Ohm's Law. - Page 1
Forum section: electronics / beginners

author: yakacm
author_rank: Newbie
timestamp: July 04, 2016, 12:42:11 pm
content: I have a question about Ohm's law which is driving me nuts, I have just got back into electronics having not studies it in 30 years. So my question is, if we have 50 milliamps, and a 2k resistor Ohm's law tells us that we have 100 volts, a 3k resistor 150v. Now lets say we are driving the circuit with a 9V battery, which is easily capable of producing 50 millimaps, how can it possibly produce 100v or 150v? And why does increasing the resistance increase the voltage, I would have thought the opposite? Am I looking at it too literally, or does it mean to move 50 milliamps thru 2k resistor you need 100 volts?

author: sleemanj
author_rank: Super Contributor
timestamp: July 04, 2016, 12:53:07 pm
content: Quote from: yakacm on July 04, 2016, 12:42:11 pmYes, the equation must balance.

author: MosherIV
author_rank: Super Contributor
timestamp: July 04, 2016, 12:55:44 pm
content: Hi"50 milliamps, and a 2k resistor Ohm's law tells us that we have 100 volts" - by ohms law - true"Now lets say we are driving the circuit with a 9V battery, which is easily capable of producing 50 millimaps, how can it possibly produce 100v or 150v?"The answer is the 9V battery cannot pass 0.05A through 2KOhms. 9V will only allow 0.0045 (4.5mA) to pass through 9V(Note : this would be difficult to prove by measurement because the shunt resistor of the Ammeter will take a burden voltage.)"Am I looking at it too literally, or does it mean to move 50 milliamps thru 2k resistor you need 100 volts?"Correct.

author: yakacm
author_rank: Newbie
timestamp: July 04, 2016, 02:28:27 pm
content: Thanks chaps.

author: rstofer
author_rank: Super Contributor
timestamp: July 04, 2016, 02:46:23 pm
content: Ohm's law has three variables of which any 2 are fixed and the 3rd is derived. If you pick Ohms and Volts, Amps is derived. So, you can't drive 50 mA through a 3k resistor with a 9V battery. You would be trying to pick all 3 variables and you can't do that.So, picking Ohms and Volts, your 9V battery would only drive 3 mA through a 3k resistor.

author: uncle_bob
author_rank: Unknown
timestamp: July 04, 2016, 06:14:30 pm
content: HiAbout the only other thing that people trip up on:It's always the voltage *across* the resistor. You can have 2837V on one end and 2836V on the other end. That only counts as 1V for ohms law. The 2836V is just there to confuse the issue.Bob

author: IanB
author_rank: Super Contributor
timestamp: July 04, 2016, 06:29:17 pm
content: Quote from: yakacm on July 04, 2016, 12:42:11 pmWhen you are introduced to Ohm's law in a physics lesson at school, you don't get told what it is. Instead, you get given some resistors, a volt meter, an ammeter, a battery or power supply, and you measure volts, amps and ohms for different combinations. Then you take all your measurements and plot them on a graph. Only after that does the teacher then explain how to interpret the graph and what Ohm's law says.This is the missing piece I strongly recommend you to add to your studies: practical experiments. It's very hard to get a good understanding just by reading books. You need to learn by doing. Then you can ask yourself questions and try to find the answers by doing the experiments.

author: BobsURuncle
author_rank: Regular Contributor
timestamp: July 04, 2016, 07:26:31 pm
content: Quote from: yakacm on July 04, 2016, 12:42:11 pmIt can't.A battery via electrochemistry is capable of producing a voltage across its terminals - not some predetermined current. We call this a voltage source. The voltage is determined by the design of the battery. The current that flows from a battery depends on the battery's voltage and the resistance to current flow that it sees in the circuit (there are practical limits but we are discussing an ideal battery). A 9V battery will supply 50 mA only when it sees 180 ohms of resistance.There are devices (not batteries) that can provide a predetermined amount of current. We call these current sources. In that case the current is dependent on the design of the device. The voltage at the devices output will depend on the resistance through which the current flows. That is to say the current source will adjust its voltage to accommodate a change in the circuit resistance so that it can maintain a certain current. (again there are practical limitations but I am talking about ideal current source)

