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Figure 7 shows a transformer - AQA - A-Level Physics - Question 5 - 2022 - Paper 2
Question 5
Figure 7 shows a transformer. Figure 8 shows a cross-section through the transformer core. Thin iron sheets are separated by material M. Figure 9 shows a schematic... show full transcript
Worked Solution & Example Answer:Figure 7 shows a transformer - AQA - A-Level Physics - Question 5 - 2022 - Paper 2
Step 1
Explain the functions of the core and the secondary coil.
Answer
The core of a transformer serves a crucial role in improving the efficiency of energy transfer between the coils. It provides a path for the magnetic flux generated by the primary coil, effectively linking the primary and secondary coils. This allows for greater magnetic coupling, thereby increasing the amount of energy transferred from the primary to the secondary coil.
The secondary coil, which acts as a conductor, receives the magnetic field induced by the primary coil. The voltage induced in the secondary coil is proportional to the number of turns and the rate of change of magnetic flux, as dictated by Faraday’s law of electromagnetic induction.
Step 2
Explain how the efficiency of the transformer is increased by constructing the core in this way.
Answer
The use of thin iron sheets, separated by material M, minimizes eddy currents within the core. Eddy currents are loops of electric current induced within the core due to changing magnetic fields, causing energy losses in the form of heat. By laminating the core, the cross-sectional area for these currents is reduced, leading to lower resistive losses. Furthermore, the material M acts as an insulator that further reduces energy waste and enhances overall transformer efficiency.
Step 3
Suggest why engineers have chosen 132 kV for this system.
Answer
Engineers often select higher transmission voltages, like 132 kV, to minimize energy losses over long distances. Higher voltage allows for a lower current for the same power transmitted, which reduces resistive losses in the transmission lines, as power losses are proportional to the square of the current (I²R losses). Thus, using 132 kV helps improve the overall system efficiency.
Step 4
Calculate the current in the 25 kV line from the power transformer.
Answer
To find the current in the 25 kV line, we first calculate the power transferred using the formula:
Here, the industrial consumers use 72 MW of power.
Therefore, using the 25 kV line:
Thus, the current in the 25 kV line is approximately 2880 A.