06-17-2012, 10:27 AM
Rich
Transformers are there (that is, step- up or step-down, not pulse types) to transfer power.
They consist of two windings, primary and secondary. The tranfer is through magnetic flux.
Now, if you have no load in the secondary, the primary will present itself as almost pure inductance. This is why the lamp won't glow, or will glow very little, because the reactive impedance is way higher than active one (210 and 240 Ohm).
Now, if you introduce a load, the current in the secondary creates the flux opposite to that created by the primary (Lenz law). This reduces the inductance, and the impedance becomes more active than reactive.
In the ultimate case of a short in the secondary (connected leads, or simply a turn short), the flux of the sec. will prctically consate the one from primary, and the inductance will disappear, leaving us with pure resistance of the primary. That is those 210 and 240 Oms in series.
So, if your reactive impedance is, say, at 1H is 360 ohms, and your active is 450, the total is roughly 550. Short the output and it will drop back to 450. This calculation is for illustration only, typically the inductive is a larger portion.
So, even if the lamp glows with the output open, it will glow more with it shorted. If there is a short in any winding, itwill glow at the maximum no matter whether the secondary is shorted or not.
Of course it is better to test with a generator and a scope at higher frequencies, but this should work too. Better to test with a smaller lamp as the active impedance will limit the current and a large bulb may not glow noticeably, whereas a 5-10 W one will.
Transformers are there (that is, step- up or step-down, not pulse types) to transfer power.
They consist of two windings, primary and secondary. The tranfer is through magnetic flux.
Now, if you have no load in the secondary, the primary will present itself as almost pure inductance. This is why the lamp won't glow, or will glow very little, because the reactive impedance is way higher than active one (210 and 240 Ohm).
Now, if you introduce a load, the current in the secondary creates the flux opposite to that created by the primary (Lenz law). This reduces the inductance, and the impedance becomes more active than reactive.
In the ultimate case of a short in the secondary (connected leads, or simply a turn short), the flux of the sec. will prctically consate the one from primary, and the inductance will disappear, leaving us with pure resistance of the primary. That is those 210 and 240 Oms in series.
So, if your reactive impedance is, say, at 1H is 360 ohms, and your active is 450, the total is roughly 550. Short the output and it will drop back to 450. This calculation is for illustration only, typically the inductive is a larger portion.
So, even if the lamp glows with the output open, it will glow more with it shorted. If there is a short in any winding, itwill glow at the maximum no matter whether the secondary is shorted or not.
Of course it is better to test with a generator and a scope at higher frequencies, but this should work too. Better to test with a smaller lamp as the active impedance will limit the current and a large bulb may not glow noticeably, whereas a 5-10 W one will.
