A dumb question about ohm's law - Page 1
Forum section: electronics / beginners

author: sairfan1
author_rank: Frequent Contributor
timestamp: May 09, 2025, 06:53:46 pm
content: I'm sure my understanding towards it are wrong and I'm missing something big here.I wanted to calculate voltage drop across resistor, ohm law says V = I x RValues for the formula areI = 0.500 AR = 10,000Where I'm using V = 3.3v 0.5Am power supplyV = 0.500 x 10,000 = 5,000What I wanted to know, what will be voltage if i connect ground pin and positive pin after resistor, here load is not present.Sorry I'm not in my lab to see what actually volt meter shows.

author: mariush
author_rank: Super Contributor
timestamp: May 09, 2025, 07:02:32 pm
content: The resistor will be a "load" on the power supply.Since it's a 10k resistor, there's gonna be V = I x R => I = V / R = 3.3 / 10,000 = 0.00033 A or 0.033 mA of current through the resistor.The power supply is capable of 0.5A, doesn't mean the whole 0.5A will go through the resistor.Simulated circuit :https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgoqoQFMBaMMAKDAxRFwBZwNCukXmBR4oIctmjYu08VRhIAagHsANgBcAhgHM6LAE5cEndgOwnBvKmHiQWAd2OcezrtgGsdb7B5+WFRys+c0szKBYgA

author: TimFox
author_rank: Super Contributor
timestamp: May 09, 2025, 07:23:33 pm
content: Similarly, if the power supply were “constant current” instead of the more common “constant voltage”, there would be a maximum voltage (called the compliance voltage) that the supply could deliver into the load.For a 0.5 A current supply, capable of only 3.3 V, into more than 6.6\$\Omega\$ load, the voltage limit prevents the current from reaching 0.5 A.Furthermore, you can’t build a supply that delivers both a specific voltage and current into an arbitrary load resistance: one or the other value can be achieved, depending on the resistance or non-linear load (e.g. diode).

author: sairfan1
author_rank: Frequent Contributor
timestamp: May 09, 2025, 08:01:58 pm
content: For example a resistor is connected to 3.3v 0.5A power supply, other end of resistor is connected to uC ADC pin, now how will i calculate voltage through 10K resistor at uC ADC pin.

author: benj38
author_rank: Frequent Contributor
timestamp: May 09, 2025, 08:09:20 pm
content: @sairfan1, to summarize the first two replies, when you have an equation like Ohms's law of V = I x R, it means that you can chose the value of two of the variables, and nature will chose the value of the third to satisfy the equation. Your resistor is 10,000 ohms, so that is one value picked. You can then either pick the voltage and nature will decide the current such that the equality holds, or you can pick the current and nature will decide the voltage.Power supplies come in two basic flavors: constant voltage, and constant current. The average power supply (like most regulated wall-warts), which your 3.3V 0.5A supply is most probably one of, supplies a constant voltage (in your case 3.3V), and the current specification of 0.5A is themaximumcurrent it can provide. Theactualcurrent it will provide depends on the load resistance and Ohm's law. Thus, the smallest resistor you are allowed to connect to your supply is 6.6 Ohms; anythingsmallerand you overload the supply since it requires more than 0.5A to maintain the 3.3V. As a result, the voltage supplied will no longer be 3.3V but lower, and/or the unit will overheat and fail or a protection mechanism will trigger.Lab power supplies can also operate in a constant current mode, where you chose the output current, and the voltage they produce varies based on the load resistance according to Ohm's law. Obviously, in this mode they will have a maximum voltage that they can provide (similar to how the constant voltage supply has a maximum current it can supply), and if the load resistance istoo highthey will overload and not be able to supply the current you selected, because this would require a voltage higher than they can provide.

author: BeBuLamar
author_rank: Super Contributor
timestamp: May 09, 2025, 08:16:24 pm
content: But without any calculation if you connect a 10,000 Ohms resistor to the 3.3V power supply which can deliver 0.5A max the voltage drop across the resistor is 3.3V.

author: benj38
author_rank: Frequent Contributor
timestamp: May 09, 2025, 08:30:05 pm
content: Quote from: sairfan1 on May 09, 2025, 08:01:58 pmThe ADC pin will see 3.3V.This scenario though is different than what your original question implied. The reason is that your 10,000 Ohms resistor isnotwhat is connected to your 3.3V supply!What is connect to your supply is this resistorin serieswith the input circuitry connected to the ADC pininsidethe uC!Most microprocessors have an input circuitry resistance of around 10 Mega-Ohms!Overall, the supply sees a load resistance of 10 Mega-Ohms plus 10,000 Ohms, and this requires a super tiny amount of current (well below the maximum available current of 0.5A), so the supply can easily maintain the 3.3V.The voltage the ADC pin sees is thus almost exactly 3.3V. You can figure the exact voltage using ohm's law to calculate the current flowing I = 3.3V / (10^6 + 10000), and then the voltage at the ADC pin is I x 10^6.Note that the above calculation is just math. Your supply most certainly does not provide 3.3000000000000V, but something much less accurate, so for all practical intents the voltage that the ADC pin sees is the supply voltage since the 10000 Ohms resistor has almost no effect on the current flowing.BTW, "voltage through" is not a good way to think of this; what goes "through" the resistor is current.

