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Electricity - HEA 11.7 HEA11.7 HEA 11.7 HEA11.7 HEA TING EFFECT OF ELECTRIC CURRENTTING EFFECT OF ELECTRIC CURRENTTING EFFECT OF ELECTRIC CURRENTTING EFFECT OF ELECTRIC CURRENTTING EFFECT OF ELECTRIC CURRENT

Grade 10CBSE

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

🔑Concepts

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The Heating Effect of Electric Current: When an electric current is passed through a high resistance wire (like nichrome), the resistance wire becomes very hot and produces heat. This is known as the heating effect of electric current.

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

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Electric Power (PP): The rate at which electrical energy is consumed or dissipated in an electric circuit. The SI unit of power is Watt (WW).

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Practical Applications: (1) Electric Heating Appliances like iron, toaster, and heater use nichrome. (2) Electric Bulb: The filament is made of Tungsten because of its high melting point (3380∘C3380^{\circ}C). (3) Electric Fuse: A safety device made of an alloy with a low melting point to protect circuits from overcurrent.

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Commercial Unit of Electrical Energy: The commercial unit is kilowatt-hour (kWhkWh), commonly known as a 'unit'. 1 kWh=3.6×106 J1 \text{ kWh} = 3.6 \times 10^6 \text{ J}.

📐Formulae

H=VItH = VIt

H=I2RtH = I^2Rt

P=VIP = VI

P=I2RP = I^2R

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

E=P×tE = P \times t

💡Examples

Problem 1:

An electric iron of resistance 20 Ω20 \, \Omega takes a current of 5 A5 \, A. Calculate the heat developed in 30 seconds30 \text{ seconds}.

Solution:

Given: I=5 AI = 5 \, A, R=20 ΩR = 20 \, \Omega, t=30 st = 30 \, s. Using Joule's Law of Heating: H=I2RtH = I^2Rt H=(5)2×20×30H = (5)^2 \times 20 \times 30 H=25×20×30H = 25 \times 20 \times 30 H=15000 JH = 15000 \, J

Explanation:

The heat produced is calculated by substituting the values into the formula H=I2RtH = I^2Rt. The resulting value is in Joules (JJ).

Problem 2:

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

Solution:

Given: V=220 VV = 220 \, V, I=0.50 AI = 0.50 \, A. Power PP is given by: P=VIP = VI P=220×0.50P = 220 \times 0.50 P=110 WP = 110 \, W

Explanation:

Power is the product of potential difference and current. Substituting the given values gives the power in Watts (WW).

Problem 3:

Calculate the difference in heat produced if a current of 10 A10 \, A flows through a 5 Ω5 \, \Omega resistor for 2 s2 \, s versus 1 s1 \, s.

Solution:

Heat for 2 s2 \, s: H2=102×5×2=1000 JH_2 = 10^2 \times 5 \times 2 = 1000 \, J Heat for 1 s1 \, s: H1=102×5×1=500 JH_1 = 10^2 \times 5 \times 1 = 500 \, J Difference: 1000−500500\begin{array}{r} 1000 \\ -500 \\ \hline 500 \end{array}

Explanation:

We calculate the heat produced for both time intervals using H=I2RtH = I^2Rt and find the difference using subtraction.