Why does increasing voltage result in a lower current?

[Mark04]

Ok, something about Ohm's law confuses me. People always say high voltage means low current, and low voltage means high current. For example, the car industry is considering putting 48V batteries in their cars for less current and less power loss. So my question is this:

When I set up a basic circuit of a 1V battery and a 1 Ohm resistor, I calculate 1A. However, when I replace the 1V battery with a 2V battery, I get 2A. To me that says more voltage, more current. Can someone PLEASE explain this to me? Thank you!

 

[The Bob]

Ok, something about Ohm's law confuses me. People always say high voltage means low current, and low voltage means high current. For example, the car industry is considering putting 48V batteries in their cars for less current and less power loss. So my question is this:

When I set up a basic circuit of a 1V battery and a 1 Ohm resistor, I calculate 1A. However, when I replace the 1V battery with a 2V battery, I get 2A. To me that says more voltage, more current. Can someone PLEASE explain this to me? Thank you!

Ohm's law working in Potential Divider circuits. Also, 1 ohm resistors are not used very much, althought I know they are because I had to use them. My point is that the resistance is normally quite large because of EMF.

Anyway, look at a light bulb. As you increase the voltage, the current decreases because the resistance increases. You have to remeber that as votage increases, it is likely for the resistance to increase, decreaseing the current.

The Bob (2004 ©)

 

[broegger]

According to Ohm's law voltage and current are proportional for a given resistance:

$$U = RI$$​

If you increase $$U$$, $$I$$ increases correspondingly. When people say that "high voltage means low current" it is probably because they are interested in the power output:

$$P = UI$$​

If you increase the voltage by a factor of 2, you can decrease the current by the same factor and still get the same power output $$P$$.

 

[Integral]

High and low are relative terms. In the same simple circuit, a higher voltage means a higher current. If someone tries to tell you different, feel free to laugh.

I have also heard of the car industry moving to new voltage standards, but had not heard that it was to reduce current, I would guess that it may have to do with the fact that 24V is an industry standard for control systems, perhaps they wish to tap the wide availability of those devices. ..Thats just a guess I have no direct knowledge about it.

EDIT:
If you want the same power out, then if you increase the circuit voltage, you must increase the circuit resistance.

P= I * E
Suppose you want 60W output (maybe the head lights)
If E = 12V; I=5A

if E = 48V; I=1.25A

We have

$$P = I^2 R$$

so for I = 5A; R=2.4ohm
for I = 1.25A ; R=38.4 ohm.

So for the same power out if you increase the voltage you must increase the circuit resistance and reduce the circuit current. So the phrase you left out of the initial statement "reduce voltage, increase current" for constant power.

 

[The Bob]

High and low are relative terms. In the same simple circuit, a higher voltage means a higher current. If someone tries to tell you different, feel free to laugh.

What about potential divider circuits? That is what happens. The voltage increases and the current decreases. This alsi happens, proportionally in a light bulb. Am I wrong?

The Bob (2004 ©)

 

[Integral]

You need to read the phrase "simple circiut" in my post very literally.

 

[The Bob]

You need to read the phrase "simple circiut" in my post very literally.

Fair enough. I apologise.

The Bob (2004 ©)

 

[Integral]

As far as I know Ohms law holds for potential dividers. Could you elaborate?

 

[kanato]

I think he's referring to the voltage that you get between the resistors: $$V = V_{in} \frac{R_2}{R_1+R_2}$$. If you place a load on the circuit at that point, then more current flows through the resistor R_1, which lowers the voltage across R_2 which is what V is measuring.

 

[The Bob]

I think he's referring to the voltage that you get between the resistors: $$V = V_{in} \frac{R_2}{R_1+R_2}$$. If you place a load on the circuit at that point, then more current flows through the resistor R_1, which lowers the voltage across R_2 which is what V is measuring.

This is the general idea. Imagine the potential divider has a variable resistor and a normal resistor. As the resistance increases then the current decreases across that resistor and so the voltage increases. The same is true vice versa.

