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Electricity - Heating effect of electric current

Grade 10CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

The heating effect of electric current occurs when an electric current passes through a conductor of high resistance, causing it to become hot and produce heat. This involves the conversion of electrical energy into thermal energy.

According to Joule's Law of Heating, the heat (HH) produced in a resistor is: (1) directly proportional to the square of current (I2I^2) for a given resistance, (2) directly proportional to the resistance (RR) for a given current, and (3) directly proportional to the time (tt) for which the current flows through the resistor.

Practical applications of the heating effect include domestic appliances like electric irons, toasters, ovens, and heaters which use alloys like Nichrome due to their high resistivity and high melting point.

In an electric bulb, the heating effect is used to produce light. The filament is usually made of Tungsten because it has a very high melting point of 3380C3380^{\circ}C and does not oxidize easily at high temperatures.

An electric fuse is a safety device based on the heating effect of current. It consists of a piece of wire made of an alloy with a low melting point. If the current exceeds a safe limit, the fuse wire melts and breaks the circuit.

Electric Power (PP) is the rate at which electrical energy is consumed in a circuit. Its SI unit is the Watt (WW).

📐Formulae

H=VItH = VIt

H=I2RtH = I^2Rt

P=VIP = VI

P=I2RP = I^2R

P=V2RP = \frac{V^2}{R}

1 kWh=3.6×106 J1 \text{ kWh} = 3.6 \times 10^6 \text{ J}

💡Examples

Problem 1:

An electric iron of resistance 20Ω20 \, \Omega takes a current of 5 A5 \text{ A}. Calculate the heat developed in 30 s30 \text{ s}.

Solution:

Given: I=5 AI = 5 \text{ A}, R=20ΩR = 20 \, \Omega, t=30 st = 30 \text{ s}. Using Joule's Law: H=I2Rt=(5)2×20×30=25×20×30=15000 JH = I^2Rt = (5)^2 \times 20 \times 30 = 25 \times 20 \times 30 = 15000 \text{ J}.

Explanation:

The heat produced is calculated by substituting the values of current, resistance, and time into the formula H=I2RtH = I^2Rt. The final result is in Joules (JJ).

Problem 2:

An electric bulb is connected to a 220 V220 \text{ V} generator. The current is 0.50 A0.50 \text{ A}. What is the power of the bulb?

Solution:

Given: V=220 VV = 220 \text{ V}, I=0.50 AI = 0.50 \text{ A}. Power P=VI=220×0.50=110 WP = VI = 220 \times 0.50 = 110 \text{ W}.

Explanation:

Power is the product of potential difference and current. Multiplying the given voltage and current gives the power rating in Watts (WW).

Heating effect of electric current - Revision Notes & Key Formulas | CBSE Class 10 Science