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Summary Electric Power
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Providing short notes for electric power. It includes concepts, formulas, solved examples , etc
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ELECTRICAL POWER :Energy liberated per second in a device is called its power. The electrical power P delivered or
consumed by an electrical device is given by P = VI, where V = Potential difference across the device and
= Current.
If the current enters the higher potential point of the device then electric power is consumed by it (i.e.
acts as load). If the current enters the lower potential point then the device supplies power (i.e. acts as
source).
V.dq
Power =
dt
= V
P = V
If power is constant then energy = P t
If power is variable then Energy = pdt
Power consumed by a resistor
V2
P = 2R = V = .
R
When a current is passed through a resistor energy is wasted in overcoming the resistance of the
wire. This energy is converted into heat.
V2
W = Vt = 2 Rt = t
R
The heat generated (in joules) when a current of ampere flows through a resistance of R ohm for t
second is given by:
I2 Rt
H = 2 Rt Joule = Calorie
4.2
1 unit of electrical energy = 1 Kilowatt hour = 1 KWh = 3.6 x 106 Joule.
Example . If bulb rating is 100 watt and 220 V then determine
(a) Resistance of filament
(b) Current through filament
(c) If bulb operate at 110 volt power supply then find power consumed by bulb.
Solution : Bulb rating is 100 W and 220 V bulb means when 220 V potential difference is applied
between the two ends then the power consumed is 100 W
Here V = 220 Volt
P = 100 W
V2
= 100 So R = 484
R
Since Resistance depends only on material hence it is constant for bulb
V 220 5
= Amp.
R 22 22 11
power consumed at 110 V
110 110
power consumed = = 25 W
484
, ** Thermal energy produced, Q = i2 Rt in Joules
i 2 Rt
Or Q = in cal.
4.2
As H i 2 , heating effect of current is common to both A.C and D.C.
Joule’s effect is irreversible.
W Pt E it i 2 Rt E 2t
Heat energy produced due to the electric current H =
J J J J RJ
H mst
Where s 4200J /Kg0C
where J is mechanical equivalent of heat.
FUSE WIRE
The fuse wire for an electric circuit is chosen keeping in view the value of safe current through the circuit.
The fuse wire should have high resistance per unit length and low melting point.
However the melting point of the material of fuse wire should be above the temperature that will,be reached
on the passage of the current through the circuit
A fuse wire is made of alloys of lead (Pb) and tin (Sn).
Length of fuse wire is immaterial.
The material of the filament of a heater should have high resistivity and high melting point.
The temperature of the wire increases to such a value at which, the heat produced per second equals heat lost
per second due to radiation from the surface of wire
I2 2 H 2 r I2 r3
r
H = heat lost per second per unit area due to radiation.
EX. : A fuse wire with radius of 0.2mm blows off with a current of 5 Amp. The fuse wire of same
material, but of radius 0.3mm will blow off with a current of
3 5 3
1) 5 Amp 2) Amp
2 2
27
3) 5 Amp 4) 5 Amp
8
3 3
i1 r1 2 0.2 2
i 2 r2
Sol. i2 r 3 ;
0.3
27
i2 5 Amp
8
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