The Bob (2004 ©)

EDIT: That was condesending of me. That is the idea, not just the general idea. Even with thermistors and LDRs, the principle is the same.

 

[Integral]

Hummm...

No matter what I do to R₂ in a voltage divider, all of the currents and voltages can be found using Ohm's Law. If a load is connected in parallel to R₂ I can still apply Ohm's Law, if I know the current draw of the load I can find its effective resistance, etc. So in my mind this is not an example of a non ohmic device. None ohmic device are light bulbs, a motor, or many active solid state devices, which I believe includes thermistors. In non steady state conditions Ohm's law cannot be applied to this class of device. But it can be applied to purely resistive networks no matter how involved.

 

[The Bob]

Hummm...

No matter what I do to R₂ in a voltage divider, all of the currents and voltages can be found using Ohm's Law. If a load is connected in parallel to R₂ I can still apply Ohm's Law, if I know the current draw of the load I can find its effective resistance, etc. So in my mind this is not an example of a non ohmic device. None ohmic device are light bulbs, a motor, or many active solid state devices, which I believe includes thermistors. In non steady state conditions Ohm's law cannot be applied to this class of device. But it can be applied to purely resistive networks no matter how involved.

I was saying that a potential divider is ohmic. I was also saying that a light bulb is ohmic. One of these must be wrong though and I will go with the potential divider is ohmic.

The Bob (2004 ©)

EDIT: I changed my mind. A potential divider circuit is not ohmic and a light bulb is. The resistance of a light bulb changes and so the voltage and current are porportional but only to the resisitance as it changes. It is still ohmic.

A potential divider circuit is not ohmic because as the resistance of one resistor changes these affects happen:
1. The current over that resistor decreases
2. The voltage over the other resistor increases
This is why I say it is not ohmic because the output come at different places and at different amounts (volts) depending on the resistance. Also, depending on the resistor that changes then the output will change as well.

 

[kanato]

I was saying that a potential divider is ohmic. I was also saying that a light bulb is ohmic. One of these must be wrong though and I will go with the potential divider is ohmic.

The Bob (2004 ©)

EDIT: I changed my mind. A potential divider circuit is not ohmic and a light bulb is. The resistance of a light bulb changes and so the voltage and current are porportional but only to the resisitance as it changes. It is still ohmic.

A potential divider circuit is not ohmic because as the resistance of one resistor changes these affects happen:
1. The current over that resistor decreases
2. The voltage over the other resistor increases
This is why I say it is not ohmic because the output come at different places and at different amounts (volts) depending on the resistance. Also, depending on the resistor that changes then the output will change as well.

The light bulb is not ohmic because the resistance is variable, and current dependent (temperature dependent, really, but the equilibrium temperature of the filament will depend on the current running thru it, so..). I mean, you can take any set of numbers (such as voltage/current measurements), and divide them and get an associated set of resistances, and then say the are proportional but some other factor changes each time as well. But Ohm's law expects that proportionality to be constant.

The voltage divider is an example of a non-ideal voltage source.. ie. a voltage source with internal resistance (regular batteries are another example). It would appear to be non-ohmic because for resistances in the range of R_1 and R_2, if you halve the resistance, the current does not quite double. But if careful measurements are made to determine the resistance in the voltage source, it can be treated by Ohm's law with constant resistances.

 

[The Bob]

The light bulb is not ohmic because the resistance is variable, and current dependent (temperature dependent, really, but the equilibrium temperature of the filament will depend on the current running thru it, so..). I mean, you can take any set of numbers (such as voltage/current measurements), and divide them and get an associated set of resistances, and then say the are proportional but some other factor changes each time as well. But Ohm's law expects that proportionality to be constant.

This is where my school teacher got me. I didn't believe her that it was ohmic because the resistance is not constant. It is proportional to the voltage but it is not constant. I'm going to believe what I think in future.

