12 2 4 6 8 t(h) V T Von T Voff For example, assume that a 12 volt alkaline battery is operating an old black and white TV at 12 volts. When potential of battery drops to 10 volts in three hours, TV goes switched off. Now, we assume that energy of source “dropped to zero in three
  • 4. 6 hours” for that TV even if we can operate a device that works at 10 volts. Solved Problem 0.1 See the energy time graph given below. A genetically modified tree (having life of 120 days) is kept in laboratory for its survival and further repro- duction under controlled environment. Tree needs 1.5 Joule energy per second for surviving (threshold energy for surviving). According to this graph find when tree will star dying? Explain your answer. Solution 1 2 2 4 6 8 10 12 −2 −4 −6 t(d) E(J)
  • 5. 0.1. ENERGY 7 From this graph it is clear that after six days from now (t = 0), energy supplied to tree would decrease below 1.5 Joules. The energy below 1.5 Joules is useless for tree. Hence it will start dying. In other words, we can say that energy of source is “zero” for the “life of tree” at and after t = 6 days from now. Solved Problem 0.2 14 kilogram coal is burned to generate energy. It gives constant energy 12J (i.e. per second release of energy) for six hours and after that energy generation decayed linearly to zero in next 24 hours. Find whether it is energy source or power source. Solution First we find energy generated by the source in infinite time duration. From the given problem, energy generation lost in 24 hours. So, total energy generated by this source is E = 12 × 6 × 3600 + 1 2 × 12 × 24 × 3600
  • 6. 8 6 12 6 12 18 24 30 36 −6 −12 −18 t(h) E(J) It gives total energy released by source is 7.78 × 105 J. After thirty hours, source does not release any energy. Hence we can say in infinite time being, energy released by this source is E = 7.78 × 105 J This is finite value greater than zero but less than infinity. Again, average power in infinite time being is Pavg = lim t→∞ E t = lim t→∞ 7.78 × 105 J t = 7.78 × 105 J ∞ = 0
  • 7. 0.1. ENERGY 9 So, average power of this source for infinite time being is zero. Hence it satisfy the conditions 0 < 7.78 × 105 J < ∞; Pavg = 0 Hence this source is an energy source. 0.1.2 Power Source A power source has capacity to supply a fixed amount of energy for a infinite period of time. Power source provides non zero finite power for time infinite, hence its average power for time being is non zero finite value. Pavg = lim t=∞ P Et t = ∞ ∞ = 1 > 0 Here, X Et = ∞ is energy of the power source. In other form 0 < Pavg < ∞ A power source can not emit infinite energy, if do so, it exhausted energy of energy source. For continuous sustained power supply, a power source must have infinite
  • 8. 10 energy for infinite duration. E = ∞ As concept of energy and power source are reversely related, hence a source can be either energy source or can be power source but not both at a same time. Solved Problem 0.3 Assume that an energy source is emitting energy in packets (as a continuous train of energy packets) as shown below. Find whether source is energy source or power source. Solution t A This is train of energy packet being emitted from the source. There is no drop in energy packets at any instant of time. So, energy is not zero at any point. Assume
  • 9. 0.1. ENERGY 11 each packet has energy E′ then total energy of all packets emitted by the source in infinite time is E = lim n→∞ E′ × n = E × ∞ = ∞ Now, average power is given by Pavg = lim t→∞ E t = ∞ ∞ = 1 > 0 Mathematically we can also induce that average power of energy being emitted from the source will be equal to one packet of this train of energy packets (P′ > 0). Therefore, Pavg is finite. 0 < Pavg < ∞ Hence the source from which this train of energy packets being emitted is a power source. Neither Power Nor Energy Source If a source has infinite energy and infinite power then it is neither power source nor energy source. Mathematically E = ∞; Pavg = ∞