Answer
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Hint: - When an electric current is passed through a conductor, it generates heat.
- For an electric bulb, we use heat to generate light.
- Heat is generated according to Joule’s law of heating.
- More resistance means more heat.
- Again, resistance is inversely proportional to the cross-sectional area of the wire.
Complete step by step solution:
The filament of the torch bulb should be a conductor. That’s why when the current will pass through it, it will generate heat. And from that heat energy light will be generated.
According to Joule’s law of heating,
The heat generated at any resistance, \[H = {I^2}Rt\]
Where $I$ is the current through the resistance $R$ and $t$ is the time.
Again, $R = \rho \dfrac{L}{A}$ where$R$ is the resistivity of the material, $L$ is the length of the wire and $A$ is the cross-sectional area of the wire.
So, to generate more heat energy resistance should be more and to get more resistance cross-sectional area of the wire should be less.
So, to get more heat we should use a thin wire.
The filament of the torch bulb is a thin wire.
The correct option is (D).
Note:
- Same concept works for electric iron or electric heaters or electric fuses also.
- And to create more resistance the length of the wire should me more. That’s why the filament has a spiral (spring-like) shape.
- Basically, filament is a very thin spiral wire.
- For an electric bulb, we use heat to generate light.
- Heat is generated according to Joule’s law of heating.
- More resistance means more heat.
- Again, resistance is inversely proportional to the cross-sectional area of the wire.
Complete step by step solution:
The filament of the torch bulb should be a conductor. That’s why when the current will pass through it, it will generate heat. And from that heat energy light will be generated.
According to Joule’s law of heating,
The heat generated at any resistance, \[H = {I^2}Rt\]
Where $I$ is the current through the resistance $R$ and $t$ is the time.
Again, $R = \rho \dfrac{L}{A}$ where$R$ is the resistivity of the material, $L$ is the length of the wire and $A$ is the cross-sectional area of the wire.
So, to generate more heat energy resistance should be more and to get more resistance cross-sectional area of the wire should be less.
So, to get more heat we should use a thin wire.
The filament of the torch bulb is a thin wire.
The correct option is (D).
Note:
- Same concept works for electric iron or electric heaters or electric fuses also.
- And to create more resistance the length of the wire should me more. That’s why the filament has a spiral (spring-like) shape.
- Basically, filament is a very thin spiral wire.
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