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I have a simple Christmas light LED pack. It features 3xAAA battery compartment, an on off switch, a 10 ohm resistor in series and 9 white LEDs wired in parallel along about an 80cm two wire lead.

Obviously with just the 10ohm resistor, the batteries do not last terribly long. So I thought I would replace the 10ohm resistor with a 150ohm resistor. When I bridge the 150 ohm resistor across the open switch and 10 ohm resistor (i.e. connect +V to the 150ohm resistor then to the LED which in turn connects to -V) I observed a current draw that I did not expect.

The current draw that I observed was about ~9mA which was quite different from what I calculated ~30mA.

I have used two multimeters and obtained very similar measurements. Following is the original circuit and the modification I used to measure the current.

enter image description here

Here are my readings using two different multimeters.

Measurement Multimeter 1 Multimeter 2 Units
Voltage (+ve terminal to -ve terminal) 4.05 4.03 Volts
Resistance 148.8 149.4 ohms
Current 9.43 8.9 mA

However, using the formula V = IR or reworked I = V/R. I would have expected that I should be 4 / 150 = 0.0267 mA or 27mA, not the ~9mA I measured.

The resistor I used has the following bands: Brown, green, black, black, brown. And the 3 AAA batteries have been powering the LED's for probably 3-4 hours already via the 10 ohm resistor hence the voltage of about 4V remaining.

So my question is why is my calculation so different from the observation by a factor of about 3x? What am I missing / doing wrong?

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  • \$\begingroup\$ Don't forget that a milliamp-meter has also resistance ... or a "burden" voltage. view.officeapps.live.com/op/… electronics-tutorials.ws/blog/ammeter.html \$\endgroup\$
    – Antonio51
    Commented Dec 22, 2021 at 13:36
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    \$\begingroup\$ You calculation is not correct. LEDs drop volage should be considered. And, by the way, connecting LEDs in parallel without resistors is bad practice. LED parameters may be slightly different which cause the different currents running through and difference in brightness \$\endgroup\$
    – user263983
    Commented Dec 22, 2021 at 13:39
  • \$\begingroup\$ @user263983 is correct. Also, you trying to apply ohms law while you measure open-circuit voltage but closed-circuit current. If you measure the voltage across the multimeter terminals with a 2nd multimeter, you'll at least see that ohms law holds. \$\endgroup\$
    – RJR
    Commented Dec 22, 2021 at 13:53
  • \$\begingroup\$ LEDs have a forward voltage drop that needs to be considered. You should measure the voltage drop of each LED using the diode setting on your multimeter. You will likely see around 2.5V - 2.7V. Different colours will have different forward voltage drops. \$\endgroup\$
    – Steve771
    Commented Dec 22, 2021 at 13:58
  • \$\begingroup\$ Thanks for those replies. Unfortunately the display is a sealed unit. I cannot open it without destroying it (it is transparent but glued closed). The only access I have is the leads into the battery box. As such, I cannot do much about how the LEDs are wired up. Sadly, I cannot get the diode measurement to work on either. On one, it always shows 0 - despite changing the battery and carefully following the instructions (the LED doesn't light up). On the other, the LED glows faintly, but I cannot interpret the display which simply reads 1 and the docs are in Chinese. \$\endgroup\$
    – GMc
    Commented Dec 23, 2021 at 1:58

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