Transformers and power usage costs - Page 1
Forum section: electronics / beginners

author: SimonF
author_rank: Contributor
timestamp: February 25, 2022, 04:56:36 pm
content: New to electronics and just learned how a transformer works. We'll, not quite to the level I would have liked, but halfway there. As with nearly everything you learn, the learning usually stimulates more questions than it answers.Learning about primary/secondary isolation made me think "there's no reason a plugged in transformer wouldn't pass current through the primary regardless of whether anything is turned on on the secondary side". So I googled it and that turned out to be the case, but that only started more questions.Assumption 1. I don't know how power companies charge for electricity usage, but I always assumed there would be some power-in minus power-out calculation so that you only pay for what you use up rather than what passes through. Otherwise power companies will be able to charge several consumers for essentially the same parcel of current. Say some current goes through an incandescent light, for instance. Some will be lost to heat, but some will go clean through (and perhaps into the lamp of the neighbour/s?).If assumption 1 is true wouldn't using a transformer mean you pay only for losses to heat on the primary side rather than current used to produce work on the secondary? The secondary is isolated, so you aren't directly spending current from the grid because it doesn't flow through the secondary, right?If assumption 1 is wrong does it mean you pay for maximum current on the primary regardless of work done on the secondary side, and even if you have 'energy efficient appliances' running from the secondary side it doesn't matter? Because the primary side will just run at some standard VA whether or not anything happens on the secondary?

author: TimFox
author_rank: Super Contributor
timestamp: February 25, 2022, 04:59:55 pm
content: If you apply AC voltage to the primary of a transformer with no load on the secondary, the input looks like a large inductance ("magnetizing inductance") in series with a relatively small resistance ("copper loss").Small transformers may have too small a magnetizing inductance for optimum efficiency due to cost and weight constraints, but with high-power transformers the inductance should be large and the current through it relatively small (compared to the maximum current rating).The simple transformer model comprises a magnetizing inductance, leakage inductance (due to coupling less than unity, an interesting topic on its own), the ratio between primary and secondary, and the loss resistances.

author: Ground_Loop
author_rank: Frequent Contributor
timestamp: February 25, 2022, 05:48:30 pm
content: The magnetic field flux caused by the primary winding induces a voltage on the secondary winding that will cause a current to flow once the external circuit is completed. What you pay for is the actual real energy that that enters your home as measured by the meter outside your home. All of the losses incurred in the transmission equipment between the generating source and your home are well understood by the utility company and baked into the rate you pay for energy.

author: edavid
author_rank: Super Contributor
timestamp: February 25, 2022, 05:50:38 pm
content: Quote from: SimonF on February 25, 2022, 04:56:36 pmThis is wrong. For an ideal transformer, when the secondary is not loaded, there's no primary current. For a real transformer there is a small amount of leakage - perhaps 0.5W-1W of power waste for an old fashioned wall wart.QuoteHouses are wired in parallel, not series, so there is no current "passing through".QuotePrimary current is proportional to secondary current.

author: m k
author_rank: Super Contributor
timestamp: February 25, 2022, 06:52:35 pm
content: Quote from: SimonF on February 25, 2022, 04:56:36 pmElectricity is a middle man.You pay what your energy efficient appliances do with that electricity.You are billed what your energy meter says.It's voltage * current * time.It's also energy over that transformer.By doing so it changes its shape but it's not disconnected between sides.There are some temporal differences but that's simply because it's real and not timeless.

author: Benta
author_rank: Super Contributor
timestamp: February 25, 2022, 06:57:21 pm
content: Quote from: edavid on February 25, 2022, 05:50:38 pm

author: dmills
author_rank: Super Contributor
timestamp: February 25, 2022, 07:16:21 pm
content: Quote from: TimFox on February 25, 2022, 04:59:55 pmAnd (Germaine to the subject of energy costs) because the magnetising inductance is an inductance the magnetising current is 90 degrees out of phase with the applied voltage, so while current flows the real power (ignoring the copper losses) averages to zero over the cycle.In most places domestic power consumption is measured based on real power (Irms * Vrms * cos theta) where in this case theta is 90 degrees...Power companies have the concept of Real power and apparent power (as well as several other things), but for DOMESTIC metering it is usually real power that matters, industrial power pricing can get WAY more complicated.

author: Siwastaja
author_rank: Super Contributor
timestamp: February 25, 2022, 07:30:32 pm
content: Quote from: edavid on February 25, 2022, 05:50:38 pmI know you want to simplify, but maybe this is going too far from truth.You could say, with no load on the secondary, primaryreal poweris near zero. So you don't need to pay much.But, current still flows! This is so-called reactive current, causing reactive power. What reactive means, then? Put simply, it means, the transformer "consumes" current - and hence, energy, from the grid, but does not put the energy to the load (because there is no load in our case, remember!), and neither converts it to heat, but instead, stores it internally, and then pushes it back to the grid! This process repeats 50 times a second (in Australia, where the OP is).Because this energy is only borrowed and put back, it does not (usually) cost you*. Except that whenever you transfer energy, you are going to have some losses: in grid itself, in your house wiring, and last but not least, in the transformer itself! As a result, this reactive current causes some very actual loss, albeit small. That loss would be the part you called "leakage".*) terms and conditions may apply.I know this might be totally overwhelming to a beginner, but the key understanding here is that the transformer transfers energy from primary to secondary, but it also tends to "borrow" and "put back" quite significant amounts of energy by just being connected into the grid.

author: Benta
author_rank: Super Contributor
timestamp: February 25, 2022, 07:44:01 pm
content: Depends on your definition of "ideal transformer". If "ideal" means infinite primary inductance (which is what I assumed in my post), the statement is true.But I think we've lost the OP at this point.

author: zeke.
author_rank: Contributor
timestamp: March 01, 2022, 11:29:08 pm
content: Best way to explain this will be with pictures for a beginner. Figure 1 is what an ideal transformer will look like (let’s just consider an ideal transformer, suitable for this situation). Figure 2 is the circuit equivalent of Figure 1. The resistance and inductance on the secondary can be analyzed on the primary circuit because of the magnetic field produced by the primary which connects to the secondary through the iron core of the transformer (mutual flux). We can place the resistance and inductance from the secondary onto the primary circuit by multiplying them by the turns ratio (let’s not get into details for this now but that’s what it is).Let’s consider two scenarios for a transformer. Figure 3, no load on the secondary winding. Figure 4, secondary winding is loaded. Both Figure 3 & 4 can have the same primary voltage but the current will differ because of the total impedance (resistance and reactance) on the circuit. Figure 3 (this is what you googled for primary side pulling current regardless of the secondary) the primary will have something call the “magnetic branch” which will always draw current (Rc has a large resistance so not much current or real power will be pulled). This is just a part of the losses of a transformer. As I mentioned before, you have the resistance and inductance of the secondary in the primary circuit, however the end of the circuit is opened so no current will pass through it. Figure 4 now connects the opened side of the circuit because of the load on the secondary. The total impedance of the circuit will be much less than Figure 3’s total impedance which is why more apparent power (real and reactive, you only pay for real) will be consumed.Lets just forget about inductance for now, If you add the secondary winding resistance and load resistance in series this will be in parallel to Rc. Add the parallel connection, this resistance will be in series with the primary winding resistance which will be less than Figures 3 total resistance. Magnetic branch resistance || load + secondary resistance < magnetic branch resistance.There was a good amount of hand waving going on but I think it’s appropriate just to get a high level understanding of a transformer.