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Physics: Electricity and Magnetism - Electricity Generation, Transmission, and Power Grids

Grade 8IB

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

🔑Concepts

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Electromagnetic Induction: The process of generating an electric current by moving a conductor through a magnetic field or by changing the magnetic field around a conductor. This is the fundamental principle behind electric generators.

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Electric Generators: Devices that convert mechanical energy into electrical energy. In most power plants, a turbine spins a coil of wire inside a magnetic field to induce an Alternating Current (AC).

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The National Grid: A network of cables and transformers that connects power stations to consumers. It ensures a reliable supply of electricity across a wide area.

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Step-up Transformers: Located at power stations, these increase the voltage (VV) and decrease the current (II). Since power loss in cables is proportional to I2I^2, reducing the current significantly reduces energy wasted as heat.

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Step-down Transformers: Located near homes and factories, these decrease the voltage to safer levels (e.g., 230V230V or 110V110V) and increase the current for consumer use.

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Renewable and Non-renewable Energy: Electricity can be generated using non-renewable sources (coal, gas, nuclear) or renewable sources (wind, solar, hydroelectric). Most methods (except solar) involve spinning a turbine.

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Power Loss in Transmission: When electricity flows through a wire, resistance causes heating. The power lost is calculated using the formula Ploss=I2RP_{loss} = I^2R. This is why high-voltage transmission is essential for efficiency.

📐Formulae

P=V×IP = V \times I

Ploss=I2×RP_{loss} = I^2 \times R

VpVs=NpNs\frac{V_p}{V_s} = \frac{N_p}{N_s}

Vp×Ip=Vs×IsV_p \times I_p = V_s \times I_s

💡Examples

Problem 1:

A transformer has 200200 turns on its primary coil and 40004000 turns on its secondary coil. If the input primary voltage is 230V230V, calculate the output secondary voltage.

Solution:

Using the transformer equation: VpVs=NpNs\frac{V_p}{V_s} = \frac{N_p}{N_s} Rearranging for VsV_s: Vs=Vp×NsNpV_s = V_p \times \frac{N_s}{N_p} Vs=230×4000200V_s = 230 \times \frac{4000}{200} Vs=230×20=4600VV_s = 230 \times 20 = 4600V

Explanation:

This is a step-up transformer because the number of turns on the secondary coil is greater than on the primary coil, resulting in a higher output voltage.

Problem 2:

A power line has a resistance of 5Ω5 \Omega. Compare the power loss when transmitting 10,000W10,000W of power at 100V100V versus 1,000V1,000V.

Solution:

Case 1 (100V100V): I=PV=10000100=100AI = \frac{P}{V} = \frac{10000}{100} = 100A Ploss=I2R=1002×5=10000×5=50,000WP_{loss} = I^2R = 100^2 \times 5 = 10000 \times 5 = 50,000W Case 2 (1,000V1,000V): I=PV=100001000=10AI = \frac{P}{V} = \frac{10000}{1000} = 10A Ploss=I2R=102×5=100×5=500WP_{loss} = I^2R = 10^2 \times 5 = 100 \times 5 = 500W

Explanation:

By increasing the voltage by a factor of 1010, the current decreased by a factor of 1010, which reduced the power loss by a factor of 100100 (10210^2). This demonstrates why high voltage is used in power grids.