Ask Me Anything...I Have An Answer
- Thread starter LuvDaKimChi
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certainly
that's what I am beginning to conclude
seems like the only possible answer
however
if the load is small enough that the transformer is not saturated, THEN you CAN get twice as much power on the output of a step-up transformer as you would if you did not have the transformer.
EXAMPLE:
A step-up transformer doubles voltage from 120 to 240 volts and is capable of transforming 10 amps on the input to 5 amps on the output.
If the load on the transformer is 120 ohms, then without the transformer the current through the load would be 1 amp and WITH the transformer the current through the load will be two amps.
This is not a free lunch because watts into the transformer (V*I) equals the watts coming out
because the amperage on the primary side will always be double that of the amperage on the secondary side (assuming the transformer has doubled the voltage)
concluding, a transformer changes the amperage on both the primary and secondary side (assuming there is a load)
correct ?
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This statement is correct.
Not really. High voltage DC can be transmitted with less loss than high-voltage AC - especially if the wires have to be insulated, as in undersea cables. DC requires more capital expense to transform (you have to convert it to AC and then transform it -- transformers require AC). Also DC generators (and DC motors) have bothersome things like commutators/ brushes -- alternators don't.)
Correct.
This does not sound like the conclusion an expert on the subject would necessarily make.
A cautious "yes" is the best response, here. But it's not because the relationship will no longer be valid. Here's the deal.
In your 2:1 transformer, you have N turns of wire in the primary and 2N turns of wire in the secondary. (More turns = more volts.) If you apply a voltage of 120 volts to the primary coil, you will have 240 volts in the secondary. If you have a load of 100 ohms connected across the terminals of the secondary coil, a current of 240 / 100 = 2.4 amps will flow in the secondary coil. A current of 4.8 amps will flow in the primary coil. This will consume electricity, the primary being connected to the power station) at the rate of 120 x 4.8 = 576 watts.
(Note that, in the secondary, 240 x 2.4 = 576 watts also. However, the transformer is not 100% efficient, so if the current in the secondary is 2.4 amps at 240 volts, the supply current in the primary will not be 4.8 amps, but will actually be 4.9 or 5.0 amps, at 120 volts.)
Now, if the secondary coil has an open circuit, i.e the terminals of the secondary coil are not connected to anything, that means the resistance of the secondary circuit is now infinite ohms. Therefore, the current in the secondary coil is 240 / infinity, which is zero.
The way transformers work, because the resistance of the secondary coil is infinite, the resistance of the primary coil also becomes infinite.
(Actually, the resistance of the coils in a transformer is called impedance, rather than resistance. Depending where you are starting from in your level of knowledge, the explanation of impedance and electromagnetic induction might be too much. Suffice it to say that if there is infinite resistance in the secondary, there will be infinite resistance in the primary.)
So, if there is zero current in the secondary, there will be zero current in the primary. The current that flows through the 2nd coil is determined by the resistance of whatever is connected to the 2nd coil. The current in the primary coil depends on whatever is the current in the 2nd, and is twice the current in the 2nd (plus a little bit).
As everybody knows, if you double zero, you still get zero. Told you it was simple.
It's been a while so help me on something that has slipped my feeble mind.
You have your step up transformer described above connected to the 120 volts and an open circuit connected to secondary. I have a perfect voltmeter (infinite ohms) connected to the secondary ... what voltage will I read ?
Not a trick question - I can't remember.
I do know that lightening will cut off parts of the amplitude causing major disruption
FM does not have this problem
ok
so on the assumption there is no such thing as stupid question (ahem) I shall continue
you stating the difference of current is to be found in the windings of the coils ( that is wound around magnets )
so what is the current in the conducting line between the primary coils and the secondary coils ? Is it the same as the current in the line from the transformer to the resistance?
A perfect voltmeter in the sense of having infinite ohms is imaginary only. Every voltmeter, in order to perform the act of measuring the voltage between two points in a circuit, draws off some current. The higher the resistance of the voltmeter, the smaller the drawn-off current. But the measurement always has to be energized. All you can do is minimize the draw-off, by increasing the resistance (impedance) of the voltmeter.
You can have an active voltmeter, which has its own power supply, which can theoretically behave as if it has infinite resistance. But the way that works is to put back the drawn-off current from its own power supply. The fact remains that in order to measure the voltage across the terminals of a coil, there must be at least a tiny current flowing in the coil.
While not stupid, this is a bad question for you to be asking.
If there is such a thing as a conducting line between the primary coils and the secondary coils, it is a ground wire. It has nothing to do with the normal operation of the transformer.
