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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 two main functions: it provides a magnetic path that significantly increases the linkage of the magnetic flux between the primary and secondary coils, thereby enhancing the efficiency of the transformer. The core's material, usually made of iron, allows for more effective conduction of magnetic fields.
The secondary coil, which is the output side of the transformer, generates an induced electromotive force (emf) when the magnetic flux changes. This induced voltage is determined by the ratio of the number of turns in the primary coil to that in the secondary coil.
Step 2
Explain how the efficiency of the transformer is increased by constructing the core in this way.
Answer
The transformer efficiency is enhanced by using thin iron sheets separated by a material, M, which minimizes losses due to eddy currents. Eddy currents, if not minimized, can generate unwanted heating and reduce overall efficiency. By laminating the iron, the area available for these currents is reduced, thus allowing more of the induced magnetic flux to contribute to the transformer's operation rather than being lost as heat.
Step 3
Suggest why engineers have chosen 132 kV for this system.
Answer
Engineers often choose a voltage like 132 kV for long-distance transmission because it strikes a balance between transmission losses and safety. Higher voltages reduce current for the same power transfer, which in turn decreases resistive losses in the transmission lines, making the system more efficient while maintaining manageable insulation and construction costs.
Step 4
Calculate the current in the 25 kV line from the power station.
Answer
To calculate the current in the 25 kV line, we first determine the power on that line using the formula:
Where:
- P is the power (72 MW)
- V is the voltage (25 kV)
- I is the current
First, we convert 72 MW to Watts:
Now rearranging the formula to find I:
I = rac{P}{V} = rac{72 imes 10^6}{25 imes 10^3}
Calculating this gives us:
Thus, the current in the 25 kV line is 2880 A.