Voltage & current

[Qojoe]

Hi everyone, ive been doing research online and cant find a simple example to help me visualize what im trying to figure out. From the reading ive been doing, if you increase the voltage in a circuit the current is is lower. Can someone explain to me how that works. I hope its not a silly question but i need help visualizing this.

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[ron]

Generally, a device will draw a particular amount of power.

Since Power equals Voltage times current (P=VI), as you raise the voltage, the current has to decrease if the power drawn by the device doesn't change.

https://en.wikipedia.org/wiki/Electric_power

 

[drcampbell]

If you increase the voltage without changing anything else the current will increase.

If you select a different voltage to accomplish the same task -- stepping up the voltage for long-distance transmission, or configuring a motor for 480v instead of 240v, for example -- the power consumption will be the same, which will demand less current at the higher voltage.

Ohm's law doesn't contain loopholes.

 

[kwired]

Hi everyone, ive been doing research online and cant find a simple example to help me visualize what im trying to figure out. From the reading ive been doing, if you increase the voltage in a circuit the current is is lower. Can someone explain to me how that works. I hope its not a silly question but i need help visualizing this.

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There is more to this then just increasing voltage and watching current drop. Resistance type loads current is directly proportional to applied voltage.

Where you run into the situation you are trying to figure out is with power distribution or with "dual voltage" rated equipment, where the power rating remains constant regardless of what input voltage may be.

Take a pair of 600 watt 120 volt rated heating elements and incorporate them into some appliance.

600 watts/120 volts = 5 amps

Place two of those elements in parallel across a 120 volt source and you draw 5 amps each with a net of 10 amps delivered from the source and have overall power of 1200 watts.
Place those same two elements in series across 240 volt source and you still have 5 amps @ 120 volts across each individual element and a power rating of 1200 watts.

They do the same thing with dual volt motors, dual volt transformer windings, etc. Same coils carry same current, have same voltage across them, just are connected in series or parallel depending on input voltage being doubled or not, leaving same overall power output.

 

[Strathead]

There are many ways to look at this. First off it is math. Two concepts that HAVE to be grasped to understand electrical theory are "inversely proportional" and "directly proportional". These terms are absolute. Inversely proportional means that as one goes up the other goes down an equal amount so if I have 4 of one item and 6 of another item and I only have room for 10, if I add two of the first item, I have to take away two of the second item to maintain balance. Directly proportional means that as one goes up or down the other goes up or down mathematically the same. so if the ratio is 4 to 1, then when the one goes up to 2, the 4 goes up to 8.

So all that said, current and voltage are inversely proportional. For a visualization, Think of a water spigot. If I fill a water tank, with the spigot all the way open, at 10 pounds of pressure, it fills up in x amount of time. If I then increase the pressure to 20 PSI, nothing else changes, then you should be able to conceive that the tank will fill up twice as fast or 2x because twice as much water will come out of the faucet in the same time frame. Generally water pressure equates to voltage, and water flow equates to current flow.

Hope this helps.

If not, then I suggest you take a course on Algebra. It is required for Union apprenticeships.

 

[K8MHZ]

Hi everyone, ive been doing research online and cant find a simple example to help me visualize what im trying to figure out. From the reading ive been doing, if you increase the voltage in a circuit the current is is lower. Can someone explain to me how that works. I hope its not a silly question but i need help visualizing this.

I have read some of the answers you were given. So, with that being said, the true way to visualize a circuit is to build one.

Set up a simple circuit with a 12 volt light bulb. Measure the current with 12.0 volts applied. Then raise the voltage to 14.0 volts and measure the current.

The current will go up, not down.

Now you need to try to understand why you have been told that when voltage goes up, current goes down while a working model shows just the opposite.

 

[K8MHZ]

If not, then I suggest you take a course on Algebra. It is required for Union apprenticeships.

Our local doesn't require a course on algebra, but if you haven't had at least one, you won't even understand the questions on the application exam.

There are always more applicants than seats in class, so the best scores from the mostly math based application exam are used to determine who gets in the class that year.

 

[kwired]

I have read some of the answers you were given. So, with that being said, the true way to visualize a circuit is to build one.

Set up a simple circuit with a 12 volt light bulb. Measure the current with 12.0 volts applied. Then raise the voltage to 14.0 volts and measure the current.

The current will go up, not down.

Now you need to try to understand why you have been told that when voltage goes up, current goes down while a working model shows just the opposite.

