It can be shown that a square wave can be represented by a sum of sine waves as given by the formula below
v = v0sinωot + v1sinω1t + v2sinω2t + v3sinω3t +...
Solar cells have an output that is dc. The pd across a cell depends on the incident radiation and the efficiency of the solar cells – Hubble telescope cells have an efficiency of about 30% in full sunlight. Solar cells for use on buildings etc. normally have an efficiency of about 15%. The dc power from the solar cells has to be transmitted to the mains which is 240V ac. At your disposal you have electronic black boxes. A possible solution is sketched out in Figure 2.1
The image did not show, so I have attached it here http://www.physics.ox.ac.uk/olympiad/Downloads/PastPapers/BPhO_Round_2_Paper_2014.pdf
It is question number 2.
What is the black box for? Suggest components for the black box.
Some electrical energy may be lost as heat. How can you maximise the efficiency of the electrical energy transfer from the solar panels to the 240 V ac of the mains?
Explain why electrical energy is transferred by the grid using three or four wires.
It has been suggested that one way of storing electrical energy would be to store it in a rotating flywheel. A simple flywheel should store electrical energy from a 1MW power supply for 20 minutes. Suggest a design for this flywheel.
The Attempt at a Solution
F[/B]or the first part, I am guessing the black box is to convert the dc to ac, and also to ensure you have a 240V supply even though sunlight intensity changes. To change dc to ac, I think you would have a gate that periodically turns the cd current on/off at a frequency of 50Hz, and a combination of an inductor and capacitor to turn the dc into ac. Although I am not sure how you could ensure you always have a 240 V output if you have a variable voltage input. Is there a component that deals with this?
With regards to the efficiency, I am guessing that it is just about having a circuit with as little resistance as possible to reduce the heating effect in the wires. For the same reason, you use a greater number of thinner cables in the grid- less power wasted, but this time it is because the greater wire resistance means that a smaller current flows, and the power lost is proportional to I^2 so you reduce power loss.
I have no idea for the last part. Could I have a hint?
Unfortunately there are no answers to this...
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