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#### kapilp

##### Member level 3
Hi, I've purchased a USB Power Adaptor (240-110V AC input, 5V/1A output), similar to the one used for Apple products. The adaptor even though rated for 1A, provides a maximum of 100mA near 5V (greater than 4.5V). When I connect a 10E/2.5W resistor across the terminals, the current meter shows 450mA (fine!), but the voltage reading says 3.5V. This doesn't make any sense to me according to Ohm's Law. Can somebody explain this please ..?

Thanks.

What voltage reading on the termnals do you get from same voltage meter when no load is connected?

@prototyp ....... 4.98V...
shouldnt it normally be around 5.5V?

No, around 5V sounds good to me.

How long is the cables? Are they very thin?

I'm using a USB cable with a mini USB plug on one side that goes in my device ... should that matter?

try lighter load i.e bigger resistors and the voltage will not decrease....
you need a adapter with actual higher current rating....

@hussain
I understand the overloading part, but cant correlate it with Ohms law... 10Ohms*450mA = 3.5V dosent make sense.

That gives 7.8 ohms. Your resistor has probably lower resistance than specified. Not far from 10 ohms anyway.

......no, .. its 10ohms ...

it is because the power of your adapter is not as it is written....
since power P = IV hence voltage decrease as the current increase...

kapilp

### kapilp

Points: 2
It probably has to do with the voltage ripple which increases when you pull more current so the dc multimeter can't measure that correctly.
This is also a switching power supply so any ripple will be in a frequency of many KHz.
An oscilloscope would show you that the shape of the output voltage is not a straight DC line.

Alex

kapilp

### kapilp

Points: 2
@alexan_e .... checked it with an oscilloscope from Hz to MHz range, there's hardly any ripple in the output (0.1Vp-p) ... Is there a possibility that the adaptor internally compensated/filtered the ripple, but ends up reducing the output as well..?

@hussain .... I agree with the P=IV part (thats what I mean by overloading), but i still dont get the fact that the system wont obey Ohm's law.

Last edited:

As long as you're not able to measure voltage and current simultaneously, you shouldn't blame ohms law. The power supply may have a foldback characteristic, supplying more current when ampmeter is series connected to the resistor. Or the instrument is inaccurate.

The "1A" rating seems to be pretty void anyway. You may want to determine the actual characteristic with a variable resistor or a set of fixed ones.

i agree with FVM you shouldn't blame Ohm's law..
if your adapter have actual ratings 5v/1A then even at 1A the voltage would have been stable at 5V (almost)...

it will follow the Ohms law when you operate it within its actual (not written) power range...

@FvM an Hussain, .. I'm not blaming Ohm's Law for anything. I've checked the adaptor with a variable resistor, noted the characteristics and then posted in this forum. I'd already come to the conclusion that the device cannot supply the stated power and that the maximum capability seems to be somewhere around 1.5W. Then I rechecked it with the above stated fixed resistor (on suggestion of the vendor who only had an ammeter) and got confused. All I want to know is why is the Ohm's law not satisfied above the threshold.. ? When it crosses the threshold, why does the current not drop, instead of the voltage .. ? Sorry if I put it wrongly in the first place ...

Thanks

Ahhh....your location is Mumbai and naturally the adaptor you bought should be a sub standared one available for 25-35 Rupees. The marking on these will certainly differ from the one available as actual output. Note that there are even those which gives out 8 or 9 volts instaed of 5. Beware of such volatges as to protect your USB device. Measure the volatge and ampheres and then use. OR go for a genuine product from a reputed dealer. Cheers

All I want to know is why is the Ohm's law not satisfied above the threshold.. ?

The OHM's law is valid always, as long as you have 2 of the three variables of the equation (current, voltage, resistance) then the third will be exactly the equation result, there is no way to have a different result so you are obviously measuring something wrong or the circuit data change between the current and voltage measurement.

Alex

@pranam77 ... agree with you, its of chinese make and was available cheap ... so got it for testing ... it actually cost me 150 Rs. ... but is thankfully not a linear one as you mention ... those are the ones that come for 25 Rs. ... damaging stuff ....

---------- Post added at 04:59 ---------- Previous post was at 04:44 ----------

@ alexan_e .. the voltage and current readings are taken simultaneously. I am attaching the schematic of the test setup, tell me if I am wrong here ...

adaptor output ------ ammeter -------- voltmeter+
/
\ R=10ohms
/​

sorry for no schematic

There is no way except to buy a new one

@ alexan_e .. the voltage and current readings are taken simultaneously. I am attaching the schematic of the test setup, tell me if I am wrong here ...

adaptor output ------ ammeter -------- voltmeter+
/
\ R=10ohms
/​

sorry for no schematic

When you use use an ampere meter and the load in series you create a voltage divider, the total voltage will be Vammeter+Vload but since you have connected the voltmeter to the load only you read a lower value.
Try to connect the voltmeter at the power supply connector, before the ammeter to read the full voltage.

Alex

---------- Post added at 10:19 ---------- Previous post was at 10:03 ----------

This will not help about the error you have with the voltage across the 10ohm resistor and the current through it, you will just measure the exact output voltage of your psu.
I wonder if the measurement will change if you add a small capacitor across power supply output or the load.

Alex

kapilp

Points: 2