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new to the site and to electronics.

I just started learning about ohm's law and Kirchhoff's voltage law and trying to confirm values I am reading.

I have a simple circuit with a 9 V battery, 100 ohm resistor and red LED, the datasheet tells me forward voltage is 2.2 to 2.4. Checking with a multimeter the battery is 9.2 V and the resistor is 99.6 ohm.

First issue is voltage, when I measure the LED it has 2.33 V and the resistor has 5.8 V, this is 8.13 V not 9.2?

Then for current I believe we first get voltage (9.2 - 2.33 = 6.87) then 6.87/99.6 = 68.96 mA but when I measure I get 56.8 mA?

I don't understand why the numbers don't add up, can anyone help explain please?

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When you add your resistor and LED to a feeble 9 volt supply/battery, it's likely that the terminal voltage drops so, measure the voltage across the supply when you have your circuit (resistor and LED) connected.

So, you should probably trust the implied 8.13 volts under load. But, you have to remember that the multimeter measuring amps might add a few ohms of series resistance too.

So, with 5.8 volts across the resistor you should have a current of 58.23 mA but, when you insert the meter this is slightly less at 56.8 mA. The 56.8 mA would imply a resistance of more like 102.1 Ω. So, your meter's current shunt resistor is about 2.5 Ω.

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    \$\begingroup\$ Thank you so much, this really helps me understand a bit mnore, much appreciated. \$\endgroup\$ – ccandy Jan 19 at 13:53
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Welcome to the board, and to the world of electronics! I think you're going about things in a good way, by actually making circuits and testing them to see if they do what you expect.

What's most likely happening is that when the battery connects to the circuit, and delivers a current of about 60mA, the voltage at its terminals goes down. This is because the battery itself also has some resistance internally (which gets more as it discharges). You can confirm this by measurement. (Measure battery voltage with the circuit connected, and then the voltage across resistor and LED).

A few remarks :

  1. 60mA is a lot of current for most LEDs - around 20mA is a typical maximum. I hope you don't burn your LED out (if you do it will go out and no current will flow).
  2. We generally don't need to specify voltages and currents to quite such a degree of precision. In this kind of circuit, working to with +/-1mA would be more than enough.
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    \$\begingroup\$ Thanks, I was surprised about the current, I just followed an example and yep I have burnt some LEDs already :( But all part of the learning, the little comment about +/- 1mA really help with general understanding so thanks again. \$\endgroup\$ – ccandy Jan 19 at 13:54

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