author: edy
author_rank: Super Contributor
timestamp: July 06, 2016, 03:07:53 am
content: Quote from: yakacm on July 04, 2016, 12:42:11 pmTo get 50 mA of current to flow through a 2k resistor, you need 100 V difference in potential between the 2 ends of the resistor. A more resistant resistor (like 3k ohm resistor) will require a BIGGER difference in potential if you still want to get 50 mA of current through it. In this case, 150 V difference in potential is needed to drive the same 50 mA of current through it.Keep in mind that 1 AMP current is one Coulombs of charge passing a point in a wire per second. So the more resistant something is, the harder it will impede the flow of charge. So to keep the flow at 50 mA, you need more "PUSH" on that charge, which is why you have to increase the VOLTAGE if you make your wire more and more resistant.Regarding the 9V battery, that is the potential difference between the 2 ends. It's set. The only thing you can now vary is the current flow between the 2 ends based on the resistance of the path you put in between. For example, if you short the battery through a 10 ohm resistor, you will have a 0.9 A current flow which will likely burn up the resistor/wire/battery very quickly. The power being dissipated by the resistor is 8.1 watts (9V x 0.9 A).If you stick a 1k resistor across the ends of a 9V battery, you will have made the resistance so high that only 9mA of current can now flow. The power dissipated is 0.081 Watts which is certainly below a 1/4 W resistor rating. So nothing will blow up but your resistor may get a little warm.So to summarize, if you want to push the SAME amount of current through a circuit (coulombs of charge) you will need more potential energy difference across it to overcome the resistance that is suppressing your current flow.In real circuits, there are limits to how much current your source can supply before it starts to dip, how much voltage can be generated, etc. Your source will determine how much it will be able to handle. Some are very good at keeping the current flow very stable, even if you put more and more resistance across that drops voltage. Voltage sources keep a stable voltage whether you pull a little or a lot of current through the circuit. For example, imagine a power supply and you stick a load on it... if it is a constant voltage source, it will keep the voltage stable even when your load is higher. Up to a point, it will keep up.... eventually if you surpass the limit of the power supply, the voltage will start to drop. It cannot overcome the load. Short-circuiting an voltage source, it will try to keep the voltage difference the same between the 2 points and end up producing a huge current.V=IR... if you keep V constant, then as R goes up, "I" must come down. Eventually "I" becomes 0 at very high R (load)... no current flows. If you short-circuit, effectively make R=0, then "I" shoots up and you get very high current.On the other hand, a constant current source will try to keep the same current no matter what the load is. If you have a very small resistance, or short-circuit, the rated current will flow through the wire. However, if you start to put bigger and bigger loads or resistance, the current source will keep trying to push the same current through (to keep stable) and you will generate higher and higher voltages across the circuit. At some point, if you break the circuit and there is an air-gap, there will be a plasma arc spark across the gap because your current source still wants to keep the same current flowing... even when there is no wire! The voltage across the gap is now huge.V=IR. This time we keep I constant. If we increase R, the load, then V also goes up. At some point if R is huge (say an air-gap) then V is also huge. If R=0 (short circuit) then V=0.

author: yakacm
author_rank: Newbie
timestamp: July 06, 2016, 10:08:46 am
content: Thanks for all you replies guys, yup I was thinking about it all wrong, obviously volts and amps, are 2 different ways of measuring the same thing, IE Electricity. A 9v battery would only be able to push a small amount of current thru a 2k resistor, I get it now. I was thinking it was cause and effect, like 50ma thru 2k ohm would create 100v, that's what I couldn't get my head around, the volts is the power and the amps is the current, so 9V only has enough power to push a certain amount of current thru the resistance. I was confused as to why with greater resistance the voltage increased, obviously the voltage doesn't increase, it's that more power is needed to overcome the resistance, go it. Thanks again everyone who replied, much appreciated. I do have bread boards and resistors and a multi meter so I am able to experiment, sometimes it is hard to set the circuit up so that you get the answer you are looking for, perhaps hard isn't the right word, it's not immediately obvious how to set the circuit up.

author: edy
author_rank: Super Contributor
timestamp: July 06, 2016, 11:01:31 am
content: Just remember, you measure voltage in parallel, and current in series. Be sure that your multimeter is able to handle the current you expect or it will blow up!

author: BobsURuncle
author_rank: Regular Contributor
timestamp: July 06, 2016, 05:40:13 pm
content: Quote from: yakacm on July 06, 2016, 10:08:46 amYes, now you are getting it. I suspected there was a cause and effect confusion. A piece of advise though....It may seem nitpicky, but it is a necessity. Right from the start be sure you have a good understanding of electrostatics and the definitions of energy, power, potential energy, charge, current, resistance and their units: joules, watts volts, coulombs, amps, ohms and how they are interrelated. Then when you are thinking or talking about them be precise in your terminology or you will get confused as things get a little more complicated.For example, volts are not power. Volts are potential energy. 1 volt is the amount of energy (1 joule) it would take to move 1 coulomb of charge from one terminal to the other. Power is energy delivered or absorbed per unit of time. This battery when open circuit has 9V of potential energy across its terminals but delivers no power; it does have stored energy (in this case 15Ah = 54,000 joules of energy when fully charged). If connected to a load and if it delivers 50mA of current then it is delivering 0.45 watts of power or 0.45 joules of energy per second.Stated another way, things, including batteries, don't posses power. Power is a measure of the amount of work done per unit of time. It may have lots of stored energy but that doesn't mean it is doing any work. Things do posses energy and therefore are capable of delivering an indeterminate amount of power, i.e. doing work. These distinctions are important, especially when you get into AC circuits or hybrid systems e.g. mechanical and electrical.Be sure you can do the ohms law calculations with common EE units (e.g. amps) or SI units(e.g. coulombs/sec.)edit: Oops! In this phrase "(in this case 15Ah = 54,000 joules of energy when fully charged)" I got the 15Ah from a different thread at a different site.

author: IanB
author_rank: Super Contributor
timestamp: July 06, 2016, 07:58:28 pm
content: Quote from: yakacm on July 06, 2016, 10:08:46 amSometimes this is the right way to look at it, depending on the context. For example, you have a long extension lead delivering 10 A to a load at the other end. The extension lead has a total resistance of 0.5 ohms, so you will create a voltage drop of 5 V over the wire. (Pushing 10 A through 0.5 ohms creates a voltage of 5 V.)

author: BobsURuncle
author_rank: Regular Contributor
timestamp: July 06, 2016, 08:33:54 pm
content: Quote from: IanB on July 06, 2016, 07:58:28 pmWhat do you think is "pushing" the current?

author: IanB
author_rank: Super Contributor
timestamp: July 06, 2016, 11:24:34 pm
content: Quote from: BobsURuncle on July 06, 2016, 08:33:54 pmPink unicorns

author: uncle_bob
author_rank: Unknown
timestamp: July 07, 2016, 05:58:30 pm
content: Quote from: BobsURuncle on July 06, 2016, 08:33:54 pmHiUmmm .... errr ... the Poynting vector?Bob