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Electricity (Basic)
> Electric Power
Electric Power
The electric power is given by:
P
=
V
·
I
Where
V
is the voltage and
I
is the current.
Corresponding units:
watt(W) = volt (V) · ampere (A)
The power can also be determined by using the following formulas:
P
=
I
^{2}
·
R
↔
R
=
P
/
I
^{2}
↔
I
= √(
P
/
R
)
P
=
V
^{2}
/
R
↔
R
=
V
^{2}
/
P
↔
V
= √(
P
·
R
)
More on Electric Power
Electric Power is defined as the rate at which work is done by the source of e.m.f. in maintaining the current in the electric circuit. The practical unit of power is kilowatt and horse power; where 1 kilowatt = 100 watt and 1 H.P = 746 watt.
If resistances (i.e. electrical appliances) are connected in series, the current through each resistance is same. Then power of an electrical appliance, P α R and P α V (as V = IR), it means that in
series combination of resistances, the potential difference and power consumed will be more in larger resistance
.
If resistances (
i.e.
electrical appliances) are connected in parallel, the potential difference across each appliance is the same. Then P α 1/R and I α 1/R (as V = IR), which means in
parallel combination of resistances, the current and power consumed will be more in smaller resistance.
For a given voltage
V,
if the resistance is changed from
R
to (
R/n
) and the power consumption changes from
P
to
nP
, then according to
P
= V
^{2}
/
R
, we have:
P
= V
^{2}
/(R/n)) = n(V
^{2}
/R) = nP, where
R
= R/n
and
P
= nP
When the appliances of power
P
_{1}
, P
_{2}
, P
_{3}
…
P
_{n}
are in series with a voltage source, the effective power consumed (
P
_{s}
) is given by:
1/
P
_{s}
= 1/P
_{1}
+1/P
_{2}
+1/P
_{3}
+…+1/P
_{n}
For
n
appliances, each of resistance
R
, are connected in series with a voltage source
V,
the power dissipated
P
_{s}
is then given as:
(1)
P
_{s}
=
V
^{2}
/n
R
When the appliances of power
P
_{1}
, P
_{2}
, P
_{3}
…
P
_{n}
are in parallel with a voltage source, the effective power consumed (
P
_{p}
) is then given by:
P
_{s}
= P
_{1}
+P
_{2}
+P
_{3}
+…+P
_{n}
For
n
appliances, each of equal resistance
R
, are connected in parallel with a voltage source
V
, the power dissipated is then given as:
(2)
P
_{p}
=
V
^{2}
/(
R
/n) = n
V
^{2}
/
R
From (1) and (2) we have
P
_{p}
/ P
_{s}
=
n
^{2}
or simply written as:
P
_{p}
= n
^{2}
P
_{s}
.
According the formulas given above, we can explain that:
In
series
grouping of bulbs across a given source of voltage, the bulb of greater wattage will give less brightness and will have lesser resistance potential across it but same current, whereas in
parallel
grouping of bulbs across a given source of voltage, the bulb of greater wattage will give more brightness and will allow more current to pass through it, but have lesser resistance and same potential difference across it.
Electric Energy
The electric energy is defined as the total work done or energy supplied by the source of e.m.f. in maintaining the current in an electric circuit for a given time:
Electric energy = electric power
×
time = P
×
t
Thus the formula for electric energy is given by:
Electric energy =
P
×
t
=
V
×
I
×
t
=
I
^{2 }
×
R
×
t
=
V
^{2}
t / R
S.I unit of electric energy is joule
(denoted by J), where
1joule = 1watt × 1 second = 1volt × 1ampere × 1second
Commercial unit of electric energy
is kilowatthour (
kWh
), where
1
kWh = 1000
W
h = 3.6
×
10
^{6}
J =
one unit
of electricity consumed
.
The number of units of electricity consumed is n = (total wattage × time in hour)/1000
The cost of consumption of electricity in a house = no. of units of electricity consumed × amount for one unit of electricity.
Maximum power Theorem
It states that the output power of a source of current is maximum, when the internal resistance of the source is equal to the external resistance in the circuit. So, if
R
is the external resistance of the circuit and
r
is the internal resistance of the source of current (i.e. a battery) then the output power is maximum, when R =
r.
This theorem is applicable to all types of source of e.m.f. and is related with the output power and NOT with the power dissipated.
If
E
is the applied e.m.f. of the source of e.m.f.
i.e
. a battery of internal resistance
r
and R is the external resistance then current in the circuit is given as:
I = E/(R + r)
At the maximum output power,
R = r
, so we have:
I =
E
/(r + r) =
E
/(2r)
and
maximum output power:
P
_{max}
= I
^{2}
r =
E
^{2}
/(4r)
When the battery is short circuited, power is zero. In this case, the entire power of the battery is dissipated inside the battery due to its internal resistance. So the power dissipated inside the battery is given as:
P = (
E
/r)
^{2}
× r =
E
^{2}
/r
Efficiency of a source of e.m.f.
