Using Ohm's law in a simple circuit with a DC power adapter

Question

I'm a bit confused on using Ohm's law so I'm going to ask about it using an example. I have a simple circuit with power provided by a generic DC power adapter and a 1 kΩ resistor.

1. Say the adapter provides 1 A and I need to calculate the voltage after the resistor. Using V = I · R, I is the current which goes through the resistor and NOT the current provided by the adapter (1 A), correct? I guess I need to put a multi-meter in the middle to find the current first.
2. What if my circuit is just a wire connecting + to - with close to zero resistance, essentially shorting the circuit? I guess at that point the current through the wire IS the maximum current provided by the adapter. Ohm's law says V = I · R = 1 · 0.1 = 0.1 V; that doesn't make sense.
3. Say the adapter provides 1 A and 6 V. Ohm's law then says 6 = 1 · R which means R = 6 Ω. Now what does this represent? Is there a 6 Ω resistor in the adapter?
4. Say the adapter provides 6 V and I need to find the current in my circuit, so I = V/R = 6/1000 = 6 mA?
5. Now say my adapter provides 6 V but only gives me 0.5 mA, which means there is a 6/0.0005 = 12000 Ω resistor inside the adapter. Now my answer in [4] doesn't make sense anymore, the adapter is only providing 0.5 mA, how can the current be 6 mA in the circuit?

Answer 1

1. No. The current through the resistor and the current out of the adapter is one and the same current.

2. Yes, but also no. Your R is 1k, not 1 ohm, so V = IR= 1 x 1k = 1000V. In order for your adapter to be pushing 1A through a 1k resistor it must also be capable of supplying 1000V.

3. No. Just stop right there. You cannot specify both the current and the voltage that your adapter supplies. You can specify one, and put a limit on the other, but the resistance in the circuit will determine the actual value (as long as it doesn't exceed the limit).

4. Yes! No reservations - you have this one completely correct!

5. Yes(ish). If your adapter supplies 6V, but it's only capable of supplying a maximum of 0.5mA then it might have a 12k resistance inside (or it might be more complicated than that).
   If it does have a 12k inside, then when you attach the external 1k you'll effectively have 13k connected to a 6V source, resulting in 0.462mA flowing and 0.462V across the 1k resistor.
   Otherwise, if it limits the current to 0.5mA by more complicated means than a simple series resistor, then there's 0.5mA through the 1k resistor and the adapter is only supplying 0.5V.

Answer 2

1. Say the adaptor provides 1 A and I need to calculate the voltage after the resistor. Using V = I · R, I is the current which goes through the resistor and NOT the current provided by the adaptor (1 A), correct? I guess I need to put a multi-meter in the middle to find the current first.

The adaptor provides up to 1 A, and you need to calculate the voltage across the resistor. If the resistor draws less than the supply can provide (6 V / 1k = 6 mA, so that's OK), then the voltage across the resistor is 6 V. A multimeter on the mA range in series with the resistor would let you measure that current.

2. What if my circuit is just a wire connecting + to - with close to zero resistance, essentially shorting the circuit? I guess at that point the current through the wire IS the maximum current provided by the adaptor. Ohm's law says V = I · R = 1 · 0.1 = 0.1 V; that doesn't make sense.

Who knows what the adapter is going to do? It's not a simple physics entity like a resistor or battery. It's been designed to be more clever. With luck, it will protect itself against the short circuit you're putting on it. How it does that is in the gift of its designers. It may deliver its rated current continuously. It may 'fold back' once it has sensed the overload, and only deliver a lower current until the overload is removed. It may switch off, and recycle on again every second. It may deliver several amps and catch fire. It may deliver several amps, and blow a replaceable fuse. It may deliver several amps and heat a self-resetting PolyFuse, which will limit the current to a few mA, needing you to disconnect to allow it to cool and reset. It may deliver several amps, cook for a minute, open a non-resettable thermal fuse, and never work again. I've met all of these behaviours in the field.

3. Say the adaptor provides 1 A and 6 V. Ohm's law then says 6 = 1 · R which means R = 6 Ω. Now what does this represent? Is there a 6 Ω resistor in the adaptor?

Almost certainly not. 1 A is the safe current it can provide. Depending on what overload behaviour (see point 2) your supply has, it may be a limit current, or a trip current, or a fusing current. A limit or trip current will tend to use a small resistor, maybe 100 mΩ in series with the output, with a comparator to sense the voltage across it.

4. Say the adaptor provides 6 V and I need to find the current in my circuit, so I = V/R = 6/1000 = 6 mA?

Absolutely correct.

5. Now say my adaptor provides 6 V but only gives me 0.5 mA, which means there is a 6/0.0005 = 12000 Ω resistor inside the adaptor. Now my answer in [4] doesn't make sense anymore, the adaptor is only providing 0.5 mA, how can the current be 6 mA in the circuit?

If the adapter is implemented as a voltage source in series with a resistor (a reasonable model for a battery, almost never a suitable model for a power supply), then it will limit the current. So if we short circuit the supply, it will now source 0.5 mA into its internal resistance. If we put 1 kΩ on the supply, then the 13 kΩ total will draw 0.46 mA, and drop only 0.46 V across the external 1k resistor. That's why power supplies are not implemented like that.

