Beginner - Silly questions - Hope someone can help

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fuqur2dom

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Hi,

I am just starting out in the wonderful field of electrical fun. I have some questions that I can't seem to find the answers to, maybe they are just too simple.

1) The relationship between voltage and current. Is it possible to have more current (10amps) with lower voltage (6 volts). The reason I ask is I have read about the static charge machine that fire huge sparks, extremely high voltage but very low current. The person went onto explain that "either very high voltage with low current, or very high current with low voltage" will create these effects.

2) With a simple 9V battery, why do I get a reading of 9 volts, but not 9 amps when no resistors are in the test.

3) This one goes back to my first question. Bench power supplies - I see different ranges on them, but take one that has a range of 0-30V and 0-10 Amps. Does this mean I could have 1 volt with 10amps?

Thanks for any help, I know they are silly questions, I almost feel I know the answer but am somehow thrown off by the post I read.

Thanks,

-D
 
There are two concepts that surround your three questions. I don't know how much you know about either concept yet. So I'll just mention them, and let you let us know what you are trying to learn.

The first concept is called, "Ohm's Law." It applies to everything we touch, though sometimes it is more complicated than at other times. The Law says that "Voltage is equal to Resistance times Current." In the more complicated versions, you replace "Resistance" with "Impedance." In your questions, you did not mention resistance. But I think that if you allow for the possibility that the resistance value could be anything (i.e., very high, very low, or in between), you can see how the numerical values of voltage and current can be close to each other, and that either one could be much higher than the other.

The second concept is the definition of power. Here again, there are simple applications and more complex ones. But basically, Power equals Voltage times Current. That is the basis of ptonsparky's second answer.
 
D, to expand a bit farther, when you have a fixed impedance (basically AC resistance; it's DC resistance plus the effects of capacitance and inductance, collectively known as reactance), increasing voltage will result in an increase in current.

However, when discussing a desired fixed power level, which is a product of voltage and current, you have to decrease current when increasing voltage, or vice versa. This is usually done by altering the impedance of the load to suit the voltage.


So, to address your questions:

1. To maintain a given power level, yes. However, a high current (regardless of the voltage) can do more damage than a high voltage alone, which is why static electricity does little or no equipment damage, except of course for sensitive electronics.


2. Ohm's Law (which states that one volt will produce a current of one amp through an impedance of one ohm) indicates that a load of one ohm is the impedance that will result in 9 amps with a supply of 9 volts, as long as the supply is capable of supplying that much current.

Try placing six 1.5v D-cells in series next time you try that experiment.


3. The output voltage is set by the voltage control, while the resultant current will be purely a function of the impedance of the load you connect between the terminals. The exception would be a power supply that has either current limiting or a constant-current function.
 
Thanks for all the feedback

Thanks for all the feedback

Thanks so much for all the responses. I am getting closer to understanding this. I am slightly more confused though at the same time.

As I understood ohm's law, I thought it was : E=IR.

1 Volt = (1 amp)(1 ohms)

So if I am understanding this correctly we can actually have:

1 Volt = (10 amps)(some value opposite to resistance) to still equal 1 volt of force?

LarryFine, I also put 6 D-cells in series and got a reading of 9.5 volts and 2.5 amps, very similar to the 9v battery. Am I taking an incorrect reading?

I know I'm missing a lot with ohm's law. Can anyone recommend some good reading material on these principals?

Lastly, what makes a battery not able to deliver the current for it's voltage: 9amps from a 9volt battery? Is it internal resistance?

Thanks again for all the help!

-D
 
fuqur2dom said:
Lastly, what makes a battery not able to deliver the current for it's voltage: 9amps from a 9volt battery? Is it internal resistance?


-D
Click to expand...

That would make sense to me.

I guess the car battery rated for 700-amps (at 12-volts) has pretty low internal resistance?
 
There is no reason to expect a 9V battery to deliver 9A. Voltage is an indication of electron energy, and current is a measure of charge flow (in coulombs). Sort of like apples and oranges.