author: TimFox
author_rank: Super Contributor
timestamp: May 09, 2025, 08:44:00 pm
content: Yes: better to say "voltage across" and "current through" a component.

author: Rick Law
author_rank: Super Contributor
timestamp: May 09, 2025, 09:41:54 pm
content: Quote from: sairfan1 on May 09, 2025, 08:01:58 pmI think there is some confusion due to either wording or understanding. Your "voltage through 10K resistor at uC ADC pin" causes confusion. Your ADC (or DMM) should be connected in parallel to the resistor -- that is, the ADCV+is connect to theV+ends of the resistor and the ADC's ground (or DMM's ground/common) is connect tothe other endof the resistor.The wording "voltage through 10K resistor at uC ADC pin" implies you may be connecting them in series instead of parallel. In this case, you wont measure correctly.

author: TimFox
author_rank: Super Contributor
timestamp: May 09, 2025, 09:47:13 pm
content: A useful practice when using a voltage source, series resistor, and an arbitrary load circuit (not necessarily linear) is to use a "load line".Start with the current vs. voltage curve of the load circuit (e.g., a diode), with voltage on the x-axis and current on the y-axis.For the resistor, add a straight line on the same graph: it goes through the no-load output voltage and zero current (lower right in the first quadrant) and zero voltage and the short-circuit currentVsource/Rseriesat the upper left.The intersection of the line and the curve defines the operating point for voltage across and current through the load.(This is usually done for active devices where there are multiple curves for different values of control parameter, gate-source voltage or base current.)

author: pcprogrammer
author_rank: Super Contributor
timestamp: May 10, 2025, 06:19:55 am
content: Quote from: benj38 on May 09, 2025, 08:30:05 pmWhere do you get this value from?I'm no expert, but I think it is a lot lower than that. Maybe in the Mega Ohms range. The datasheets I looked at for STM32 devices are not so clear on this.And there is another issue to consider, the switch resistance and capacitance of the sample and hold circuit used in the ADC. Adding a large series resistance will cause it to slowly charge and the capacitor might not reach the full input voltage during the sample time.

author: iMo
author_rank: Super Contributor
timestamp: May 10, 2025, 06:33:29 am
content: Usually a resistor wired to an ADC input (of a typical MCU) should be lower than 10k. The lower the better due to the fact there is the sample and hold capacitor which needs to be charged fully within a specified time (as mentioned above, the C is around 10-20pF, the charging times X us typically).Also the internal "resistance" of an logic device input is usually limited by the ESD clamping diodes reverse current, say X nA. The diode's forward current depends on the input voltage (and could be X mA).

author: MrAl
author_rank: Super Contributor
timestamp: May 10, 2025, 06:43:32 am
content: Quote from: sairfan1 on May 09, 2025, 06:53:46 pmHi,It sounds like you've got Ohm's Law down pretty well. There are other factors though that appear when we do an experiment in real life.When we power something from any power supply, we have to note the limitations of the power supply. For example, if we have a power supply that can do 12 volts at 0.5 amps,when we connect a 10k resistor to the output, we won't see 10000*0.5=5000 volts. That's because the other limitation of the power supply is 12 volts, so the most current we will see is 12/10000=1.2ma. If we use a 1k resistor we will see at most 12/1000=12ma. In other words, the 12v spec takes precedence.Now if we connect a very low value resistor like 10 Ohms, when we do the calculation 12/10=1.2 amps we can note that 1.2 amps is greater than the spec of 0.5 amps, so NOW the current spec of 0.5 amps takes over. The most we can get is 0.5 amp, so the power supply voltage will cut down to 0.5*10=5 volts. That's a long as the power supply does not cut out completely because some power supplies cut out when the current gets too high. You also have to be careful in case the power supply is not protected for overloads because the power supply could blow out.Think about this a little then do several calculations until you are confident you understand all this well. You can practice with a free circuit simulator program.

author: benj38
author_rank: Frequent Contributor
timestamp: May 10, 2025, 10:02:40 am
content: Quote from: pcprogrammer on May 10, 2025, 06:19:55 amOOPS. I meant Mega, not Tera. I guess that this is what happens when you read the economy column before you go on EEVBlogIn fact, 10 MOhms is a bit on the low side. For example, the DC input resistance of the ATmega328 ADC is 100 MOhms.