I don't know how transformers work but the primary coil has a closed circuit. If there is 0 load on the transformer it would short circuit

  • \$\begingroup\$ Look up inductors \$\endgroup\$ Commented Mar 26, 2021 at 12:53
  • \$\begingroup\$ Do you mean no load or zero ohms load on the transistor output? A short circuit connected to transformer secondary could very well make it short and burn. \$\endgroup\$
    – Justme
    Commented Mar 26, 2021 at 12:56
  • \$\begingroup\$ @Justme The primary coil is directly connected to Live and Neutral. That's what I'm talking about. I have no idea how transformers work btw \$\endgroup\$
    – Sam Pan
    Commented Mar 26, 2021 at 13:00
  • 1
    \$\begingroup\$ By the way, "short" in the "short circuit" means "wire without impedance", which a coil is not. \$\endgroup\$
    – AlexVB
    Commented Mar 26, 2021 at 13:06
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    \$\begingroup\$ @SamPan "I have no idea how transformers work btw" doesn't this mean that you have to find out how they work before asking why they don't burn? \$\endgroup\$
    – Maple
    Commented Mar 26, 2021 at 15:35

1 Answer 1


If there is 0 load on the transformer it would short circuit

This would only be true if you connected the transformer to DC source (and the coil would consist of an ideal conductor with zero resistance). But normally a transformer will be connected to AC source and then you have to take in mind the reactive resistance of the coil which is \$X_L = 2π·f·L = L·ω\$. With voltage V you get a current of \$I = \frac{V}{X_L}\$. Voltage and current are out of phase by 90 degrees, so this (ideal) transfomer doesn't dissipate any energy.

  • \$\begingroup\$ Could you explain that last statement a little better? You gave a formula for I, so it seems like there is some current. So why does it dissipate no energy? And what does current and voltage being out of phase have to do with that? \$\endgroup\$
    – Tim
    Commented Jun 5, 2021 at 19:13
  • \$\begingroup\$ Yes there is a current. But with an ideal inductance this current leads to a build up of a magnetic field in one half wave of the current. So there is a flow of energy from the source into this magnet field. In the next half wave the magnetic field collapses and the energy stored in the magnetic field flows back to the source. With an ideal coil of zero ohmic resistance the exchange of this energy is lossless. \$\endgroup\$
    – Elec1
    Commented Jun 6, 2021 at 21:18

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