Now take two identical 12 volt bulbs and put them in series across a 24 volt source. You doubled the voltage, doubled the watts, but current remained the same as with one bulb on 12 volts. Voltage across each individual lamp also remained at 12 volts - watts of each individual lamp remained the same as well.

This is where confusion from people that ask questions like OP did come from - differences in applications. OP may very well be wondering why a motor rated 20 amps at 120 volts is rated for 10 amps at 240 volts. Answer is you are getting same VA output but you are reconnecting the windings within that motor to divide the current differently for each voltage input.

 

[Carultch]

I have read some of the answers you were given. So, with that being said, the true way to visualize a circuit is to build one.

Set up a simple circuit with a 12 volt light bulb. Measure the current with 12.0 volts applied. Then raise the voltage to 14.0 volts and measure the current.

The current will go up, not down.

Now you need to try to understand why you have been told that when voltage goes up, current goes down while a working model shows just the opposite.

In this case, the resistance of the bulb is constant. Not the power consumption of the bulb. V=I*R is your equation, which shows that Voltage directly determines I (current). Half the voltage, half the current, quarter the power, and the bulb glows much dimmer. Double the voltage (if rated for safe operation), double the current, quadruple the power, and the bulb glows brighter. A bulb cannot automatically change the geometry of the filament to compensate for a change in voltage, in order to preserve power.

Now suppose you still have a 12V lightbulb, and you set up a 2-to-1 power converter to power it at its intended 12 V operation from a 24V source. The current in the circuit from the source to the converter (the primary circuit) will be half of that delivered to the lightbulb (the secondary circuit), assuming an ideal power converter.

The bulb still operates at its intended power that it uses at 12V. Because it itself is operating at 12V. The converter is changing initially 24V to 12V, like an "electrical gear", which allows the source to get better "leverage" on supplying the load a longer distance. Similarly, mechanical gears do a trade of torque and rotation rate, also quantities associated with power, and keep the power nearly the same.

The power converter is not necessarily a transformer, although this is the principle of a transformer's operation. Transformers only work with AC. DC-to-DC converters do the equivalent job for DC.

 

[K8MHZ]

Now take two identical 12 volt bulbs and put them in series across a 24 volt source. You doubled the voltage, doubled the watts, but current remained the same as with one bulb on 12 volts. Voltage across each individual lamp also remained at 12 volts - watts of each individual lamp remained the same as well.

The above is not an accurate experiment. The load is changing, not the voltage.

This is where confusion from people that ask questions like OP did come from - differences in applications. OP may very well be wondering why a motor rated 20 amps at 120 volts is rated for 10 amps at 240 volts. Answer is you are getting same VA output but you are reconnecting the windings within that motor to divide the current differently for each voltage input.

I think the confusion lies in the fact that the examples given as answers are not applicable to the question as asked.

Unless the load changes, current will rise with the voltage. If the current drops as the voltage is applied, the load must have been changed, too. The same is true for the opposite, if the voltage drops and the current rises, like in an overloaded motor, that, too, is a result of a change in the load. In the motor's case, it's a lack of REMF, an interesting subject in itself.

So, now the OP has to understand that if all he looks at is voltage and current and ignores the changes in the load, it will appear that sometimes current goes up with voltage, and sometimes it goes down, which would seem to be inconsistent with physical laws.

 

[K8MHZ]

Now suppose you still have a 12V lightbulb, and you set up a 2-to-1 power converter to power it at its intended 12 V operation from a 24V source. The current in the circuit from the source to the converter (the primary circuit) will be half of that delivered to the lightbulb (the secondary circuit), assuming an ideal power converter.

Very well.

We have the set up, and it's powered up with 24 volts to the converter. The converter is drawing 1.2 amps.

Now I raise the voltage from 24 to 26 volts. Did the current go up, or did it go down?

 

[iwire]

Unless the load changes, current will rise with the voltage. If the current drops as the voltage is applied, the load must have been changed, too.

Try that with a circuit supplying say LEDs and you will find the current drops as the voltage rises. In this case the 'load' changes itself.

The real problem with trying to give a simple answer to the OP is that there is not one. So much depends on the characteristics of the load there is no one size fits all answer.

 

[kwired]

The above is not an accurate experiment. The load is changing, not the voltage.



I think the confusion lies in the fact that the examples given as answers are not applicable to the question as asked.

Unless the load changes, current will rise with the voltage. If the current drops as the voltage is applied, the load must have been changed, too. The same is true for the opposite, if the voltage drops and the current rises, like in an overloaded motor, that, too, is a result of a change in the load. In the motor's case, it's a lack of REMF, an interesting subject in itself.

