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Electricity Part 3

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Published in: Physics
1,642 Views

For class 10

Anita Y / Gurgaon

7 years of teaching experience

Qualification: Masters in physics

Teaches: Mathematics, Physics

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  1. Electricity part 3 for class 10 ( heatinq effects and electric power) BV Anita Yadav M.sc Physics Heating Effect of Electric Current When electric current is supplied to a purely resistive conductor, the energy of electric current is dissipated entirely in the form of heat and as a result, resistor gets heated. The heating of resistor because of dissipation of electrical energy is commonly known as Heating Effect of Electric Current. Some examples are as follows: When electric energy is supplied to an electric bulb, the filament gets heated because of which it gives light. The heating of electric bulb happens because of heating effect of electric current. When an electric iron is connected to an electric circuit, the element of electric iron gets heated because of dissipation of electric energy, which heats the electric iron. The element of electric iron is a purely resistive conductor. This happens because of heating effect of electric current. Cause of heating effect of electric current: Electric current generates heat to overcome the resistance offered by the conductor through which it passes. Higher the resistance, the electric current will generate higher amount of heat. Thus, generation of heat by electric current while passing through a conductor is an inevitable consequence. This heating effect is used in many appliances, such as electric iron, electric heater, electric geyser, etc. Joule's Law of Heating: Let; an electric current I is flowing through a resistor having resistance equal to R. The potential difference through the resistor is equal to V. The charge Q flows through the circuit for the time t. Thus, work done in moving of charge Q of potential difference V = VQ
  2. Since, this charge Q flows through the circuit for time t Therefor; power input (P) to the circuit can be given by following equation: joule's law of heating formula We know, electric current I = Q/t Substituting Q/t = I in equation (i), we get; (ii) P = VI Since the electric energy is supplied for time t, thus after multiplying both sides of equation (ii) by time t, we get or, H = Vit or, H = 12Rt p x t = VI x t = Vlt (iii) Thus, for steady current l, the heat produced (H) in time t is equal to Vlt (iv) We know; according to Ohm's law; V = IR By substituting this value of V in equation (iv), we get; H = IR xlt (v)
  3. The expression (v) is known as Joule's Law of Heating, which states that heat produced in a resistor is directly proportional to the square of current given to the resistor, directly proportional to the resistance for a given current and directly proportional to the time for which the current is flowing through the resistor. Example 1: If an electric heater consumes electricity at the rate of 500W and the potential difference between the two terminals of electric circuit is 250V, calculate the electric current and resistance or, 500 W = 250 V x I or, I = 500 W + 250V=2A or, R = 250 2 A = 125 Q through the circuit. Solution: Given, power input (P) = Potential difference (V) = 250 V Electric current (l) =? 500 w Resistance (R) through the circuit =? We know that power (P) = VI We know, resistance R = V/l Practical Application of Heating Effect of Electric Current & Electric Power
  4. For exploiting the heating effect of electric current, the element of appliances must have high melting point to retain more heat. The heating effect of electric current is used in the following applications: Electric Bulb: In an electric bulb, the filament of bulb gives light because of heating effect of electricity. The filament of bulb is generally made of tungsten metal; having melting point equal to 33800C. Electric iron: The element of electric iron is made of alloys having high melting point. Electric heater and geyser work on the same mechanism. Electric fuse: Electric fuse is used to protect the electric appliances from high voltage; if any. Electric fuse is made of metal or alloy of metals, such as aluminium, copper, iron, lead, etc. In the case of flow of higher voltage than specified, fuse wire melts and protects the electric appliances. Fuse of IA, 2A, 3A, 5A, IOA, etc. are used for domestic purpose. Suppose, if an electric heater consumes 1000W at 220V. Then electric current in circuit I = P/V or, I = 1000 W- 220 V = 4.5 A Thus, in this case a fuse of 5A should be used to protect the electric heater in the case of flow of higher voltage. Electric Power: Sl unit of electric power is watt (W). IW = 1 volt x 1 ampere = IV x IA 1 kilo watt or 1kW = 1000 W Consumption of electricity (electric energy) is generally measured in kilo watt. Unit of electric energy is kilo watt hour (kWh) 1 kWh = 1000 watt X 1 hour = 1000 w x 3600 s
  5. EXAMPLES:- Example 1: If the potential difference is 220V and the power of bulb is 1 IOW, what is the electric current flowing in the circuit? Solution: Given, Potential difference, V = 220V, Power of bulb Electric current (l) =? We know that P = VI Example 2: If the power of an electric heater is 1000W and electricity of 240 V is flowing through it, find the electric current in the electric heater. Solution: Given, Power (P) = IOOOW, Potential difference (V) = 240V, Electric current (l) =? We know that P = VI or, 1000 W = 240 v x I or, 1kWh = or, 110 W = 220 V xl or, 110 w + 220 v = 0.5 A or, I = 1000 W + 240 V = 4.16 A 3.6 x 106 watt second - - 3.6 x 106 J

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