# Voltage across resistor if current is limited?

Discussion in 'General Electronics Chat' started by geratheg, Aug 17, 2014.

1. ### geratheg Thread Starter Member

Jul 11, 2014
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3
If you have a power supply and you limit its current, while wiring two resistors of different values in parallel to the power supply, would this limited current make it such that the voltage across the larger resistor is 0V because current tends to take the path of least resistance?

I'm curious and don't have this kind of power supply to test but here are 3 cases and what I'm thinking:

Schematic provided in attached pictures.
Voltage: 12V in each case.
I'm thinking like this: The smaller resistor is the path of least resistance, so current would tend to want to go through the smaller resistor before going through the larger resistor.

Case 1) Current is limited to 0.8 Amps (Less than enough current to get a 12V drop across smaller resistor)
My answers: Voltage across 15 Ohm resistor is 0V, voltage across 10 Ohm resistor is 8V.

Case 2) Current is limited to 1.2 Amps (Enough current to get a 12 V drop across smaller resistor)
My answer: Voltage across 15 Ohm resistor is 0V, voltage across 10 Ohm resistor is 12V.

Case 3) Current is limited to 1.6 Amps (More than enough current to get a 12 V drop across smaller resistor)
My answer: Voltage across 15 Ohm resistor is 6V, voltage across 10 Ohm resistor is 12V.

Are these answers correct theoretically? If any is wrong, please correct me with the correct answer to each case, number the cases so I know which case you're correcting me on.

On another note, unrelated to the questions here I'm thinking that when there is a short circuit, the reason the voltage across a resistor in this circuit is 0V is because there is not enough current to go through the resistor since infinite current is flowing to the path of no resistance. So no current really remains to flow through the resistor because in this traffic of infinite current there is no "stop" due to no resistance so current prefers to take this path of least resistance. Is this correct?

Thanks!

• ###### Resistor Current.jpg
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3. ### ericgibbs AAC Fanatic!

Jan 29, 2010
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hi,
The current will 'divide' into the two resistance paths, the current in each path will be inversely proportional to its resistance.
E

4. ### crutschow Expert

Mar 14, 2008
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It's a misnomer to say that "current takes the path of least resistance" (in a parallel circuit) and that statement has confused many people. (I'd like to know who started that saying because it has created a lot of mischief ). The correct saying is "more current takes the path of least (or lesser) resistance" with current inversely proportional to resistance. Thus if you have a 1 ohm resistor in parallel with 2 ohms, twice as much current flows through the 1 ohm resistor compared to the 2 ohm resistor. This is independent of the value of the voltage or current applied to the pair.

If the resistors are in series then both carry the same current and the voltage drop across each resistor is proportional to the resistance. Thus if the same 1 ohm and 2 ohm resistors are in series, the voltage drop across the 2 ohm resistor is always twice the voltage drop across the 1 ohm resistor for any application of voltage or current. The sum of their two voltage drops is, of course, equal to the applied voltage across the pair.

5. ### geratheg Thread Starter Member

Jul 11, 2014
107
3
I understand. I know current divides, but that is if there is sufficient current. I'm more curious as to what would happen if current was limited. I've seen a power supply that can supply a voltage but limit the maximum current. I'm just curious what the voltage drops would be if all the current that's available are as in the 3 cases I listed, those currents would be the max.

I think the path of least resistance holds for an ideal no resistance short circuit, and was meant for that theoretical situation and not one where there is resistance present like in reality. And the only time I use this idea of path of least resistance is when there is a short circuit with no resistor present other than the wire itself.
As for the 3 cases specified, I'm asking just out of curiousity.

Apr 5, 2008
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Hello,

In the three descibed situations, we can not speak about the use of a voltage source.
In those cases we speak about a current source.
The current will ALWAYS split over the two resistors.
The 10 Ohms resistor will take 15 / 25 th of the current and the 15 Ohms resistor will take 10 / 25 th of the current.

The total resistence seen by the current source is the parallel value of both resistors.
This value is 1 / (1/10 + 1/15) = 6 Ohms.
The voltage in case 1 is 0.8 Amp * 6 Ohms = 4.8 Volts.
The current in the 10 Ohm resistor is 4.8 Volts / 10 Ohms = 0.48 Amp.
The current in the 15 Ohm resistor is 4.8 Volts / 15 Ohms = 0.32 Amp.

