Where Does Extra Voltage in a Step-Up Transformer Come From?

[learning_phys]

In a step-up transformer, since you can increase the voltage by increasing the number of turns on the second circut, where does this extra voltage come from? why does this not violate conservation of energy?

 

[alphysicist]

Hi learning_phys,

In a step-up transformer, since you can increase the voltage by increasing the number of turns on the second circut, where does this extra voltage come from? why does this not violate conservation of energy?

It does not violate conservation of energy because voltage is not energy.

Instead, if the voltage is increased a certain amount (let's say the voltage is doubled), and if there are no energy losses in the transformer, then what else (besides the voltage) can you relate on each side of the transformer? In other words, if you're getting a larger voltage, what else is changing to satisfy conservation of energy?

 

[learning_phys]

P=IV

so I1V1=I2V2

if V2>V1, then I1>I2?

 

[mgb_phys]

Exactly, the transformer doubles the voltage and halves the current
(actually a little less than half because there are some losses)

 

[learning_phys]

what is the benefit of losing current for gaining voltage?

 

[mgb_phys]

Generally because current flowing through a resistor generates heat.
Consider a power line taking electricity from a power station to a city.
You need to transfer a fixed amount of power (P=IV) and the power lines have a certain resistance, the higher V you can use, the lower I and so the lower losses.

So transformers are used at the power stations to step-up the supply to a high voltage and then it is stepped-down to 110V or 220V when it reaches your house.

 

[learning_phys]

but the higher the voltage, the greater the power (P=IV) so no matter if you use a higher voltage, the equation for power dissipated will still be the same since I1V1=I2V2

 

[mgb_phys]

No the power is the same.
P = IV and V= IR so P = I^2R or P = V^2/R
It's expensive to make R small ( you need bigger or more conductive wires ) so the easiest way to transfer a certain amount of POWER is to use as high a voltage as possible.

This is the main reason for using AC, transformers very efficently change voltage so you can easily step up to 500KV for a transmission line and then back down to 220V for use.

 

[learning_phys]

so you're saying the power dissipated in the wire is P=V^2/R?

then the larger the voltage, the greater the power dissipation... wouldn't that mean we would want a smaller voltage to have less power dissipation?

 

[mgb_phys]

Yes the ideal case is no volts and no amps - but that doesn't run your TV!

Suppose you need to supply 1MW to a small town and your power line is 100 Ohm.
You could send it at 110V with a current of 10,000Amps but that would lose all the power in heating the transmssion line.
Or you could step it up to 1MV and use a current of 1Amp, you would then lose only 100W in the resistor.

Typically you run the major grid at 400 kV * 1kA, and 11 kV * 150A for local supplies
Overall the UK grid loses about 800MW (1 reactor) worth of power.

 

[learning_phys]

power loss from the step up voltage is still: P=IV=1A*1MV=1MW

I don't see how increasing the voltage and decreasing the current helps because P_loss=I1V1=I2V2

 

[stewartcs]

power loss from the step up voltage is still: P=IV=1A*1MV=1MW

I don't see how increasing the voltage and decreasing the current helps because P_loss=I1V1=I2V2

The only power lost in this case is in the wires and is equal to $$P = I^2R$$, hence, the the lower the current the lower the losses. So, if the voltage is doubled by a step-up transformer, the current is cut in half due to the conservation of energy principle (assuming an ideal transformer of course).

The total power from the source is unchanged, hence your confusion.

CS

 

[mgb_phys]

power loss from the step up voltage is still: P=IV=1A*1MV=1MW

No the total power USED is 1MV * 1A = 1MW, but you want to use it all anyway

The power LOST is P = IV but in this case V is the voltage difference across the resistor (the power line) if the resistance is 100R and the current is 1A then from V=IR the voltage drop across the wire is only 100V, you still get 999,900V at the other end.

 

[learning_phys]

i'm still confused, i thought we are apply 1MV across the wires

why do we calculate the voltage for the wire? Where does the 1MV play into this?

 

[stewartcs]

i'm still confused, i thought we are apply 1MV across the wires

why do we calculate the voltage for the wire? Where does the 1MV play into this?

For example: Let's say that I want to distribute 1 million volt-amps. I could use 1000 volts and 1000 amps. However, the I x R drop in the wire would be horrendous. So instead, I use a step-up transformer to pump the voltage up to 500,000 volts, and send 2 amps. I still send 1 million volt-amps, but now a 2 ohm resistance drops only 4 volts in the cable. A 4 volt loss in 500,000 volts is minuscule.

The voltage drop in the wires is just V = IR.

CS

 

[learning_phys]

if the voltage drop in the wires is just 4 volts, how does the other 500kV get transferred?

and how does this favor AC?

 

[mgb_phys]

Voltage doesn't get transferred, you have a voltage difference between two points.

Perhaps it's simpler to think of a battery circuit.
Imagine a battery with 10V, you are connecting it to a heater and you want 10Watts from the heater, so you need 1Amp to flow (P=IV)
But the wire to the heater has a resistance of 1R so you lose 1V across the wire (V=IR) and so the heater has only 9V across it and so only gets 9W.

If you used a much bigger battery with 100V but supplying only 0.1A you would be producing the same amount of power, but this time the voltage drop across the resistor/wire is only 0.1V (V=IR) leaving 99.9V at 0.1A at the other end, you would then have 9.99W for the heater.

 

[learning_phys]

ahhh makes sense now. where does alternating current come into play?

 

[mgb_phys]

The most efficent way to change voltages is with a transformer, transformers need AC.

 

[Gear300]

In a step-up transformer, since you can increase the voltage by increasing the number of turns on the second circut, where does this extra voltage come from? why does this not violate conservation of energy?

Yeah, I actually saw this as confusing too for some time. Try not to think of there being 2 circuits consisting of a primary and secondary circuit. Just think of the transformer as an element of one circuit. That implies that the resistance and other elements of a primary/secondary circuit affects the counterpart circuit.

 

[Gear300]

ahhh makes sense now. where does alternating current come into play?

If the current didn't change, there wouldn't be an induced voltage. Alternating current is constantly changing, so it is an advantage in this situation.

 

[norm]

Question: Low voltage lighting transformer. If a voltmeter (not true RMS) is reading .2, is there a way to determine if the transformer is good without using a true RMS voltmeter?