I have a question about Ohm’s law. I know the basics, I=V/R and so on. The question I have is, what limits the amount of current with a low resistance value? Say I have a 9 volt battery, with a 1 OHM resistor between the terminals. Now, according to Ohm’s law, I=V/R = 9 Amps of current. I know this cannot be right. I also notice the numbers of amps written on the side of the battery, does this mean that no matter how low the resistance, you are only getting that many amps out of the battery? If so, with a lower resistance, are you getting no current? Or are you getting closer and closer to the max current able to be supplied by the battery?
In this case, Ohm’s law is correct, but the battery will
only supply whatever it’s maximum listed current is. Even
with a direct short it would supply no more current.
If the battery was powerful enough to supply 9amps it would.
Lower resistance -> More current. V=IR, I and R are inversely proportional/have an inverse relationship, what ever you want to call it. You are getting closer and closer to the max current able to be supplied by the battery.
Exception - shorting a circuit (take a power supply and directly connect the positive and negative terminals, you’ll most likely find that it overloads and you no longer get any current out of it).
DISCLAIMER: I am just a student and not THAT knowledgeable about this topic, just sort of.
I kind of figured that was the answer. The only other question I have now is, in all the books I read about Ohm’s Law, never once was there any disclaimer or anything of the type “NOTE: Ohm’s Law only works up to the maximum amount of current able to be supplied by your voltage source”
The power source is considered to have an internal resistance. There’s a limit to the amount of current that any power source can produce, whether it’s a limit in the size of magnetic fields produced (or a limit in the coupling between the magnetic field and generator windings) in a generator/alternator or a limit in how fast a chemical reaction can occur in a battery. Internal resistance is a way of representing this.
Ohm’s law is always correct assuming that V is always constant for any given value of R, and therefore will always graph as a straight line. The books are right because they always assume a no-load condition. I guess the concept of loading is just too much to throw at some one who is just beginning to learn ohm’s law?
9 volts is the source’s no load terminal voltage. As voltage sources are called upon to supply current to a load, their terminal voltage drops.
When current is drawn out of a voltage source, not only does it have to overcome the resistance of the external circuit (R[sub]L[/sub]), it also has to overcome its own internal resistance (R[sub]i[/sub]). Since, my child, you have shown awareness of ohm’s law, I am going to make a giant leap here and assume that you are also aware of Kirchhoff’s laws (specifically the 2nd). Kirchhoff sez:
Check out this page for a helpful diagram. The first diagram shows the entire 4V source dropped across the resistive load. This diagram is only accurate if we assume that the load is the single source of resistance in the circuit (or at least 99+% of it, depending on how thin that hair is that you want to split). The second diagram is more accurate in addressing your question, but since I can’t hack into that web site and alter the diagram a little, you’ll have to do a little imagining.
First, let’s change the source to 9V to better stay with your question. Now imagine (we’re still on the 2nd diagram here) that the larger resistance to the right is the load. The small resistance up top can be thought of as the battery’s own internal resistance (even though it is internal to the source, it still works in series with the load). Let’s also say that the battery’s internal resistance ::reaches into the air and grabs a number:: is 3 ohms. The source can almost certainly supply a reasonable load of, say, 1,000 (1K) ohms with enough current to (1) operate the load and (2) obey ohm’s law without having to take R[sub]i[/sub] into consideration. Compare:
Ignoring R[sub]i[/sub]: 9V ÷ 1000 ohms = .009 amps (9mA)
Including R[sub]i[/sub]: 9V ÷ 1003 ohms = .00897 amps (easily rounded to .009 amps or 9mA)
If your application does not allow you to round off as I did (for instance if the load absolutely must have .009 amps at all times to operate properly), then whatever circuit you are designing requires a constant current source and not a simple chemical battery. However for the vast majority of applications where we use 9V batteries, loading is permissable (indeed expected) and therefore we are allowed to ignore the errors introduced by the rounding off of numbers.
Now let’s substitute your 1 ohm load for the 1K and reapply both ohm’s & Kirchhoff’s laws to find the currents & voltages. Remember, the total circuit resistance (R[sub]L[/sub]+R[sub]i[/sub]) is now 4 ohms:
Ignoring R[sub]i[/sub]: 9V ÷ 1 ohm = 9 amps (9A)
Including R[sub]i[/sub]: 9V ÷ 4 ohms = 2.25 amps
Now you see why we can not ignore R[sub]i[/sub] when the load is heavy (when the current demand is high). Now apply Kirchhoff’s law to find the voltage dropped across the load. Or if you are still a little squeamish about Kirchhoff, re-arrange Ohm’s law to find the drop:
Voltage across the load (V[sub]L[/sub]) = I × R[sub]L[/sub]. Substituting the numbers: V[sub]L[/sub] = 2.25A × 1 ohm = 2.25V.
Yes, you will measure only 2.25 volts across a 1 ohm load. The voltage measured at the battery terminals will also be 2.25V, so where did the missing 6.75V go? It’s dropped across the 3 ohm R[sub]i[/sub]. You can’t measure it because it’s inside the battery! We call this the battery’s terminal voltage under load.
Taking your example one step further, let’s short the battery’s terminals (R[sub]i[/sub] = zero). According to Ohm’s law, this results in infinite current throught the short. Viola! Energy crisis solved with a single 9V battery! Not so, unfortunately. This is the battery’s short circuit current and is a way of indirectly measuring (and thereby confirming the existance of) R[sub]i[/sub].
Finally, it should be noted that R[sub]i[/sub] increases as the battery ages and the chemicals are depleted. That is why batteries supply less and less current over time, even when their no-load terminal voltage may still measure close to 9V.
EE’s use models to construct circuits with idealised components.
Every component part of an electrical circuit has its main design properties, for instance a resistor ‘resists’ the flow of current, but in addition can have a significant capacitance or, in the case of wire-wound resistors, significant inductance.
These secondary qualities of the component are usually very small in value and can often be ignored except in certain applications such as high frequencies etc.
If a component is used within its design parameters this is fine, so your 9volt battery is designed with a maximum output of xxxmilliamps, go beyond that, such as into short circuit conditions, and other factors come into play.
In the battery case it is said to have a pure voltage source with a resistor in series with it, and to make this easier to visualise on a diagram a dotted line box will be drawn around them to indicate the ‘physical’ boundaries of the battery.
This modelling is used a great deal, such as in Thevenin circuits and is a very convenient way of determining conditons affecting a component that has been connected to output terminals.
The whole circuit can be reduced to a voltage/current source with a resistor in circuit(or capacitor or inductor) with just two terminals for the load.
The whole idea of modelling is to make things easier to understand, take time to look up Thevenin, voltage and current sources.