The efficiency of a source of emf. is defined as the ratio of output power (
i.e.
the power across the external resistance of the circuit, to the input power (i.e. power drawn out of the source of e.m.f.). So
Where V = potential drop across the external resistance R,
E= E.M.F. of the source of current,
I= current in the circuit.
If
r
is the internal resistance of source of
e.m.f.,
then
V = IR and E = I(R +
r
)
or
When the power obtained from the source is maximum, then R =
r.
In this situation we have:
Thus the maximum efficiency of a source of
e.m.f.
is 50%.It means for a cell, only half of the total power drawn from the cell is utilized for useful purposes whereas the other half is dissipated inside the cell.
Example 1:
An elevator must lift 1000 kg a distance of 100 m at a velocity of 4 m/s. What is the average power the elevator exerts during this trip?
Solution:
The work done by the elevator over the 100 meters is easily calculable:
W
=
mgh
= (1000)(9.8)(100) = 9.8×10
^{5}
Joules.
The total time of the trip can be calculated from the velocity of the elevator:
t = x/v = 100m/ 4m/s = 25s
.
Thus the average power is given by:
P
= W/t = 9.8×10
^{5}
/ 25s = 3.9×10
^{4 }
Watts, or 39 kW.
Example 2:
An object in free fall is said to have reached
terminal velocity
if the air resistance becomes strong enough to counteract all gravitational acceleration, causing the object to fall at a constant speed. The exact value of the terminal velocity varies according to the shape of the object, but can be estimated for many objects at 100 m/s. When a 10 kg object has reached terminal velocity, how much power does the air resistance exert on the object?
Solution:
To solve this problem we will use the equation
P
=
Fv
cos
θ
, Instead of the usual power equation, as we are given the velocity of the object. We merely need to calculate the force exerted on the object by the air resistance, and the angle between the force and the velocity of the object. Since the object has reached a constant speed, the net force on it must be zero. Since there are only two forces acting on the object, gravity and air resistance, the air resistance must be equal in magnitude and opposite in direction as the force of gravity. Thus
F
_{a}
= 
F
_{G}
=
mg
= 98 N, pointing upwards. Thus the force applied by air resistance is antiparallel to the velocity of the object. Thus:
P
=
Fv
cos
θ
= (98)(100)(cos180) =  9800 W
Example 3:
The power of a pump motor is 4KW. How much water in kg/minute can it raise a height of 20m? (g = 10 m/s
^{2}
)
Solution:
Given power of motor P = 4KW =4000 W
If mass of water raised in one second = m kg.
Total work done in lifting water,W = mgh
Power P = W/t, but t = 1 minute = 60 sec.
4000 = mgh/60
4000 = (m
×
10
×
20)/60
m = 1200 kg.
Example 4
: When water is flowing through a pipe then its velocity changes by 5%, find the change in the power of water?
Solution:
Power = Force
×
Velocity = Rate of change of momentum
×
velocity = {(mass/time)
×
velocity}
x
velocity = {(adv)
×
v}
×
v =adv
^{3}
where 'a' is area of cross section, 'd' is the density of water and 'v' is the velocity of flow of water.
Therefore, Power of water is directly proportional to the cube of velocity of water so let
P = Kv
^{3}
(k is a constant and is equal to 'ad'.)
Taking log on both sides
log P = 3log v + log k
Differentiating on both sides
dP/P = 3.dv/v
percentage change in power, dP/P
×
100 = 3
×
5% = 15%.
Example 5
: The kinetic energy of rushing out water from a dam is used in rotating a turbine. The pipe through which water is rushing is 2.4 meters and its speed is 12 m/sec. Assuming that whole of kinetic energy of the water is used in rotating the turbine, calculate the current produced if efficiency of the dynamo is 60% and the station transmits power at240 kV. Density of water = 10
^{3}
kg/m
^{3}
.
Solution:
Given that
r = radius of pipe = 1.2m, average speed of water v = 12 m/s
V = 240 kV = 240
×
10
^{3}
volt, density of water p = 10
^{3}
kg/m
^{3}
.
Now, kinetic energy of rushing water per second i.e.
Power P = (1/2)(mass flowing per sec)
×
v
^{2}
= (1/2)pr
^{2}
(l/t) rv
^{2}
= (1/2)pr
^{2}
rv
^{3}
= (1/2) 3.14
×
(1.2)
^{2 }
×
10
^{3}
×
(12)
^{3}
watt
= 3.9
x
10
^{6}
watt
Current in the transmission cables is given by:
current = output power/voltage
= (60% of power P)/(240
×
1000)
= [(60/100)
×
3.9
×
10
^{6}
]/(240
×
1000) = 9.75 A
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