Electronic – How to calculate current and voltage draw of a single resistor circuit

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If I have a single 270 ohm resistor connected in a circuit to a 3.3V 50mA power supply, how can I measure the voltage and current for the resistor? I'm familiar with Ohm's law V=IR, where V is the voltage differential, but what is the differential for a resistor?

I'm trying to work out why when I put a blue LED (3V@20mA) into the circuit, the voltage I actually measure is less than 3V (more like 2.4V).

Update

I think the thing that is confusing me is the discrepency between the theoretical values, and the measured values.

Theoretical values:

I = V/R
I = 3.3 / 270
I = 12.222mA

Measured values:

3.28V
267 ohm
11.7mA

But then if I plug the measured values back into Ohm's Law, they don't equate?!

I = 3.28 / 267
I = 12.28mA

V = IR
V = 0.0117 * 267
V = 3.12V

Best Answer

You have to start with a closed circuit, so that there can flow current. Do you have the resistor connected between the + and the - of the power supply? Then the voltage difference is 3.3 V. And you use Ohm to calculate the current.

The LED. Did you place that in series with the resistor? Which is how a LED circuit is built: the resistor makes sure that there's not too much current through the LED. Always use one.

If the LED's voltage would be 3 V then the difference between your supply voltage and the LED's voltage would be across the resistor. Kirchhoff is to blame for that. Kirchhoff's Voltage Law (KVL) says that the total of the voltages in a closed loop is zero. So we'll have 1.1 mA through LED and resistor. The 3 V was specified at 20 mA, so we're an end below that. It's normal for a LED to have a lower voltage at low currents.

But note that the 1.1 mA was true for 3 V LED voltage. We're apparently at 2.4 V, so the difference is now 0.9 V, and the current 3.3 mA. If you decrease the resistor value so that the current increases, you'll notice that the LED's voltage will increase as well.

How do you calculate the value? (Here we go again)

\$ R = \dfrac{\Delta V}{I} = \dfrac{3.3 V - 3 V}{20 mA} = 15 \Omega \$

edit re your update of the question
Welcome to the real, imperfect world. What you have at hand is measurement error. This is an important issue in engineering, and handling it properly can be a painstaking process.

You're giving your numbers in three significant digits, that's probably what the multimeter gives you. A multimeter's precision is most of the time expressed as a percentage (relative error) + a "count" (absolute error). A hobby quality meter may for instance have 2 % precision +/- 1 count. The 2 % should be clear: a 100 V reading may actually represent anything between 98 V and 102 V. The 1 count is an error in the last digit. A 5 may actually be a 4 or 6. That's an absolute error and doesn't depend on the value the meter gives you. If you measure 100 V then 1 count represents 1 %, if you read 900 V (same number of digits!) then 1 count is 0.11 %.

Let's presume you have a decent multimeter with 1 % +/- 1 digit precision. Then worst case your values may become

3.28 V - 1 count = 3.27 V, - 1 % = 3.237 V
267 Ω + 1 count = 268 Ω, + 1 % = 270.7 Ω
11.7 mA + 1 count = 11.8 mA, +1 % = 11.92 mA.

3.237 V / 270.7 Ω = 11.96 mA, which agrees well with the 11.92 mA we calculated for worst case. If your multimeter has a 1.5 % precision the calculated current will fall perfectly within the measured value's error range.

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