You can calculate the other cases yourself.

Bertus

7. ### crutschow Expert

Mar 14, 2008
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I think I see the problem in your reasoning. You appear to misunderstand current limiting in a power supply. If the current is limited than the output voltage drops as determined by the load equivalent resistance value and equals I*R, independent of what the power supply voltage setting is. The voltage setting voltage only appears at the output when the current is below the current limit. You can't limit the current and have the output voltage remain constant since that would violate Ohm's law. Whatever limit current there is, it still flows in inverse proportion to the parallel resistance values.

Make sense now?

8. ### geratheg Thread Starter Member

Jul 11, 2014
107
3
I think it makes sense. The way I understand it now is if the current is limited then the voltage will adjust such that it's the same voltage across both resistors, regardless of what voltage the power supply says. Is this correct?

Apr 5, 2008
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Hello,

When your powersupply is set on 12 Volts with a current limit on 0.8 Amps,
the powersupply will reduce the voltage when the resistance is below 12 Volts / 0.8 Amps = 15 Ohms.

Bertus

10. ### geratheg Thread Starter Member

Jul 11, 2014
107
3
Suppose the power supply is a 12v battery with a 0.8 amp current limit, and I know such a low current limit would be unlikely, but would the voltage reduce in this situation also?

11. ### GopherT AAC Fanatic!

Nov 23, 2012
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0.8 amps is low? Most 12 volt DC adapters run much less than 800 milliamperes.

12. ### geratheg Thread Starter Member

Jul 11, 2014
107
3
Lol, electronics is somewhat new to me. So I thought it would be low for a battery.

13. ### crutschow Expert

Mar 14, 2008
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Depends entirely on the battery. A car battery can deliver hundreds of amps while a small button cell can only deliver a few mA. But the amount of current drawn from the battery is determined by the load resistance (and any internal battery resistance) or the current limit setting if there's a current-limit circuit.

14. ### geratheg Thread Starter Member

Jul 11, 2014
107
3
Okay, I came up with that amperage for the purpose of the question.

15. ### MikeML AAC Fanatic!

Oct 2, 2009
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Stare at this. The stuff inside the dashed box is a proxy for what I call a Lab Power Supply. Think of it having two knobs; one to set the output voltage, another to set the current limit.

In this example, the voltage knob is set to 10V, and the current knob is set to 1A. The simulation shows what happens if we connect a variable resistor that goes from 0.001Ω to 25.001Ω in steps of 1Ω across the output of the supply.

I plot the voltage out of the Lab Supply V(b) Red trace vs the value of the load resistor R1 in the upper panel, and the current through R1 I(R1) Green trace in the lower panel.

Note that while R1 is less than 10Ω, the supply is the current limited mode as shown by the Green trace. The voltage across the load resistor (also the voltage out of the supply) is directly proportional to the load resistance (E=IR).

As R1 increases beyond 10Ω, the supply switches to constant voltage mode, and holds V(b) = 10V. Now the current I(R1) is inversely proportional to the resistance I=E/R...

To model the Lab Supply in LTSpice, I used the magic behavioral resistor R2, which creates the constant voltage, constant current behavior.

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16. ### geratheg Thread Starter Member

Jul 11, 2014
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That was very helpful. Thank you
So voltage does indeed reduce if there is not enough current to get a constant 10V.

I appreciate the time you took to test this since I don't have my own supply.

17. ### crutschow Expert

Mar 14, 2008
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Mike's sim inspired me to do one with two resistors in parallel as you originally posted and an alternate way to do the voltage supply with a current limit (just to be different).

I used a 3A constant-current source in parallel with a 12V constant-voltage source and a diode between them to limit the current when the output voltage goes above the voltage set value. This simulates a 12V power supply set to a 3A limit. The voltage source is actually set to 11.7V to compensate for the diode drop.

R1 steps from 1Ω to 10Ω at the same time that R2 steps from 1.5Ω to 15Ω. As you can see, the output current stays at 3A at the lower resistor values until the output voltage rises to 12V. The sum of the two resistor currents equals 3A at the limit and track each other down to the minimum current value of 1.2A and 0.8A at 10 and 15 ohms respectively.

18. ### geratheg Thread Starter Member

Jul 11, 2014
107
3
Thank you for posting that!