fundamental voltage misunderstanding

[Designer69]

the reason I posted this question is because I got confused with current consumption of a simple heater.

the heater is rated 500W at 240VAC but they changed the connected voltage to 120VAC and said the current will reduce essentially in half. Also, only half the rated power of the heater will be produced.

I was initially thinking current would double because it's still going to try to produce 500W but I guess if the power output gets reduced then current would too.

 

[Smart $]

the reason I posted this question is because I got confused with current consumption of a simple heater.

the heater is rated 500W at 240VAC but they changed the connected voltage to 120VAC and said the current will reduce essentially in half. Also, only half the rated power of the heater will be produced.

I was initially thinking current would double because it's still going to try to produce 500W but I guess if the power output gets reduced then current would too.

Current directly proportional vs. inversely proportional depends on the load type. Many motors attempt constant power consumption, for example. Heaters do not. [Clarification: resistance-type heaters]

BTW, halving the voltage halves the current (approximately) to the heater... so power is actually 1/4 of, likely less than 1/4 of full rating.

 

[kwired]

the reason I posted this question is because I got confused with current consumption of a simple heater.

the heater is rated 500W at 240VAC but they changed the connected voltage to 120VAC and said the current will reduce essentially in half. Also, only half the rated power of the heater will be produced.

I was initially thinking current would double because it's still going to try to produce 500W but I guess if the power output gets reduced then current would too.

Whether or not the heater power changes depends on if it has multiple elements and whether you can connect them in series for high volts and parallel for low volts. Simple baseboard heaters typically aren't designed for dual voltage, but some industrial heating elements sometimes are.

If it is designed that this is possible then you still have same wattage either voltage and double the current at low voltage.

If you simply apply half volts to the element without any other changes then current is reduced by half also and power ends ups 1/4 of what it originally was.

 

[ActionDave]

Here are some old posts I keep in my bookmarks from one of my favourite members. Bless you Larry wherever you are.

The whole higher-voltage-means-lower-current idea is easy to misunderstand.

That only applies in the manufacturing of electrical equipment. For a given amount of power, as in watts, a higher voltage allows a lesser current to do the same work. mathematically, volts x amps = watts.

A heater manufacturer can design a heater to deliver 1000 watts of heat while running on 120v, and another one to deliver 1000 watts while running on 240v. They do this by making the heater for 240v element four times the resistance of the 120v element.

To maintain a given power level, when you apply twice the voltage, you must halve the current. For a given resistance, when you double the applied voltage, the resultant current also doubles, which would result in quadruple the current.

If you merely doubled the resistance, applying twice the voltage would result in the same current, doubling the power. Therefore, in order to produce the same power with twice the voltage, it takes four times the resistance to cut the current in half.

That's why electrical equipment has to be constructed to operate on a specified voltage (or voltage range). The resistance has to be made to allow just the right amount of current to flow in order to produce the power, which electrical equipment converts into work.

In the real world, however, you have to remember which factors are the constants and which are the variables. For a given piece of equipment, or a human body, the resistance is what it is, depending on everything that affects resistance. The resultant current will depend on the applied voltage.

Ohm's Law states that one volt will push one amp through one ohm. The formula can be stated three ways, depending on the unknown factor:

E = I x R to find voltage
I = E / R to find current
R = E / I to find resistance

In our case, we will know the resistance and the voltage. The unknown will be the current, so we use I = E / R.

Let's start with 120v and 60 ohms: I = E / R = 120 / 60 = 2 amps.

Now, double the voltage: I = E / R = 240 / 60 = 4 amps.

Note that, because P = E x I, notice that as the voltage doubles, and the resultant current doubles, the power quadruples: 120 x 2 = 240w. 240 x 4 = 960w.

Your error points out exactly what I was explaining. In order to conserve a given power level, the equipment has to be designed for the available voltage, so we wouldn't have a 'given resistance' for equipment meant for a different supply voltage.

To keep the same power level when doubling the supply voltage, the load must have four times the resistance, so again, we can't call it 'a given resistance'.

P = E x I = 120 x 2 = 240w. I = P / E = 240 / 240 = 1a.

Now we have a halving of the current when we doubled the voltage, but that's because we intentionally altered the resistance of the load to suit the new supply voltage:

R = E / I = 120 / 2 = 60 ohms. R = E / I = 240 / 1 = 240 ohms.

Twice the voltage; half the current; same power; four times the resistance.

 

[Besoeker]

Ohm's law specifically refers to voltage DROP across a resistive component. Not the voltage of the full circuit.

Not really relevant.
It is a simple relationship between Amps, Volts and Ohms. Where the vots come from is neither here nor there.

 

[Besoeker]

Many motors attempt constant power consumption.

The load they drive determines power consumption.

 

[Smart $]

The load they drive determines power consumption.

Not completely. At zero volts, consumption is zero no matter the load. Yet with volts greater than zero the consumption is... well, could be zero or nominal power or perhaps something else. The whole while, the load could be unchanged. Which leads me to formally declare the statement was made with the express intention that all other parameters were to be unchanged.

 

[Carultch]

Not really relevant.
It is a simple relationship between Amps, Volts and Ohms. Where the vots come from is neither here nor there.

I think it is relevant. You have to keep track of what Volts, what Amps, and what Ohms specifically apply in a particular application of Ohm's law.

In this example, the confusion is that more voltage means more current according to Ohm's law. Which would make someone think that given a fixed transmission line resistance, more voltage on the primary transmission line, would mean more current through it, and therefore more power (V^2/R) dissipated in it. But this isn't the case. Because that particular resistance is only part of a series circuit. The other part of the circuit, is the primary coil of the transformer.

In my schematic, it is (V0-V1) that is the voltage across the power transmission line resistance. And power loss in the transmission line is (V0-V1)^2/Rt. Not V0 itself, the voltage of the source.