There is a 12V (max 8A), voltage regulated power supply which is directly connected to a pure resistive lamp as shown in Figure1 below(sorry for primitive drawings): enter image description here

I'm planning to monitor the power consumption of by a DAQ device. My DAQ can handle up to 10V so if I use a 1 ohm resistor in series(Rshunt) for the current's data acquisition(between a and b) and measure the voltage across the lamp(between b and c) I was thinking to log the power consumption since the power = I*V.

My question are:

1-)Is my setup fine? Would 1 ohm resistor handle that much current?

2-)If not what kind of device would you suggest which I can log the power?

  • \$\begingroup\$ Have you worked out the power in the resistor, and then done a search to find what resistors are available? Have you considered how much voltage is left to drive the bulb if 8A was going through a 1 Ohm resistor? \$\endgroup\$
    – Icy
    Commented Nov 20, 2015 at 12:49
  • \$\begingroup\$ i see, im actually after messuring consumption from battery. is there an easy way or device to monitor power consumtion? I have daq. a device with a voltage output eould work \$\endgroup\$
    – user16307
    Commented Nov 20, 2015 at 13:00
  • \$\begingroup\$ Using a shunt resistor will work, you just need a lower value, and then amplify the voltage across it. Alternatively there are hall effect current sensors available that you could look into. \$\endgroup\$
    – Icy
    Commented Nov 20, 2015 at 13:04
  • \$\begingroup\$ what resistor value would you recommend? then I need to use an opamp as well. \$\endgroup\$
    – user16307
    Commented Nov 20, 2015 at 13:09
  • 1
    \$\begingroup\$ You do know basic electronic formulae? Power = I^2 R = 0.64W - That's a lot less than the 64W you were proposing in your original design, 1W resistors are quite common. \$\endgroup\$
    – Icy
    Commented Nov 20, 2015 at 13:34

1 Answer 1


Is my setup fine?

That depends... You didn't mention it, but I'm assuming your lamp is rated for 12V and no more than 8A? If so, then the setup is fine, but you need more information to do this successfully.

connected to a pure resistive lamp

What do you actually mean by this? All bulbs and lamps are resistive, as are motors and refrigerators. (Yes, motors are inductors, but they are still resistive). A common lamp is technically an inductor, given it is a coil of wire; however, the coil is only there to increase the resistance and adds practically no reactance. Even still, lamps are non-ohmic, which is to say, the resistance is not constant with varying voltages.

You need to know the expected current draw of the lamp. This is important in choosing a shunt resistor value. It isn't necessary, but in this sort of specific design, it can be helpful.

Would 1 ohm resistor handle that much current?

First of all, the resistance metric has nothing to do with a resistor's ability to dissipate power. Resistors have power ratings - 1/4W is fairly common. If you knew the expected current usage of the lamp, you could anticipate the current flow through the resistor; however, you could also use the 8A max of the power supply as a worst case metric.

Secondly, when you measure current with a shunt resistance, the added resistance should be virtually zero. You are not trying to add any sort of load to the circuit, just monitor it. What if the lamp was drawing 4A, then you would have a 4V drop across the 1Ohm resistor! This also means the resistor would need to be rated for at least 16W! On top of this, the added resistance will affect the operation of the lamp as you are adding reducing the available voltage drop across the lamp, which will affect its own resistance and current consumption.

Typical shunt resistors are tiny - try something like 0.001 Ohms. Now, that same 4A load would equal a 0.004V drop and only 0.016W.

With a known resistance, you can calculate the current flow by measuring the voltage across the shunt resistor; however, to notice any meaningful changes in voltage drop across the shunt resistor, you will likely need to amplify that voltage. The power used to amplify the voltage should be external from your power source, or it will also affect the operation of the lamp circuit!

Lastly, you mention using a DAQ, but you have not specified what kind or any other information about it. This site will not recommend parts and tools to you, but people will offer you help in setting something up. There are specific devices with built in shunt resistors, some require custom external components. You have to know the specifications of things, you can't just plug and play bits and pieces of things and expect it to work.

  • \$\begingroup\$ i think i'll go for a hall effect sensor. \$\endgroup\$
    – user16307
    Commented Nov 20, 2015 at 14:50
  • \$\begingroup\$ That will just fine, as long as it is rated for measuring your current levels (some are suited for higher or lower measurements). You will still likely need external amplification, unless you are using an expensive off the shelf current monitor. \$\endgroup\$ Commented Nov 20, 2015 at 15:03

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