Transformer

[Primarygun]

Here are two questions I am concerning.

1.If we use a solid iron core instead of a laminated one, the Voltage induced will decrease.

My teacher said it is, simply by saying more energy loss as heat.

P=V^2 / R

Power distributed to the secondary coil increases.

So V increases,keeping R is constant.

But I don't think so.

[i think the current can increase instead]

2.This question is relevant to my thought.

How is the current in the secondary coil determined?

by V=IR?

I think it is by P=IV.

By then, how could any appliance in the secondary coil affect the current?

Consider a step-up transformer.

If the V in 2nd coil is very large, V^2/R will violate the law of conservation of energy.

[insane_alien]

!. the power output decreases because of eddy currents. kind of like whirlpools in water. this wastes energy. the laminate just prevents these eddy currents from getting too big/forming.

[Primarygun]

But, we can't say that the power for the 2nd coil increases, so V increases, right?

[insane_alien]

yeah

[cyeokpeng]

Here are two questions I am concerning.

2.This question is relevant to my thought.

How is the current in the secondary coil determined?

by V=IR?

I think it is by P=IV.

By then' date=' how could any appliance in the secondary coil affect the current?

Consider a step-up transformer.

If the V in 2nd coil is very large, V^2/R will violate the law of conservation of energy.[/quote']

 

The above analysis is wrong.

During the step-up process, extra "resistance" is created in the secondary coil due to impedance tranformation from the primary coil to secondary coil which is dependent on the turns ratio.

So in effect, Power = VI in the primary coil = VI in the secondary coil.

 

Correct me if I am wrong! Thanks

[Primarygun]

Here's a simplified question and my question 2.

I hope you guy may neglect the original one.

In a complete circuit, a substance is providing energy at a fixed rate, say P.

A light bulb of resistance R is connected in series.

P=VI,

We apply a desired voltage V1.( like the case generated by a transformer)

Now, what's power consumed by the bulb.

I found that I couldn't use V=IR.

Please help me.

[Douglas]

Primarygun, I found your post #1 to be confusing, so I didn't respond.

 

In post #6, I assume that "a substance" is an appliance of unknown resistance, in series with a bulb of unknow resistance, in which case, you'd have to measure the total circuit current, and the voltage drop across one device, to determine the power consumption of each device.

[swansont]

Light bulbs are often non-ohmic devices, which is why/because Ohm's law doesn't hold for them.

[Douglas]

Light bulbs are often non-ohmic devices, which is why/because Ohm's law doesn't hold for them.

If yer talking tungsten, they are ohmic, but a wilding fluctuating ohmic.....at any temperature.

[Primarygun]

In post #6, I assume that "a substance" is an appliance of unknown resistance, in series with a bulb of unknow resistance, in which case, you'd have to measure the total circuit current, and the voltage drop across one device, to determine the power consumption of each device.

And swansont, sorry please assume light bulb's resistance is relative a constant/

The current is fixed as we fix the emf.

So, how does the resistor affect the circuit?

[swansont]

I don't see why the equations don't work if you know that Ohm's law will apply.

[Primarygun]

We get energy from burning fossil fuels.

The power 50 W we could get is fixed.

P=IV.

We just supply the electrical energy produced to a complete circuit which connects with a light bulb of resistance 10 ohms.

We apply the initial EMF as 240k V as high as the power station used to apply.

By P=IV, I=1/(240k).

Now,a current I=1/240k is flowing through the bulb.

Using P=I^2R, P used up by bulb is not equal to the one from burning fossil fuels.

Why?

[swansont]

You can't put just any load on the circuit. By choosing this voltage, you have dictated the kind of load you need. Circuits are designed to match these up.

 

Impedance matching

[Janus]

We get energy from burning fossil fuels.

The power 50 W we could get is fixed.

P=IV.

We just supply the electrical energy produced to a complete circuit which connects with a light bulb of resistance 10 ohms.

We apply the initial EMF as 240k V as high as the power station used to apply.

By P=IV' date=' I=1/(240k).

Now,a current I=1/240k is flowing through the bulb.

Using P=I^2R, P used up by bulb is not equal to the one from burning fossil fuels.

Why?[/quote']

 

Power is not energy. Power is measured is in watts and energy in watt-seconds (or joules, BTUs, Kwhs etc.)

 

If you burn a certain amount of fossil fuel, you will get a certain amount of Energy which equates to supplying a certain number of watts for a certain amount of time. Supplying 50 watts for 1 sec is the same as supplying 1 watt for 50 secs.