Answer 3

1. With a constant voltage supply, the voltage across the resistor will be the same as the supply voltage as long as the current is below the current limit threshold. If the current approaches the current limit threshold, the adapter will either go in to constant current mode and supply current at the current threshold, or it will go in to fold-back current limiting and supply something less than 1 A (depends on the design of the supply).

2. Assuming the supply current limit is 1 A, if you short the supply, the supply will either go in to constant current mode and supply 1 A, or it will go in to fold-back current limiting and supply something less than 1 A. During current limiting, the voltage will drop.

3. Power adapters can have sophisticated current limiting circuitry which will limit the current by either going in to constant current mode or fold-back current limiting. The current limiting circuitry in a linear supply is controlled by a pass transistor and circuitry to limit the current.

4. If the adapter supplies 6 V and isn't current limiting, then loading the supply with a 1k resistor will draw 6 mA per Ohm's law.

5. If the adapter has a maximum current of 0.5 mA and you load the adapter with a 1k resistor, then the adapter will go in to current limiting. For a constant current limit, you will get \$ 0.5 mA \times 1k = 0.5 V \$ across the resistor. If the adapter uses fold-back current limiting, you need more information on how the fold-back mechanism works to determine the current flowing through the resistor and the voltage across the resistor.

Many power supplies are designed to have a very low output impedance to give a constant voltage output. To protect the power supply and the circuitry, many power supplies will limit the current to a certain maximum value which means the effective output impedance of the supply will increase. A basic controlled linear power supply uses a pass transistor and circuitry to keep the voltage constant as well as current limiting functionality.

Answer 4

One thing to remember is that Ohm's law, V=IR, is just a simple equation in three variables. It lets you find one value if you know the other two. If you know the voltage across a thing, and the resistance of the thing, then you can calculate the current through the thing with basic algebra.

The other thing to be aware of is that 99.99% of the power supplies you will encounter in real life are going to be constant-voltage supplies. That means they maintain a constant voltage, and allow the current to fluctuate based on the resistance presented to them. Real-world power supplies also have limits, beyond which they will start to malfunction. If you try to draw more than what it was designed for, then the voltage will start to sag, the current will drop off, the power supply will start to overheat, or some combination of the three.

Of course, constant-current power supplies exist too, but they're mostly limited to test environments.

Now, to the questions:

1. Say the adapter provides 1 A and I need to calculate the voltage after the resistor. Using V = I · R, I is the current which goes through the resistor and NOT the current provided by the adapter (1 A), correct? I guess I need to put a multi-meter in the middle to find the current first.

You didn't specify a voltage for the power supply, so I'm going to assume that it's a constant-current type. That is, it will attempt to drive 1A of current through whatever it's connected to, and will vary its voltage in order to meet that demand. In this case, the voltage will be 1A · 1kΩ = 1000V.

Voltage, like motion in General Relativity, doesn't exist in isolation. It can only be measured relative to another point. So, there's no such thing as the "voltage after the resistor". There is the voltage across the resistor, which equals the voltage across the power supply since they're connected that way.

Of course, that can get complicated quickly, so usually what happens is that we pick one point and name it "0V", and measure everything else relative to that. So, if we pick the negative terminal of the power supply to be our zero point (which is typical), then the positive terminal becomes +1000V. But it would be just as correct (if much, much rarer) to say that the positive terminal is zero, and the negative terminal is at -1000V.

2. What if my circuit is just a wire connecting + to - with close to zero resistance, essentially shorting the circuit? I guess at that point the current through the wire IS the maximum current provided by the adapter. Ohm's law says V = I · R = 1 · 0.1 = 0.1 V; that doesn't make sense.

If your power supply is a constant-current type (as I assumed from point #1), then it will regulate its output voltage down in order to maintain that 1A current. So, yes, if the wire's resistance is 0.1Ω, then the voltage will be 0.1V.

If your power supply is a constant-voltage type (let's assume 6V), then the current will skyrocket (up to a theoretical 60A) as it attempts to maintain that voltage across the minuscule resistance. A well-designed power supply will blow a fuse or trip a circuit breaker to protect itself from the short circuit. A more cheaply-designed power supply will set itself on fire.

3. Say the adapter provides 1 A and 6 V. Ohm's law then says 6 = 1 · R which means R = 6 Ω. Now what does this represent? Is there a 6 Ω resistor in the adapter?

It's impossible to have both the voltage and the current be constant, unless the resistance across the terminals is also constant. Since power supplies can be connected to varying loads, that's obviously not the case.

Is this power supply constant-voltage? Then "6V 1A" means it will vary its current output to maintain 6V across its terminals, up to a maximum of 1A. Is this power supply constant-current? Then "6V 1A" means it will provide 1A of current output, up to a maximum of 6V across its terminals.

4. Say the adapter provides 6 V and I need to find the current in my circuit, so I = V/R = 6/1000 = 6 mA?

Yep, that's correct.

