B Why can you choose only V in DC but V and I in AC?

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1. Aug 5, 2016

greypilgrim

Hi.

Assume we have some circuit with resistance $R$. In DC, the current $I$ linearly depends on the connected voltage $V$ by Ohm's law $I=V/R$ .

In AC, we can independently connect any voltage and current to the system by using a transformer, provided we have a power source with $P=VI$ . The resistance of the circuit doesn't even show up.

It's not clear to me why there seem to be more degrees of freedom about the connected power source in AC than in DC. Also, shouldn't DC come out as a limit from AC if the frequency goes to zero?

2. Aug 5, 2016

Merlin3189

This is strange. I don't know what you mean about independent voltage and current, but I don't think you can.

A DC power source is also a power source and P=VI is just as applicable to DC (even more so than in AC, because you also need to take phase into account in AC circuits.)

Because you chose a different formula! V=IR is just the same for AC as DC.

And indeed it does.

3. Aug 5, 2016

perplexabot

The transformer that is used for connecting an AC power supply isolates the supply from the circuit in question, thus the ac supply will not "see" the resistance of the circuit, only the impedance of the primary coil of the transformer (to some extent).

Yes, DC is AC as f->0 but the transformer will no longer be active in that case.

4. Aug 5, 2016

Merlin3189

No. The transformer also transforms the resistance of the load.

Say I have an AC 10V supply and connect it to a 10Ω resistance. Then I =V/R = 10/10 = 1 A.
(Edit here) Now connect that resistance via a transformer, as below.(/edit)
If I now use a transformer to step up the voltage to 20V ( that is a turns ratio of 1:2) then the current will be I=V/R = 20/10 = 2 A.
So the power is P=VI=20x2=40 W and the primary current must be I = P/V = 40/10 = 4 A
Therefore the primary circuit sees the transformer as a resistance of R = V/I = 10 / 4 = 2.5 Ω

The transformer has transformed the resistance in proportion to the square of the turns ratio 12:22=1:4
This is a general principle.

If you connect the transformer the other way round, turns = 2:1
then the voltage is stepped down 2:1 and the secondary resistance will appear increased in the primary circuit 4:1

The current largely follows the same rule, inversely with the turns ratio. But you can't just say, I'll apply 10 V 2 A to the primary and expect 20 V 1 A to flow in the secondary irrespective of the load resistance. The catch there is, you can't get 2A to flow in the primary when a 10 V source is applied, UNLESS there is the right load on the secondary
If I have a transformer with a resistive load attached and there is 10 V across the primary with 2 A flowing in the primary, then that appears as a 10/2=5Ω resistance dissipating 20 W.
Say the transformer is again our 2:1 turns ratio.
Then the secondary voltage is 2x10=20 V.
The secondary current is 2/2=1 A
and the secondary resistance must be either R=V/I=20/1=20 Ω,
or by transformed resistance, Rs=Rpx(turns ratio)2=5Ω x 22 = 20 Ω
or by power calculation, R= V2/P = 400/20 = 20 Ω

BUT, if you now remove that load resistance, then, no current flows in the secondary and no power is dissipated in a load.
The secondary open circuit voltage stays at 20 V and the primary voltage stays at 10V, no current flows in the secondary, so (almost) no current flows in the primary. The source circuit "sees" the transformer as an open circuit - voltage, but no current and no power, an infinite resistance obeying the transformer rules.
The transformed resistance is simply Rp=Rs/(turns ratio)2=∞ /4 = ∞ Ω

So much for independently connecting I and V with a transformer.

If you now reconnect a different load resistance to the secondary, you will get new currents in primary and secondary. The currents are determined by the resistance exactly as they are in a DC circuit.

As for the transformer not working at DC, you may be right, but I think I'd want to look at the specific situation before fully accepting that ruling.

Also, note that real transformers are not ideal transformers and I am neglecting some details here to explain ideal transformer action. If you want to get real, we generally used an ideal model and add things like serial resistance of windings, magnetising flux, core losses and whatever other details as parasitic elements.

5. Aug 5, 2016

Merlin3189

Re. the primary inductance or impedance, which I ignored above.
If no secondary load is connected, then the primary looks like an inductor and when you apply an AC voltage, an AC current will flow.
In a well designed power transformer that current is usually small and ignored in the sort of calculations I'm doing above.
It is also out of phase with the primary voltage, so the product of voltage and current does not give you power.

So long as the primary voltage (AC) remains constant, this out of phase magnetising current remains constant. When a secondary load is connected a (usually much larger) secondary current flows and this causes another primary current. The actual primary current is the (vector) sum of the fixed magnetising current and this transformed load current. This new part of the primary current is in phase with the applied voltage and represents real power.

6. Aug 5, 2016

Tom.G

Maybe it's easier to look at the problem as conservation of energy.

If you have a given power consumption (load) on the transformer secondary, you have the same power entering the transformer primary. You can wind the transformer with whatever turns ratio is needed to reach a desired output voltage, but to maintain the same load power you will have to change the resistance of the load. The transformer primary is typically fed with a voltage source (mains power for instance) so the primary current will depend only on the secondary power consumed and the primary voltage.