In the transformer, the primary coil has terminals A and B. The 2ndary coil has terminals C and D. It is common to connect terminals B and D together with a wire and to connect this wire to ground. But during normal operation, no current flows along this wire, and no current flows to ground.
Same with the ground wires on your electrical appliances. The idea is to make sure that all metal things in your house, that you might touch, cannot be at a lethal voltage, even if the metal is in contact with a live wire. But during normal operation, no current flows along the ground wire. Similarly, no current flows from your washing machine to your fridge, even though they are connected together, via their connections to ground.
The two coils of a transformer are no more "connected together", by both being connected to ground, than your washing machine is connected to your fridge. There is no current flowing between terminals B and D, and there is no current in the ground wire. The transformer can operate without the ground wires -- just as fridges and dishwashers can operate without ground wires.
silly me
I asked the wrong question
I understand it is the electromagnetic field that transmits energy between the coils (not current)
my question was meant to be
"what is the current in the conducting wire between the generator and the step-up transformer (call it current 1) in relation to the current in the conducting wire coming out of the step-up transformer and heading towards the resistance? (call it current 2) "
Is Current 1 always greater than Current 2 ?
Quote Originally Posted by IM469
A transformer transforms power into a different voltage but the power going in always equals the power going out: V*I = the same number (watts) on both sides of the transformer. If you plug in a transformer that has 240 V into a 20 ohm resistor - the current will be 12 amps and the power will be 240 x 12 = 2,880 Watts. On the 120 volt side of the transformer with the same power demand the current is 2,880/120 or 24 Amps (not 6 amps).
according to IM469 the answer is yes
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Actually I am very familiar with voltmeters & loading. The idea voltmeter has infinite resistance, the ideal current meter has zero. That is not an answer to my question. Your term imaginary is what we call theoretical.
Lets re-phrase the question so that it matches your voltmeter reality. I have a step up transformer connected to 120 VAC (RMS) and double the voltage such that with a small load , the voltage is 240 volts. I remove the load and put a voltmeter with an input resistance of 10E50 ohms (that is 10 to the 50th power) across the secondary winding - what voltage do I read ?
The transformer has a primary coil and a secondary coil. The two are not connected together. The primary coil receives electrical power (voltage x current) from the mains, i.e from the power station.
The voltage supplied at the primary is dictated by the power station, and is 120 volts in the example. The voltage in the secondary is a fixed multiple of the voltage in the primary, the ratio depending one the numbers of turns in the two coils. In a 2:1 step up transformer, the primary being supplied with 120 volts, the voltage across the terminals of the 2ndary coil will be 240 volts.
The primary and secondary currents are in a 2:1 relationship, but the other way round. The relationship is: primary voltage x primary current = 2ndary voltage x 2ndary current.
The secondary coil is connected to a load, which can be a simple resistance. The thing that determines what current flows in the secondary coil is the resistance of the load. In the example, this resistor is 20 ohms. So the current in the 2ndary is 240 / 20 = 12 amps.
The current in the primary is determined by the current flowing in the secondary The primary current is double the secondary current. So, in the example, the primary current is 24 amps. Again: primary voltage x primary current = 2ndary voltage x 2ndary current. Putting it another way: 1st voltage / 2nd voltage = 2nd current / 1st current.
You can regard it that the two voltages in the transformer are determined by the mains supply to the primary, and by the fact that, in this transformer, 2ndary voltage is double the primary voltage. The two currents in the transformer are determined by the ohms of the resistor in the secondary, and by the fact that primary current is double the secondary current.
The ratio between current 1 and current 2 is dictated by the numbers of turns in the two coils. When the coils are wound 2:1, the voltage ratio will always be 2:1, and the current ratio will always be 1:2.
Who is/are "we"?
Am I missing something here? Of course, the voltmeter across the secondary will read 240 volts.
Any voltmeter, whatever its resistance, will read 240 volts, assuming it's properly calibrated. The voltage in the secondary is never anything but the mains voltage in the primary multiplied by the ratio of the numbers of turns in the coils.
Theoretical (not imaginary) is a term we (students/teachers) to describe circuit design involving mathematical results as opposed to measured results.
You may not miss anything - if a transformer has a no current in the secondary does the ratio of windings have any effect ? The coupling is a changing magnetic field driven by current but there is no current so I was leaning on zero. I guess I should have worded the question - If I have a neon pilot tube that will fire at 90 volts across the transformer - will it light up ? That was my thought behind the question.
