Advertisements
Advertisements
प्रश्न
(i) Draw a clear labelled diagram of an electric bell.
(ii) Explain in brief, its working.
(iii) What material is used for the core of an electric bell? State the reason.
Advertisements
उत्तर
(i)

(ii) An electric bell is one of the most common applications of an electromagnet.
Construction and wiring: An electric bell shown in above fig. The main parts of the bell are:
- A horse-shoe electromagnet M, having iron core,
- Soft iron armature A,
- A hammer H,
- A gong G,
- A metallic spring strip SS,
- An adjusting screw SS,
- A switch (or bell-push) K, and
- A battery.
Working and function of each part:
The working of an electric bell is based on the magnetic effect of current. When the electric circuit is closed by pressing the switch K, the current flows through the coil CC and the core of electromagnet gets magnetized and therefore it attracts the armature A as shown A, the hammer H strikes the gong G and the bell rings.
At the moment, when the armature, due to attraction, moves towards the electromagnet, the connection between the strip SS and the screw S’ breaks which stops the flow of current in the circuit. Consequently, the electromagnet loses magnetism (i.e.r it gets demagnetized) and the armature A flies back to its original position due to the spring effect of the strip SS. Now the armature again touches the screw S’, resulting in the flow of current in the circuit. The electromagnet regains its magnetism and the armature A is again attracted, so the hammer H strikes the gong G again.
This process of make and break of the circuit goes on the hammer strikes the gong repeatedly and the bell rings so long as the switch K is kept pressed.
(iii) The core of the electromagnet is made of soft iron because: Firstly, it increases the intensity of the magnetic field of the electromagnet and secondly it easily get demagnetized when no current flows in the turns of the coil of insulated copper wire wound over the core, thus helping in the smooth working of the electric bell.
संबंधित प्रश्न
Draw a labeled diagram of a step-down transformer.
Derive an expression for e.m.f. and current in terms of turns ratio
Name the transformer used in the power transmitting station of a power plant.
The input and output voltage of a transformer are 220 V and 44 V respectively. Find: the turns ratio.
State the principle of a step-up transformer. Explain, with the help of a labeled diagram, its working ?
A step-down transformer reduces the supply voltage from 220 V to 11 V and increase the current from 6 A to 100 A. Then its efficiency is
A 220 V input is supplied to a transformer. The output circuit draws a current of 2.0 A at 440 V. If the ratio of output to input power is 0.8, then the current drawn by primary winding is ______.
Magnetic flux Φ in weber in a closed circuit of resistance 10Ω varies with time Φ (sec) as Φ = 6t2 - 5t + 1. The magnitude of induced current at t = 0.25s is ______.
Mention two main sources of power loss in real transformers.