(The above is the "ideal" case of course, in the real world there are some minor losses in the transformer that absorb a bit of additional power.)

7. Aug 6, 2016

CWatters

The OP should also look up DC to DC converters. These transform one DC voltage into another DC voltage.

You can insert a DC-DC converter between the source and the load resistor in the same way you can insert a transformer in the case of AC.

If you assume the DC-DC converter and transformer are 100% efficient then it's trivial to apply conservation of energy to both sides. This works for the DC-DC converter just the same as an AC transformer.

Power In = Power out = Vout2/R

8. Aug 6, 2016

perplexabot

All I said was the transformer isolates the input supply from the circuit in question thus the impedance of the circuit connected to the secondary is NOT typically seen at the primary. Is that a wrong statement?

9. Aug 6, 2016

Merlin3189

As far as I understand the term "isolation" used in the context of a transformer, it refers to the lack of electrical connectivity between the primary and secondary. So an extra-low voltage power supply powered by the mains might use a transformer so that you can't get an electric shock from touching the output circuit and ground. If you make such a power supply without a transformer (as in some LED lamps) then even though the output voltage may be just 10V or something like that, it is still possible that the output may be at mains voltage relative to ground and can give you a shock.

For use on building sites and similar hazardous environments,isolating transformers are used to power electrical tools. These transformers are 1:1 turns ratio so do not change the voltage and the current is the same in both primary and secondary. Their only function is to reduce the shock hazard by isolating the secondary circuit from the mains (and sometimes allowing the secondary to be centre tap earthed, so that you have a +120V - 0 - -120V circuit rather than a 240V - 0 circuit and can get only a maximum 120V shock rather than 240V shock when touching one wire and earth.)

But as far as isolating the resistance /impedance of a secondary circuit from the primary circuit, IMO transformers do not do that.

A 1:1 transformer provides this electrical isolation between primary and secondary, but reflects voltage, current, resistance and impedance between the two curcuits. (Within its designed operating range.) Whatever passive circuit (RLC) you connect to the secondary, the primary will appear to be more or less that when an AC voltage of the appropriate frequency is applied. (Again, the transformer has to be properly designed for the frequency you use.)

BTW on the topic of frequency, I've given more thought to your point about DC. Any real transformer is going to fail at DC and in fact at AC frequencies significantly lower than its design frequency. The core is going to saturate and it will no longer be even an approximation to an ideal transformer. This can happen for even modest changes of frequency. Transformers designed for US 60Hz overheat on UK 50Hz and often fail, sometimes quite violently.

Putting aside power transformers for a while, another use is in loudspeaker matching. These days audio amplifiers based on semiconductors generally have a low output impedance and can be directly connected to low impedance loudspeakers. But valves (tubes) have much higher output impedance and can't sensibly be connected directly to normal low impedance speakers. They are always connected via a matching transformer whose main job is to transform or match the impedances. It has a high number of turns in the primary connected to the valve and a low number of turns in the secondary connected to the speaker. The turns ratio may be something like 20:1 (Edit: or more)so that the impedance is transformed in that ratio squared, about 400:1. Then an 8Ω speaker will appear like an 3200 Ω load. The
9V AC or so required to put 20W into an 8Ω speaker is transformed to 180V AC that the valve has to provide across the primary, so in class A you need at least a 360V DC power supply to the valve (which the transformer will safely isolate from your finger when you plug in the speakers.) The secondary current of about 2.25A you need for the speaker is transformed to only 110 mA that the valve has to provide.

Here is a link to this topic, the first I found and not necessarily the best. Try looking at Elliot sound for in depth info on transformers, both power and audio.

Sound distribution systems (often known as Tannoys - a term I hate, as I haven't seen a real Tannoy for years) don't want to send amps of current all round a building or sportsfield, because you'd need thick wires to avoid losing most of your power in the cable resistance. So they use a transformer to step up the voltage of the amplifier output to a nominal 70V. This transformer obviously reduces the output current in the same ratio, resulting in a much smaller current running round all the cables. At each speaker position there is another "70V line transformer" which steps the voltage back down and current up to drive the loudspeaker (and of course provides isolation from the tingly 70V !) What might be interesting to you is the way they are often labelled:

By selecting the right impedance for your speaker and the desired power level for the line connection, you automatically get the right turns ratio to step down the voltage to the right level for the power you want and to step up the impedance so that the line and, from there via the original step up transformer, the amplifier sees the right load impedance. The isolation saves you from shocks, which while not lethal in themselves, might make you fall off the ladder if you were working up there while they were in use. But it definitely does not isolate the load impedance - exactly the opposite in fact.

10. Aug 6, 2016

CWatters

Yes. If you halve the load resistance the secondary current doubles and due to conservation of energy the primary current doubles. To the power source it looks like the impedance of the primary circuit also halved.

11. Aug 6, 2016

CWatters

PS when analysing transformer circuits it's convenient to work forwards from source to load when thinking about voltages but backwards from load to source when thinking about current an power.