# Power Conditioners in 2021 #### TLS Guy

##### Seriously, I have no life.
In a circuit with step transformer (ie: any power supply) that would incorrect Mark:
No, you are mixed up. If you step up voltage on the secondary and keep power constant then the current will be reduced.

However, if you increase voltage with the same turns ration, then the voltage will increase on the secondary and the current to the load on the secondary WILL INCREASE.
This will increase power to the load and a fry up if the power consumption becomes out of spec.

Sorry, you are confused and not me for sure. ##### Audioholic Overlord
No, you are mixed up. If you step up voltage on the secondary and keep power constant then the current will be reduced.
The same goes in reverse, if you step down the voltage and keep power constant then the current will be Increased. I'm not saying how healthy this would be for the circuit - not very, but that's the point.

P=VI. assuming power draw is constant. Voltage goes down. Guess what has to go up.
Are you saying Ohms Law is all wrong? #### TLS Guy

##### Seriously, I have no life.
The same goes in reverse, if you step down the voltage and keep power constant then the current will be Increased. I'm not saying how healthy this would be for the circuit - not very, but that's the point.

P=VI. assuming power draw is constant. Voltage goes down. Guess what has to go up.
Are you saying Ohms Law is all wrong?
No, I'm not. Your equation assumes constant power. If you increase voltage on the primary, the voltage will go up on the secondary, it has to. If the load resistance is constant, but the voltage goes up then the current has to go up. Power on the load will go up by the square of the voltage increase divided by the resistance.

So if we have a power transformer and the primary voltage is 100 volts, and it is a 2:1 transformer,, then the voltage on the secondary will be 50 volts. If the resistance across the secondary is 50 ohms then the current will be 1 amp. Now lets keep the resistance constant and increase the primary voltage to 120 volts, the secondary voltage will be 60 volts. New lets keep the resistance 50 ohms. The current is now 60 divided by 50 which is 1.2 amps.

Now the power expended across the load in the first example is the square of the current X the resistance. That is 1 X 50 which is 50 watts. After the increase in voltage to 60 volts, which is 1.2 squared X 50, which is 1.44 X 50 which is 72 watts.

Your equation only holds true if the load resistance is increased to keep power constant. However that would not be the case in the load offered by an electronic device. The load resistance would stay to all intense and purposes constant until the heating effect of the increase in power upped the load resistance slightly.

The real take home is that if power grid voltage rises then the heat goes up by increase in V squared/R, which gives the same result as the increase in current squared X the resistance. Those are the power dynamics of the increase in voltage with constant resistance.

F

#### fmw

##### Audioholic Samurai
You have that backwards. Current is directed by voltage. If you decrease voltage current will decrease. If you increase voltage current will increase and cause a fry up.
No, I don't have it backward. Assuming no change in power dissipation, a loss of voltage requires an increase in current draw. It is current that damages electronic circuits much more so than voltage. If you get a voltage sag, the circuit will draw more current to maintain the same power dissipation. Your scenario could occur if a voltage sag were severe enough to cut the power dissipation but I've never seen that personally in audio.

A perfect example is the difference between home electrical service and commercial service. My home service measures about 118 volts. At my store the measurement is 105. If you take a radio from the house to the store and play at the same volume level, the radio circuitry will draw more current. It is what it is.

### Latest posts 