When electric current flows in a purely resistive circuit electrical energy gets fully converted into heat energy. The amount of heat produced (H) in the circuit is found to be directly proportional to (i) the square of current ($I^2$) (ii) the resistance (R) of the conductor and (iii) the time (t) for which current flows. In other words $\mathrm{H} = \mathrm{I}^2\mathrm{Rt}$. Electrical devices such an electric fuse, electric heater, electric iron etc. are all based on this effect called heating effect of electric current.

(a) List two properties of heating elements.

(b) List two properties of electric fuse.

(c) Name the principle on which an electric fuse works. Explain how a fuse wire is capable of saving electrical appliances from getting damaged due to accidently produced high currents.

OR

(c) The power of an electric heater is 1100 W. If the potential difference between the two terminals of the heater is 220 V, find the current flowing in the circuit. What will happen to an electric fuse of rating 5 A connected in this circuit?

(a) Two properties of heating elements:

1. High Resistivity: They offer high resistance to the flow of current, which helps in producing a large amount of heat ($H = I^2Rt$).
2. High Melting Point / Resistance to Oxidation: They do not oxidize (burn) or melt readily even at very high temperatures (alloys like Nichrome are commonly used for this reason).

(b) Two properties of an electric fuse:

1. Low Melting Point: The fuse wire must melt easily when the temperature rises due to excessive current.
2. High Resistance (relative to connecting wires): It should generate sufficient heat quickly when current exceeds the safe limit.

(c) Principle of Electric Fuse:

An electric fuse works on the Joule's heating effect of electric current.

Explanation of how it saves appliances:

A fuse wire is connected in series with the live wire of the circuit. When an accidentally high current flows through the circuit, the heat produced ($H = I^2Rt$) in the fuse wire increases rapidly. Because the fuse wire has a low melting point, this heat causes it to melt and break the circuit. This stops the flow of current instantly, saving the electrical appliances from getting damaged.

OR

(c) Numerical Solution:

• Current flowing in the circuit:

Given:

Power ($P$) = $1100\text{ W}$

Potential difference ($V$) = $220\text{ V}$

Using the formula:

$$P = V \times I$$

$$I = \frac{P}{V} = \frac{1100\text{ W}}{220\text{ V}} = 5\text{ A}$$

The current flowing in the circuit is $5\text{ A}$.

• What will happen to the 5 A fuse:

Since the current drawn by the electric heater ($5\text{ A}$) is exactly equal to the rating of the fuse ($5\text{ A}$), the fuse is operating at its maximum limit. Under steady conditions, it will just survive. However, any slight voltage fluctuation or minor increase in current above $5\text{ A}$ will cause the fuse wire to melt and break the circuit.