Question

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 (I²)
(ii) the resistance (R) of the conductor and
(iii) the time (t) for which current flows. In other words H = I²Rt. 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 accidentally 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?

Answer 1

(a) • High resistivity: Heating elements should have high resistivity to produce sufficient heat when current flows.
• High melting point: Heating elements should have a high melting point to withstand the heat generated without melting.
(b) • Low melting point: Fuse wires have a low melting point so they melt easily when excess current flows.
• Specific current rating: Fuses are designed to allow a specific maximum current to prevent overheating of circuits.
(c) • Principle: Electric fuse works on the principle of the heating effect of electric current.
• Function: When a high current flows through the fuse wire, it generates heat due to its resistance. If the current exceeds the fuse’s rated capacity, the heat generated melts the fuse wire, breaking the circuit and stopping the current flow. This protects electrical appliances from damage caused by high currents.

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Power formula:
\[P = V \times I,\]
where:
\( P = 1100 \, \text{W}, \)
\( V = 220 \, \text{V}, \)
\( I = ? \, \text{(current)} \).
Substituting values:
\[1100 = 220 \times I.\]
Solve for \( I \):
\[I = \frac{1100}{220} = 5 \, \text{A}.\]
Fuse rating: The current in the circuit is exactly \( 5 \, \text{A} \), which matches the fuse rating. If the current increases even slightly beyond \( 5 \, \text{A} \), the fuse will blow and disconnect the circuit to prevent damage.