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how to measure current without altering the voltage


Its all in the title but you have to know that i only have a voltmeter. (it is for a project and i cannot use an currentmeter for some reason)
Sorry about my English i am french.

My first idea was to use a resistance and apply U=R*I but this change the voltage of my solar panel and i want to do both measure at once.
i am lost.


¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
The only option is to do something like use a hall effect sensor to measure the current through a wire.

Any ammeter placed in series with a current will drop some voltage.

If you can tolerate some voltage drop, then a dedicated ammeter is an option. The voltage drop may be under 100mV
first thanks for quick reply.
the problem is that I use an automatic device to do the measure
and this device just provide a voltmeter, not an ammeter. I cannot use a hall effect sensor or this kind of things. The only thing i have is a voltmeter.
Is there a way to use a resistance and U=R*I without altering the voltage?

I had tryed this : ------------solar panel------------
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with this i can measure the current, I=2V/468=4,3*10^-3A
but the voltage is 8V without resistance and 2V with it .
that is my problem, the voltage is altered.


¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Is that 470 ohm resistor your load? If it is, then measuring the voltage across it will give you both the load current and the voltage across it.

If your load is more complex you can't easily do this.

Doesn't your meter have a current range?
In your example above you're obviously effectively measuring the short circuit current of the solar panel; 4.3mA @ 2V. The resistor is in parallell with the solar panel.
What are you trying to achieve? What is the ultimate purpose/goal of your project and you measurements?
Would you be able to build or use a measuring amplifier to convert a mV current sensing resistor or hall effect sensor to a higher Voltage signal?
It is a school project: i wanna measure the current and the voltage all 5 minutes, save the measure on my hard disk (the measure device is connected with my laptop) and use the measures to display graphics (current in function of voltage for example).

steve--> my meter doesnt have a current range.


¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
To measure current, obviously some current must flow, for that you need a circuit. If you're not powering anything in particular you can use a resistor.

If you use a resistor, the voltage across the resistor will also define the current through it.

Note though that the voltmeter will also draw some current (the same way that an ammeter drops some voltage. You can't measure voltage without changing the current, and you can't measure the current without affecting the voltage. However the effects can be small enough that they are of no consequence. In this case, it is likely of no consequence.

The fact that voltage and current are precisely in step with each other makes for a very dull graph. Perhaps you should graph voltage and power? Power generated from your panel is (I^2)xR or I * V or (V^2) / R (all will calculate the same value)

Power is in Watts, which is Joules per second, so knowing the time between measurements you can calculate the amount of energy in Joules that has been captured, and you can graph the cumulative value of this.

If you have access to several identical solar panels, you could use a different resistor value with each and see if that makes any difference to the total power generated over some period. If you select the resistors correctly you will find that different resistors seem to be "better" under differing conditions.

The last finding may suggest a conclusion about solar panels.

If you do attempt something with multiple panels, they need to be close together and oriented the same way so that they receive light at the same angle and so that (as far as practicable) they get the same light conditions.
Your circuit can give you the voltage across the 470-Ohm load, which is basically the voltage that the solar panel supplies (minus a negligible amount because of the voltmeter). If your load is parallel with the source, then the voltage shouldn't change at all. Well, ideally, that is. I find it weird that it changed from 8 to 2 volts there.

If it is indeed 2V, then your answer, 4.3mA should be right.