5. Now say my adapter provides 6 V but only gives me 0.5 mA, which means there is a 6/0.0005 = 12000 Ω resistor inside the adapter. Now my answer in [4] doesn't make sense anymore, the adapter is only providing 0.5 mA, how can the current be 6 mA in the circuit?

Again, in a constant-voltage power supply, the current listed on the sticker is only the maximum that that power supply is rated for. So, if your power supply is only rated for 0.5mA, you've overloaded it by a factor of 12, which will cause all the dire things I mentioned above.

Answer 5

The thing that will help you the most is to know a detail about how we mark power supplies. If you have a 6V 1A power supply, that does not mean it always emits 1A with 6 volts across the terminals. The power supply will fix one of those numbers, and the other is a maximum. The vast majority of power supplies are voltage sources, which means they will produce 6V and produce up to 1A. (Current sources also exist, but most of the things you and I think of as "power supplies" are voltage sources).

1. The current through the resistor is the same as the current through the power supply. Why? Because there's only one loop for current to go through. So if 1A was going through the power supply, 1A is going through the resistor. However, based on the paragraph above, it is almost certain that what you have is a 6V power supply, which will produce up to 1A. So you can use V=IR to compute the current flowing. Once you have your number, check that the number is not above 1A. If it is 1A or less, then the power supply will be operating within its specified range, and will produce 6V like normal.

2. What if you did the calculations above, and got a number more than 1 amp? To the specific case of your question 2, where you have a short, the calculations will show "infinite amps," which is quite clearly higher than 1A. If this is the case, the power supply is being operated outside of its specified range. That means it will not produce the specified voltages. In this regime, you need to consult the datasheets for this specific power supply to see what it does outside of the simple 6V 1A max regime. Typical behaviors are:

• Provide more than 1A, heat up unsafely, and either melt or catch fire
• Engage some safety circuitry in the power supply that ensures it never outputs more than what is safe. This typically means the power supply will temporarily output less than 6V.

High end power supplies tend to have the safety circuitry. Cheap wall warts.... I highly recommend not trying to drive them beyond their specified maximums.

3. 6 Ohms ends up being the minimum resistance you can apply across the power supply while staying within the specifications. If you have a lower resistance than that, then the current through the resistor at 6V is greater than 1A, and you are in one of the regimes mentioned in #2.

4. Correct. Your calculations show that 6mA will flow through the circuit.

5. If you have a 6V power supply and it is providing 0.5A, you can indeed calculate that there must be 12kOhms across it. If you have a 1kOhm resistor, a 6V power supply and it is providing 0.5mA, your hypothetical situation is wrong, because it simply cannot happen. It can only provide 0.5mA if the resistance is 12kOhms. Of course, if what you actually mean is that you have a 6V power supply that is capable of outputting a maximum of 0.5mA, then you are once again in the regime of #2.

As stated at the beginning, the trick is that a power supply that is listed as 6V and 1A typically produces 6V and up to 1A. If your circuit is configured such that an ideal 6V power supply would produce more than 1 A, you have left the specified range of that power supply. At that point, the physics of the circuit will do something, but whatever it is is typically going to be more nuanced than an ideal voltage source. The voltage may go down, as the power supply struggles to catch up. Or the power supply may happily provide more than 1A, at an increased fire hazard or shortened lifespan. These are things that can be calculated, but are typically avoided. One typically find a power supply large enough for the circuit at hand. Doing anything else just risks all sorts of bad things happening, and nothing good can come of it.

Answer 6

If your power supply specifies 9V and 1A, it does not mean that it will force 1A into any load. In reality, it is liable to vary a bit but an ideal power supply would maintain its stated voltage provided the current remained below the stated limit of 1A. The actual current will depend on the load.

If no load is connected then the voltage will be 9V and there will be no current.

If your load is a 1kΩ resistor then the current will certainly not be 1A. It would require a 1kV supply to push 1A through it (except that it would probably melt or catch fire). With a 9V supply, just 9mA will flow.

With a 9Ω load, you will actually get a current of 1A.

I have been assuming an ideal supply. A real one might have a slightly higher voltage with no load or a small load and its voltage may drop close to its current limit. What happens if you exceed the current limit? That will depend on its design. It might actually work (for a while anyway), it might blow a fuse, or it might catch fire.

Answer 7

This can be confusing for someone new to this, but you need to consider Kirchoff's Voltage Law. It states that the directed sum of voltages around any closed loop is zero. In your example, this means that the voltage across the resistor must be precisely the same as the voltage across the battery.

This generalizes so whenever you have things connected in parallel, the voltage drop across them is the same.

Ohm's law still holds, so we've constrained the voltage part of it to be fixed to 6V. Therefore, the current through the resistor must be $$I = \frac{V}{R} = \frac{6\text{V}}{1\text{kOhm}} = 6\text{mA}$$.

When you attach a battery or a power supply to a circuit, they can be thought of as voltage sources, which means they output their voltage, providing whatever current is dictated as per the load that they are connected to. The current rating is the maximum current that the power supply can provide, not the current that it always provides.

There are circuits you can make as current sources, but their voltage will vary based on the load that they are connected to.

EDIT:

qrk's answer is correct about the supply essentially turning into a current source when the limit is reached.