Electric Energy and Joule's Law

1.0Electric Energy

The total energy supplied by a source of emf in order to maintain the electric current in the circuit in a given time is called ‘electric energy’.

E = P × t = V × I × t

Unit of Electric Energy

SI unit : Joule

Here, 1 Joule = 1 volt-ampere-sec = 1 VAs

Other unit : 1 Watt-hour  = 3600 J

1 Kilo Watt Hour (kWh) = 3.6 × 10⁶ J 

Kilo Watt Hour (kWh) is a commercial unit of electric energy. If an electric device of 1 kW is used for 1hr, then the energy consumed is 1 kwh.

Long distance power transmission is done at high voltages because at high voltages, current flowing through the transmission wires is less. As a result, heat losses (H ∝ I²) will be less.

2.0Joule’s Law of Heating

The energy supplied to the circuit by a source (like a battery) in time t is given by,

E = P × t = V × I × t [ P = V × I]

This energy (E) gets dissipated in the resistor as heat. Thus, for a steady current I, the amount of heat H produced in time t is 

H = V × I × t

Applying Ohm’s law, i.e., V= I × R, we get,

$H=I^{2}Rt$

This is known as Joule’s law of heating. 

According to Joule’s law of heating, the heat produced in a resistor is directly proportional to the square of current for a given resistance, directly proportional to resistance for a given current, and directly proportional to the time for which the current flows through the resistor.

Example

1. Data for various electrical appliances is given in the table below.

The power for the hairdryer and current for the iron  are, respectively :

(1) 480W ; 0.25 A

(2) 240W ; 4A

(3) 480W ; 4A

(4) 960 ; 2A

Solution

Option (3) is correct.

Given, for the hair dryer, Voltage (V) = 240 V, Current (I) = 2 amp.

Power of hair dryer

P = VI = 240 × 2 = 480 W.

Given, for the electric iron, Power (P) = 960 W, Voltage (V) = 240 V, current for the iron,

$I=\frac{P}{V}=\frac{960W}{240V}=4A$

3.0Also Read