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Physics going on in a transformer

Former Community Member
Former Community Member

I’m trying to get a better understanding of the physics going on in a UK site transformer, those that reduce the UK mains voltage to 110V.



I’m fully aware of the basic operating principles of a transformer and the basic Power in equals V*I . Power out is the same as power in and induced voltage is given by Faradays law dependent upon the number of turns, and changing magnetic flux with time. With these we can work back from a load to determine the current draw from the secondary side of a transformer.



What I would like clarified is, depending upon the varying load on the secondary side, what at different loads is happening to the magnetic field strength in the core and the power in the primary side.



For example when I have a 1800W tool connected to my 3.3Kva 110V transformer and the power is on in the primary side but the load tool is not switched on. Obviously there is current flowing in the primary coil since there are no switches on these transformers apart from the socket switch. This will create a magnetic field which will induce an emf on the secondary windings.



If there is no load on the secondary side, is the magnetic field strength in the core and the power in the primary at its maximum even with no load on the secondary?



Or is primary input current much lower in comparison to what it could be if the transformer secondary is loaded to its maximum current output? Does putting a load on the secondary cause a feedback through the magnetic field resulting in the current increasing in the primary, hence increasing the magnetic field?



Thanks


Parents
  • It's not memory Jon, it's the switch on at peak'ish voltage is not current limited by the existing magnetic field so the back EMF, which sets the current is given by V= -L dI/dt. So to give say -350-ish volts back EMF in a very short period gives a large I, which trips the circuit. The combination of the R and L of the primary coil gives something called a time constant, which is the next bit of theory, but you should consult Hughes or a similar textbook.
Reply
  • It's not memory Jon, it's the switch on at peak'ish voltage is not current limited by the existing magnetic field so the back EMF, which sets the current is given by V= -L dI/dt. So to give say -350-ish volts back EMF in a very short period gives a large I, which trips the circuit. The combination of the R and L of the primary coil gives something called a time constant, which is the next bit of theory, but you should consult Hughes or a similar textbook.
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