Electrical Engineering: Transformer Question for Power Shower

[Sawyer888]

Hi, I'm an apprentice multiskilled engineer, thought I prefer electrical engineering

anywho, my power shower at home stopped working and upon testing there was no ouput from the transformer, 230v/24v DC and it got me thinking, I'm aware of the step up/down voltage theory etc and the windings etc.

If the the voltage is stepped down roughly ten times doesn't that in theory mean the current is stepped up ten times, but surely this can't be the case for a little electric motor in a power shower would be blown to bits if the current was ramped that high

Could someone explain to me what's happening here?

cheers

 

[skeptic2]

The current at the secondary of the transformer is ~10 times the current in the primary. This means that any resistance on the secondary gets transformed or multiplied by that factor squared to the primary.

So if your motor has a resistance of 24 ohms and draws one amp, that load will be transformed to ~2400 ohms at the primary. The current into the primary will be only about 0.1 amp.

The amount of current used is determined by the voltage divided by the resistance. Having a high current available doesn't mean it will all be used by the load.

 

[Sawyer888]

The current at the secondary of the transformer is ~10 times the current in the primary. This means that any resistance on the secondary gets transformed or multiplied by that factor squared to the primary.

So if your motor has a resistance of 24 ohms and draws one amp, that load will be transformed to ~2400 ohms at the primary. The current into the primary will be only about 0.1 amp.

The amount of current used is determined by the voltage divided by the resistance. Having a high current available doesn't mean it will all be used by the load.

So are you saying that whatever the load/resistance is on the secondary coil (24v motor) and its properties directly affect the resistance on the primary coil and therefore causes a volt drop and hence current drop on the primary because whatever resistance the motor has is proportionally stepped up onto the primary side.

That has confused me a bit because I didn't think the secondary coil and its loads etc could effect the primary side, I assumed the primary side was static and it's properties didn't change on the basis of the secondary side

 

[skeptic2]

So are you saying that whatever the load/resistance is on the secondary coil (24v motor) and its properties directly affect the resistance on the primary coil and therefore causes a volt drop and hence current drop on the primary because whatever resistance the motor has is proportionally stepped up onto the primary side.

Yes, within the limits of the transformer. If you put a short on the secondary of a transformer, the primary will also appear as a short. With an open on the secondary the primary will appear as the inductance of the primary.

The voltage drop is the voltage across the terminals of the device. Since the primary of a transformer is across 230 VAC that is its voltage drop. The load on the secondary won't affect that. It will affect how much current is flowing in the primary however.

You can picture the primary as a high inductance coil with a resistor in parallel. The value of the resistance is the value of the resistance on the secondary times the turns ratio squared. When the secondary is open the resistor in parallel with the primary is also open. As you load the secondary with less and less resistance, the resistance across the primary also drops allowing more current to flow. In reality there is no resistor across the primary. All of the primary current flows through the coil. This is just how it behaves.

 

[Sawyer888]

Cheers I think I have a better understanding of this now

nice1 laa