Electric Power

The rate at which electric energy is consumed or dissipated is termed as electric power.

Electric power (P) = Work done (W) / Time taken (t)

P = Vit/t

P = Vi

P = (iR)i

P = i2R

P = (V/R)2R

P = V2/R

SI unit of electric power is joule/second or watt (W).

Electric power consumed in a circuit or a device is 1 W if a current of 1 A flows through it when a potential difference of one volt is maintained across it.

Since watt is a very small unit of power the bigger units kilowatt (kW) megawatt (MW) are actually used in practice.

1 kilowatt (kW) = 1000 W

1 megawatt (MW) = 106 W

1 gigawatt (GW) = 109 W

For electric power, another bigger unit horse power (hp) is also used.

1 (hp) = 746 W

Since electrical energy consumed by an electrical appliance is equal to the product of power and the time for which it is used. The SI unit for the consumption of electric energy is joule but it is very small from practical point of view. Therefore, the electrical energy spent in the electric circuit is generally expressed in watt hour and kilowatt hour.

1 watt hour is the amount of electric energy which is consumed in 1 hour in an electric circuit when the electric power in the circuit is 1 watt. 1 kilowatt hour is the amount of electric energy consumed when 1 kilowatt power is used for 1 hour in an electric circuit.

1 kilowatt hour (kW h) = 1 kilowatt × 1 hour

= 1000 watt x 3600 second

= 1000 joule/second x 3600 second

= 36 × 105 joule

1 kW h = 3.6 × 106 J

The commercial unit of electric energy is kilowatt hour (kW h).

Examples

Example 1: Find the resistance of the filament of 100 W, 250 V electric bulb.

R = V2/P

= (250 x 250)/100

= 625 Ω

Example 2: Calculate the energy consumed in a 2 kW electric heater in 2 hours. Express the result in joules.

Q = Pt

= 2 kW × 2 h

= 4 kW h

= 4 × 3.6 × 106 J

= 14.4 × 106 J

Example 3: How much time will take a 2 kW immersion rod to raise the temperature of 1 litre of water from 30°C to 60°C?

Q = Pt

Q = mcθ

mcθ = Pt

Mass of 1 litre of water (m) = 1 kg

Specific heat of water (c) = 4.18 × 103 J kg-1 °C-1

Rise in temperature of water (θ) = 60 - 30 = 30 °C.

P = 2 kW = 2000 W

So,

1 x 4.18 x 103 x 30 = 2000 × t

t = 62.7 s

Example 4: How many kilowatt hour of energy will be consumed by a 2 hp motor in 10 hours?

P = 2 hp = 2 × 746 W = 1.492 kW

Q = Pt

= 1.492 kW x 10 h

= 14.92 kW h

Example 5: A potential difference of 250 V is applied across a resistance of 1000 ohm. Calculate the heat energy produced in the resistance in 10 s.

Given: V = 250 V, R = 1000 Ω, t = 10 s

Q = V2t/R

= (250 x 250 x 10)/1000

= 625 J

Example 6: Compute the heat generated while transferring 96 kC of charge in one hour through a potential difference of 50 V.

Given: V = 50 V, t = 1 h, q = 96000 C

W = qV

= 96000 C × 50 V

W = 4800000 J = 4.8 × 106 J

= 4.8 MJ

Example 7: An electric iron of resistance 25 Ω takes a current of 5 A. Calculate the heat developed in 1 minute.

Given: R = 25 Ω, i = 5 A, t = 1 min = 60 s

Heat developed, H = i2Rt

= (5 A)2 x 25 Ω x 60 s

= 37500 J

= 3.75 x 104 J