Just stick to Ohm's law and include the internal resistance.

And, never connect an ammeter across the battery terminals!
 
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fuqur2dom said:
Thanks so much for all the responses. I am getting closer to understanding this. I am slightly more confused though at the same time.

As I understood ohm's law, I thought it was : E=IR.

1 Volt = (1 amp)(1 ohms)

...

Lastly, what makes a battery not able to deliver the current for it's voltage: 9amps from a 9volt battery? Is it internal resistance?
Click to expand...

You have the correct equation, but you can rearrange it so I or R are on the left hand side. That is usually more useful because in most cases voltage is fixed. For example, your 9V battery only makes 9 volts, it doesn't vary based on resistance or amps. Likewise, house power is 120VAC or 240VAC.

To find current, I = E/R or R = E/I. When you connect something to a circuit, most items have a fixed resistance. So if you have a light bulb that is rated at .5 amps at 9V, it is equivalent to a resistor of 18 ohms. Likewise, your 60W household lamp is also about a 240 ohm resistor because we know it must draw .5 amps at 120V to make 60 watts.

You are correct that the reason you're only measuring a small amount of current from the battery is because of its internal resistance. DO NOT try the same test with your meter on AC amps by sticking it into a power outlet. You'll burn up your meter or blow its fuse, as direct measurement amp meters are usually quite limited in the number of amps they can measure (typically 10 or less and that may be in a special jack just for amps. Note - you can certainly have less than 1 ohm of resistance, so if the battery had no internal resistance, it should be able to produce millions of amps (at least until it drains). This is why electrical theory can be so difficult at first is because you learn about "ideal" components that don't exist in reality.
 
fuqur2dom said:
As I understood ohm's law, I thought it was : E=IR.

1 Volt = (1 amp)(1 ohms)

So if I am understanding this correctly we can actually have:

1 Volt = (10 amps)(some value opposite to resistance) to still equal 1 volt of force?
Click to expand...
Yes, E = I * R is correct. That is the formula for Ohm's Law. Also, I = E / R and R = E / I.

In your eaxmple, you know the voltage and the current, so you place them on one side of the equation, and the unknown on the other, giving you R = E / I; R = 1 / 10; R = 0.1 Ohm.

A power source have an impedance just as a load does; ideally, they match in transmission-line theory, but for maximum stability, the greater the load-to-source impedance ratio, the better.

The better a source can resist change due to the load, the lower its source impedance, and the less voltage sag an increase in loading will cause. Audio amplifiers use 'damping factor' to rate this ability.

LarryFine, I also put 6 D-cells in series and got a reading of 9.5 volts and 2.5 amps, very similar to the 9v battery. Am I taking an incorrect reading?
Click to expand...
Not if you're placing an ammeter in series with the circuit. However, an ammeter reduces the current slightly, because it works by measuring the voltage across a low resistor, so a slight voltage drop, and thus current drop, is caused.

In the case of 9v and 2.5a, the formula would be R = E / I; R = 9 / 2.5; R = 3.6 Ohms. Somewhere in the circuit (source, load, ammeter, wiring, etc.) is a total of 3.6 Ohms' worth of impedance.

Lastly, what makes a battery not able to deliver the current for it's voltage: 9amps from a 9volt battery? Is it internal resistance?
Click to expand...
You've got it, in addition to the rest of the circuit, as explained above. Hint: don't try this with an automotive battery.
 
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I Got It!

I Got It!

I finally got it guys. Larry, it finally clicked with your last reply.

Somehow this whole time I've been thinking in terms of - 1 ohm as a constant minimum! Never saw you can obviously have much less resistance.:roll:

I just couldn't understand what was determining the current if the voltage with a constant of 1 ohm was not adding up.

9V with .00001 resistance (if that's possible) would render high current well over the (9v * 1ohm) calculation.

Like I said in my original post, so simple it's silly.

Thanks again for all the great help. What an excellent forum!

-D
 
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