 

So let's assume that you use the 50 joules of energy gained from burning your fossil fuel to "spin up" a generator. Ignoring friction, the generator will continue to turn until a load is placed on it. You hook up your 10 ohm light bulb. The resistance of the light bulb determines the current, so:

 

[math]I= \frac{250kv}{10 \Omega}= 25ka [/math]

 

[math]P = 250kv(25ka)= 6.25 gw[/math]

 

50 joules = 50 watt-seconds so

 

[math]t= \frac{50w-ses}{6.25gw} = 0.000000008 sec. [/math]

 

meaning that the bulb will glow for 0.000000008sec before using up the energy stored in the spun up generator.

 

On the other hand:

If you burn the fossil fuel at such a rate such to allow the generator to supply a constant 50 watts, then it doesn't matter how much wattage the bulb could use, it is only supplied with 50 watts. The amperage will drop off to the point where the wattage used by the bulb equals that supplied by the generator. (As the bulb demands more current, this current causes a counter force in the generator which opposes its rotation. The fossil fuel has to supply the energy to overcome this opposition. A balance is met where the current supplied matches the rate at which the fuel is burned.) While the load determines how much currents it demands, the source determines how much current can be supplied. And if the former exceeds the later, you get the later.

 

It is for this same reason that you cannot simply string 8 AAA batteries in series to jump start a car. The AAA's have sufficient voltage (12v) but they cannot supply the needed current.

[Douglas]

Supplying 50 watts for 1 sec is the same as supplying 1 watt for 50 secs.

I don't think so............

 

one is 50W/sec

 

the other 1/50W/sec

[Janus]

I don't think so............

 

one is 50W/sec

 

the other 1/50W/sec

 

It is in terms of Energy; a 50 watt light bulb burning for 1 sec uses the same energy as a 1 watt bulb burning for 50 sec.

 

one is 50w x 1sec

 

the other is 1w x 50 sec.

 

energy is measured in watt-seconds not watts/sec.

 

An anology would be velocity and distance traveled. Power (wattage) is equivalent to velocity and energy(joules) is equivalent to distance traveled.

 

You cover the same distance if you travel at 50 meter/sec for one second as you would traveling at 1 meter/sec for 50 secs. A watt is 1 joule/sec.

[Douglas]

Yes, you're right, I was looking at it in terms of ...rate of consumption.

[cyeokpeng]

Power is not energy. Power is measured is in watts and energy in watt-seconds (or joules' date=' BTUs, Kwhs etc.)

 

If you burn a certain amount of fossil fuel, you will get a certain amount of Energy which equates to supplying a certain number of watts for a certain amount of time. Supplying 50 watts for 1 sec is the same as supplying 1 watt for 50 secs.

 

So let's assume that you use the 50 joules of energy gained from burning your fossil fuel to "spin up" a generator. Ignoring friction, the generator will continue to turn until a load is placed on it. You hook up your 10 ohm light bulb. The resistance of the light bulb determines the current, so:

 

[math']I= \frac{250kv}{10 \Omega}= 25ka [/math]

 

[math]P = 250kv(25ka)= 6.25 gw[/math]

 

50 joules = 50 watt-seconds so

 

[math]t= \frac{50w-ses}{6.25gw} = 0.000000008 sec. [/math]

 

meaning that the bulb will glow for 0.000000008sec before using up the energy stored in the spun up generator.

 

On the other hand:

If you burn the fossil fuel at such a rate such to allow the generator to supply a constant 50 watts, then it doesn't matter how much wattage the bulb could use, it is only supplied with 50 watts. The amperage will drop off to the point where the wattage used by the bulb equals that supplied by the generator. (As the bulb demands more current, this current causes a counter force in the generator which opposes its rotation. The fossil fuel has to supply the energy to overcome this opposition. A balance is met where the current supplied matches the rate at which the fuel is burned.) While the load determines how much currents it demands, the source determines how much current can be supplied. And if the former exceeds the later, you get the later.

 

It is for this same reason that you cannot simply string 8 AAA batteries in series to jump start a car. The AAA's have sufficient voltage (12v) but they cannot supply the needed current.

 

I have realized the limitation of current in embedded system design of a particular application. The design is logically correct in terms of flow diagrams, but because, it drives plenty of LEDs, including the 7-segment display and a pair of dc motors, if the choice of batteries is not correct, even though the voltage input is right, the system will not work!

[Primarygun]

Thanks guys.

We compute the Impedance matching only for maximum power?

And, why does the proof of Vp/Np=Vs/Ns involves power input=power output?