Why can't transformers work with direct current?

[jeff1evesque]

A transformer consists of a conductor (often times can be an iron core), which is used to transfer current/voltage from one circuit (sometimes power lines) to another circuit (homes- often referred to as "stepping down"). This is done by taking a wire connected to the first circuit and wrapping it around the conductor. On the opposite end of the same conductor a different wire is wrapped around it- which leads to a second circuit. Depending on the number of coil turns N around the conductor- from both wires- the voltage/current from one circuit can be stepped down (or "stepped up"). This is defined by faraday's law,

voltage generated = -N \frac{\Delta (BA)}{\Delta t}.

Question:
Why are transformers only used in alternating current, and cannot be utilized in direct current? Since the first wire will contain current, there will be an induced magnetic field in the conductor (iron-core). So if we connect a second wire on the same conductor, it will have a proportional current (based on coil turns, and faraday's equation) regardless of whether it is AC or DC. So based on my thinking, why can't DC utilize this sort of idea of a transformer- since current/voltage will exists in both wires regardless of circuit type (AC/DC).

Thanks,


JL

 

[kuruman]

Unless the current in the primary is changing, there will be no induced current in the secondary (Faraday's Law). It's not the presence of a magnetic field that induces the current, it is the change in magnetic flux brought about by changes in the current that generate it.

 

[jeff1evesque]

Unless the current in the primary is changing, there will be no induced current in the secondary (Faraday's Law). It's not the presence of a magnetic field that induces the current, it is the change in magnetic flux brought about by changes in the current that generate it.

Got it. I should have picked up on it by looking at faradays equation with the following term \Delta (BA) = \Delta \Phi = change in magnetic flux (as you said). Another question, is faraday's equation derived from other equations (or can be), or is it a definition (derived from experimentation)?

 

[kuruman]

It is an experimental result.