The voltage divider is an example of a non-ideal voltage source.. ie. a voltage source with internal resistance (regular batteries are another example). It would appear to be non-ohmic because for resistances in the range of R_1 and R_2, if you halve the resistance, the current does not quite double. But if careful measurements are made to determine the resistance in the voltage source, it can be treated by Ohm's law with constant resistances.

I think I said that it was not ohmic and explained why. The resistors on their own are ohmic but the system is not.

The Bob (2004 ©)

 

[Integral]

We use Ohms Law to design a voltage divider. Given the resistances every voltage and current in a voltage divider can be found using Ohm's Law. How can it possibly be non Ohmic? If you are using a voltage divider to drive a non Ohmic device, then the system will be non ohmic, other wise it is indeed ohmic.

 

[The Bob]

We use Ohms Law to design a voltage divider. Given the resistances every voltage and current in a voltage divider can be found using Ohm's Law. How can it possibly be non Ohmic? If you are using a voltage divider to drive a non Ohmic device, then the system will be non ohmic, other wise it is indeed ohmic.

What I am saying is that as the voltage of one of the resistors increases then the current across the other one is decreased. I am not saying that one resistor is non ohmic because that would be silly. I am saying the whole system is non ohmic in the way one resistor affects the voltage or current of the other.

The Bob (2004 ©)

 

[pervect]

I havean't read the whole thread, but it is not possible to take a combination of linear circuit elements (resistors) that obey ohms law, and make a non-linear circuit out of them.

 

[The Bob]

I havean't read the whole thread, but it is not possible to take a combination of linear circuit elements (resistors) that obey ohms law, and make a non-linear circuit out of them.

So although the principle I am saying does not follow the law because it is made of Ohmic devices and because, in my description, it needs two resistors and the opposite of each to make it non ohmic; the system is, infact, ohmic.

Right, cool.

Thanks for correcting me everyone. Again, teachers putting ideas in my head.

The Bob (2004 ©)

 

[Integral]

What I am saying is that as the voltage of one of the resistors increases then the current across the other one is decreased. I am not saying that one resistor is non ohmic because that would be silly. I am saying the whole system is non ohmic in the way one resistor affects the voltage or current of the other.

The Bob (2004 ©)

Once again, nothing happens in a voltage divider that is not predicted or calculated via ohms law. Each component is ohmic and the entire network is ohmic.

 

[The Bob]

Once again, nothing happens in a voltage divider that is not predicted or calculated via ohms law. Each component is ohmic and the entire network is ohmic.

See post 18#

The Bob (2004 ©)

 

[Davorak]

Back on topic was the original question ever answered?

Ok, something about Ohm's law confuses me. People always say high voltage means low current, and low voltage means high current. For example, the car industry is considering putting 48V batteries in their cars for less current and less power loss. So my question is this:

When I set up a basic circuit of a 1V battery and a 1 Ohm resistor, I calculate 1A. However, when I replace the 1V battery with a 2V battery, I get 2A. To me that says more voltage, more current. Can someone PLEASE explain this to me? Thank you!

The battery in cars usually have many chemical battery cells in them. For real batteries:

If the battery cells are placed in parallel then all of their currents add together to make a low voltage high current battery.

If the battery cells are placed in series through the current output will be low and the voltage will be high.

Power output should remain the same between the parallel and the series batteries.

So the when they take about high voltage means low current they are probably talking about batteries and not ohmic systems.

 

[Integral]

Actually his question has been answered.

The key issue is power. For the same power a 48V system requires .25 the current of a 12V system. This is clear in the basic power relationship. P = I E If you multiply the voltage by 4 you must cut the current by .25 to maintain the same power.

I suspect a separate effect will be that the auto industry will then be able use 24V components. With 48V as a starting point a 24V regulator will have very stable high current output which can be used to drive many devices. Outside of the auto industry, 24V is a standard control voltage. There is a huge variety of components on the market for 24V. Using this ready supply of cheap components may enable the car makers to cut costs.