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SS1: PHYSICS – 3RD TERM

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  1. Production of Electric Current | Week 1
    6 Topics
    |
    1 Quiz
  2. Electric Current | Week 2
    5 Topics
    |
    1 Quiz
  3. Electrical Resistance of a Conductor | Week 3
    5 Topics
    |
    1 Quiz
  4. Particulate Nature of Matter | Week 4
    5 Topics
    |
    1 Quiz
  5. Crystalline and Non-crystalline Substances | Week 5
    3 Topics
    |
    1 Quiz
  6. Elastic Properties of Solids | Week 6 & 7
    4 Topics
    |
    1 Quiz
  7. Fluids at Rest & in Motion | Week 8 & 9
    6 Topics
    |
    1 Quiz
  8. Solar Collector
    3 Topics
    |
    1 Quiz



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When electricity or electric current flows through a conductor, heat is generated due to the hindrance caused by the conductor to the flowing current.

The resistance of the conductor converts electrical energy into heat energy.

The quantity of heat generated depends on the value of the current. The greater the value of the current the greater the heat generated.

Applications of the Heating Effect of Current:

This heating effect is utilised in many devices including hair dryers, toasters, electric cookers, bulbs, irons, etc.

Electric Iron: An insulator, called mica, is placed between the metal part and the coil in an iron. When the iron is switched on, the coil becomes heated due to the passage of current which is then transferred to the metallic part through the mica. The metal part finally becomes hot which is used for ironing clothes.

heatng effect-electric iron
Electric Iron.

Electric Bulb: An Electric bulb contains a thick metallic wire which is made up of tungsten metal. The wire is kept in an inert environment with a neutral gas or vacuum, e.g argon. This prevents the hot filament from evaporating and enables the bulb to run at a high temperature.

The bulb becomes heated and emits light when current flows through the tungsten wire. Most of the electrical power is dissipated in the form of heat and the rest is emitted in the form of light energy.

heatng effect-electric bulb
Electric Bulb.

Electric Fuse: Sometimes electrical appliances get burnt which can lead to fires. This is due to a sudden rise of current. To avoid this accident a conducting wire with a low melting point is connected in series with the circuit.

Glass cartridge fuse
Glass cartridge fuse.

This is called a fuse wire and its purpose is to automatically open the circuit (stop the flow of current) whenever excess current might destroy the appliance it’s connected to. The wire melts due to excessive heating.

The ‘fuse rating’ is the maximum safe current permitted to flow in it before the fuse breaks.

heatng effect-electric fuse
A fuse is a piece of wire of a material with a very low melting point, that is, it melts and breaks as soon as its temperature gets higher than its melting point.

Calculating Heat Energy of Electric Current:

The quantity of heat energy generated in an electric circuit is directly proportional to the square of the current flowing in the conductor, the resistance of the conductor and the time during which the current flow.

⇒ \( \scriptsize H = I^2RT \)

where H = heat energy in joules, J
I = current in amperes, A
t = time in seconds, s

Example 1:

Calculate the quantity of heat in joules generated in an electric coil of resistance 30 ohms when a current of 6.5A flows through it for 25 minutes.

Solution:

Data given in the question:

R = 30 ohms
Current, I = 6.5A
time, t = 25 minutes = 25 * 60 seconds = 1500 seconds

Formula: ⇒ \( \scriptsize H = I^2RT \)

Substitute the values into the equation:

⇒ \( \scriptsize H = 6.5^2 \: \times \: 30 \: \times \: 1500 \)

⇒ \( \scriptsize H = 42.25 \: \times \: 45000 \)

⇒ \( \scriptsize H = 19.01 \: \times \: 10^5 \: J \)

Example 2:

Calculate the current that will produce 450 Joules of energy in a wire of resistance 35 ohms if it flows for 300 seconds.

Solution:

Data given in the question:

Quantity of heat Q = 450 J
resistance R = 35 ohms
Current I = ?
time t = 300 seconds.

Formula: ⇒ \( \scriptsize H = I^2RT \)

make I the subject of the formula;

⇒ \( \scriptsize I^2 = \normalsize \frac{H}{RT} \)

⇒ \( \scriptsize I = \normalsize \sqrt{\frac{H}{RT}} \)

Substitute the values into the equation:

⇒ \( \scriptsize I = \sqrt{\normalsize \frac{450}{35 \: \times \: 300}} \)

⇒ \( \scriptsize I = \sqrt{\normalsize \frac{450}{10500}} \)

⇒ \( \scriptsize I = \scriptsize \sqrt{0.0428}\)

⇒ \( \scriptsize I = 0.21\: A\)

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