So, now the OP has to understand that if all he looks at is voltage and current and ignores the changes in the load, it will appear that sometimes current goes up with voltage, and sometimes it goes down, which would seem to be inconsistent with physical laws.

All depends on what one is trying to understand.

A general statement of "voltage goes up current goes down" is true - if the power being delivered remains constant. With a dual voltage motor you are not changing the output power, you are simply changing winding connections from parallel components to series components or vice versa and also making a corresponding change to input voltage, voltage and current at each individual segment of those windings is still the same in either configuration.

 

[Carultch]

Very well.

We have the set up, and it's powered up with 24 volts to the converter. The converter is drawing 1.2 amps.

Now I raise the voltage from 24 to 26 volts. Did the current go up, or did it go down?

Good question.

Depends on the details of how the converter works. If the converter can compensate for this change, it would keep constant voltage on the output, and draw less current. It would draw the same power assuming the efficiency is unchanged, which would mean it would draw 1.1A.

But if it is a passive device like a transformer, it will keep the ratio of voltages constant, power the bulb brighter at 8% greater current and 17% greater power, and therefore draw more 8% more current (1.3A) on the primary. You would have to change the taps on the transformer, physically modifying the geometry of the coils involved, to compensate for the voltage change without changing the operation of the load.

 

[iwire]

Very well.

We have the set up, and it's powered up with 24 volts to the converter. The converter is drawing 1.2 amps.

Now I raise the voltage from 24 to 26 volts. Did the current go up, or did it go down?

That goes right back to the LED supply circuit I brought up, if I raise the line voltage to the driver the line current will reduce.

So the answer is it depends.

 

[K8MHZ]

The real problem with trying to give a simple answer to the OP is that there is not one. So much depends on the characteristics of the load there is no one size fits all answer.

^This.

 

[Strathead]

The above is not an accurate experiment. The load is changing, not the voltage.



I think the confusion lies in the fact that the examples given as answers are not applicable to the question as asked.

Unless the load changes, current will rise with the voltage. If the current drops as the voltage is applied, the load must have been changed, too. The same is true for the opposite, if the voltage drops and the current rises, like in an overloaded motor, that, too, is a result of a change in the load. In the motor's case, it's a lack of REMF, an interesting subject in itself.

So, now the OP has to understand that if all he looks at is voltage and current and ignores the changes in the load, it will appear that sometimes current goes up with voltage, and sometimes it goes down, which would seem to be inconsistent with physical laws.

That is because the question wasn't truly asked with all the specifics either. Just the fact that the question was asked shows that the fundamentals of math are not being applied by the OP. Note I didn't say he doesn't understand them, just that they are being applied. I think it is likely he doesn't understand them but it may just be that he hasn't made the connection.

The real answer is E=I*R P=I*E and the other 8 or so extensions of those two formulas. Algebra.

 

[kwired]

Hi everyone, ive been doing research online and cant find a simple example to help me visualize what im trying to figure out. From the reading ive been doing, if you increase the voltage in a circuit the current is is lower. Can someone explain to me how that works. I hope its not a silly question but i need help visualizing this.

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Well if we stick to what was originally asked " if you increase the voltage in a circuit the current is is lower." then that particular circuit is not a circuit with pure resistance load. It could be an inductive load or some electronically regulated load. Simple resistance load would have an increase in current if you increased voltage. It would also have an increase in power used.

I am used to having to explain the "more volts is less amps" confusion, and it is usually about the difference between supplying an piece of equipment with 240 vs 480 volts or 120 vs 240 volts, but in both cases you are normally changing some connections in those loads to allow for the change in voltage - but the overall VA of the load still remains the same.

 

[ggunn]

Hi everyone, ive been doing research online and cant find a simple example to help me visualize what im trying to figure out. From the reading ive been doing, if you increase the voltage in a circuit the current is is lower. Can someone explain to me how that works. I hope its not a silly question but i need help visualizing this.

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V = IR and P = VI; it's as simple as that.

What's not so simple is figuring out which variables are held constant and which are allowed to change. Depending on that, increasing the voltage can result in either more current or less.

 

[Chamuit]

Hi everyone, ive been doing research online and cant find a simple example to help me visualize what im trying to figure out. From the reading ive been doing, if you increase the voltage in a circuit the current is is lower. Can someone explain to me how that works. I hope its not a silly question but i need help visualizing this.

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Visual (math)

9600VA = 480V * 20A
9600VA = 240V * 40A
9600VA = 120V * 80A

20A * 480V = 9600VA
20A * 240V = 4800VA
20A * 120V = 2400VA