# Measured amperage differs greatly from my math

I've got the following circuit:

simulate this circuit – Schematic created using CircuitLab

It works. The LED glows nice and bright.

However my math doesn't match the readings I'm getting from my digital multimeter. Using Ohm's Law, I take the voltage rating of the batteries and desired amperage of the LED to figure out how much resistance I need.

This is my math:

V = I*R = 4.8V (4x AA 1.2V/2400mA in series)
I = V/R = 0.02A (1x 3mm Green LED 3.0-3.2V/20mA)
R = V/I = 240Ω (1x 220Ω & 1x 20Ω in series)


Voltage across the batteries: 5.17V
Voltage across the LED: 2.93V
Voltage across the resistors: 2.24V
Resistance across the resistors: 238Ω
Amperage on the circuit: 9.56mA


Now obviously I am getting more volts from the batteries than I anticipated but even adjusting for that I still don't get ~9.5mA of current in my math.

V = I*R = 5.2V
I = V/R = 0.022A
R = V/I = 238Ω


What am I missing?

• Your circuit (schematic) is most definitely incorrect. The LED is backwards-biased. – Eugene Sh. Jul 19 '18 at 19:24
• And yes, V/R=2.24V/238Ohm=9.412mA. From your data. – Eugene Sh. Jul 19 '18 at 19:27
• @EugeneSh. The LED produces light so I must have built the schematic in CircuitLab incorrectly. Does the LED symbol need to point to the left? I assumed the arrow followed the electron flow. – Exide Jul 19 '18 at 19:27
• The arrow should follow the current flow. – Eugene Sh. Jul 19 '18 at 19:28
• @Exide There's a difference between conventional current and electron flow. – Daniel Jul 19 '18 at 19:30

Using your original numbers (4.8V-3.1V)/0.02A = 85$\Omega$
So try 100$\Omega$.