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Electricity: Magnetic and Heating Effects - How Does a Battery Generate Electricity?

Grade 8CBSE

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

🔑Concepts

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Heating Effect of Electric Current: When an electric current II passes through a high-resistance wire (like Nichrome), the electrical energy is converted into heat energy HH.

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Joule's Law of Heating: The heat produced in a conductor is directly proportional to the square of the current (I2I^2), the resistance of the conductor (RR), and the time (tt) for which the current flows.

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Electric Fuse: A safety device that works on the heating effect of current. It consists of a wire with a low melting point that melts and breaks the circuit if the current exceeds a safe limit.

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Magnetic Effect of Electric Current: Discovered by Hans Christian Oersted, this effect states that a current-carrying conductor behaves like a magnet and creates a magnetic field around itself.

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Electromagnet: A coil of insulated wire wound around a soft iron core. It acts as a magnet only when current flows through it. Its strength can be increased by increasing the number of turns nn or the current II.

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How a Battery Generates Electricity: A battery converts chemical energy into electrical energy. Chemical reactions within the electrolyte create a potential difference VV between the two terminals (electrodes). This difference in potential drives the flow of electrons through an external circuit.

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Electric Bell: An application of the magnetic effect of current where an electromagnet is used to pull an armature to strike a gong.

📐Formulae

H=I2RtH = I^2 R t

V=I×RV = I \times R

P=V×IP = V \times I

E=P×tE = P \times t

💡Examples

Problem 1:

An electric iron has a resistance of 40 Ω40\,\Omega and takes a current of 5 A5\,\text{A}. Calculate the heat developed in 30 seconds30\,\text{seconds}.

Solution:

Given: I=5 AI = 5\,\text{A}, R=40 ΩR = 40\,\Omega, t=30 st = 30\,\text{s}. Using the formula H=I2RtH = I^2 R t, we get: H=(5)2×40×30H = (5)^2 \times 40 \times 30 H=25×40×30H = 25 \times 40 \times 30 H=30000 JH = 30000\,\text{J}

Explanation:

The heat produced is calculated by squaring the current and multiplying it by the resistance and the time interval.

Problem 2:

Two electrical appliances produce heat energies of 5675 J5675\,\text{J} and 3425 J3425\,\text{J} respectively in a circuit. Calculate the total heat energy using vertical addition.

Solution:

5675+34259100\begin{array}{r} 5675 \\ + 3425 \\ \hline 9100 \end{array} Total Heat Htotal=9100 JH_{total} = 9100\,\text{J}.

Explanation:

To find the total energy generated in the system, we sum the individual heat values produced by each appliance.

Problem 3:

A battery maintains a potential difference of 12 V12\,\text{V} across a circuit with a resistance of 3 Ω3\,\Omega. What is the current flowing through it?

Solution:

Given: V=12 VV = 12\,\text{V}, R=3 ΩR = 3\,\Omega. According to Ohm's Law: I=VRI = \frac{V}{R} I=123=4 AI = \frac{12}{3} = 4\,\text{A}

Explanation:

The current is determined by the ratio of the potential difference provided by the battery's chemical reaction to the total resistance of the circuit.