## Electric Energy

Work must be done continuously to maintain current in a conductor as the conductors offer resistance to the flow of current. The amount of work done W in carrying a charge Q through a wire of resistance R in time t is given by W = QV. Since Q = I × t, therefore, W = V × I × t, where V is the potential difference across the wire. Since by Ohm’s law, V = IR, therefore, W = I^{2}Rt.

The electric energy dissipated or consumed is directly proportional to the square of the current I, directly proportional to the resistance R and to the time t during which current flows. H = I^{2}Rt.

## Joule’s Law of Heating

The relation H = I^{2}Rt implies that heat produced in a resistor is directly proportional to the square of current for a given resistance and directly proportional to the resistance for a given current and directly proportional to the time for which the current flows through the resistor.

Example 1.

Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be :

(a) 1:2

(b) 2:1

(c) 1:4

(d) 4:1

Answer:

(c) 1:4

Explanation: The resistances of the two conducting wires of the same material and equal lengths and dimensions are equal. Let it be denoted by R. The heat produced in a resistor is 1/2 given by H = \(\frac{V^{2}}{R}\)t

When the two resistances are first connected in series, their effective resistance = 2R and when connected in parallel, their effective resistance = R/2.

When connected in series, the heat produced V^{2} would be H = \(\frac{V^{2}}{2 R}\)t and when connected in parallel, the heat produced would be

Hp = \(\frac{V^{2}}{R / 2} t=\frac{2 V^{2}}{R} t\)

Therefore, ratio of heat produced in series and parallel combinations = \(\frac{H_{s}}{H_{p}}=\frac{V^{2} t}{2 R} \div \frac{2 V^{2} t}{R}\) = 1:4

## Applications of Heating Effects of Current

The electric iron, toaster, oven, kettle, etc, are some of the electrical devices which are based on Joule’s heating. The electric heating is also used to produce light. The fuse is another application of Joule’s heating.

## The Electric Bulb

The filament of the bulb must retain as much of the heat generated as is possible so that it gets hot and emits light. A metal having high melting point such as Tungsten should be used so that it does not melt at very high temperatures. The filament should be thermally isolated. The bulbs are filled with inert gases such as nitrogen and argon so as to prolong the life of filament.

## Fuse

It protects circuits and appliances by stopping the flow of any unusually high electric current. The fuse is placed in series with the device. It consists of a piece of wire made of a metal or an alloy of appropriate melting point. If a current larger than the specified value fLows through the circuit, the temperature of the fuse wire increases which melts the fuse wire and breaks the circuit.

Example 2.

Why does the cord of an electric heater not glow while the heating element does?

Answer:

Cord of an electric heater is made up of good conductor of electricity such as copper having low resistivity whereas the heating element is made up of a material having higher resistivity and hence higher resistance. Therefore, heat produced will be much more in heating element than the cord as heat produced depends on the resistance. Hence the cord of an electric heater does not glow whereas the heating element glows.

Example 3.

Compute the heat generated while transferring 96000 coulombs of charge in one hour through a potential difference of 50 V.

Answer:

It is given that Charge q = 96000 C and t = 1 hour, V = 50 V.

Heat generated is given by H = VIt = V × \(\frac{q}{t}\) × t =

= V × q = 50 × 96000 = 4800kJ