I have a simple circuit:
The max rating (current) of the LED is 30mA.
How can I work out what the resistance of the resistor needs to be? By using ohm's law I found it to be \$3V/0.03A = 100 \Omega\$. However using software called Yenka, and trial and error I got the minimum possible resistance to be 36Ω. However, if I use a 35Ω resistor, then the LED breaks. Is the software wrong, or (more likely) am I doing something wrong?
You'll have to check the datasheet or measure it to know how much voltage drops over your LED. Let's say this is 2V. Then the voltage over the resistor is the difference between your power supply (3V) - voltage over the LED (2V) = 1V. To get 30 mA through the resistor (and thus also the LED) your resistor has to be 1V / 30 mA = 33 Ohm.
If your LED voltage is lower the current will be somewhat higher, but the LED shouldn't break!
You have to take in count the \$V_{F_{typ}}\$ of the LED.
Let's take a red LED as example:
The LED and the resistance (R) are in series. The source voltage (3V) is the summation of voltage across the LED (\$V_{F_{typ}}\$) and the voltage across the resistance (\$V_{Res}\$). To calculate R we need to know the voltage on the resistance. Let's use the maximum current, but normally is used 15mA.
\$R = \frac{V_{Res}}{I_{F_{max}}}\$
So,
\$V_{Res} = V - V_{F_{typ}}\$
\$V_{Res} = 3 - 1.7\$
\$V_{Res} = 1.3V\$
\$R = \frac{V_{Res}}{I_{F_{max}}} = \frac{1,3}{0,03} = 43,33\Omega\$
It's important calculate power dissipation in the resistor, too:
\$P_{res} = V_{res} * I_{F_{max}}\$
\$P_{res} = 1.3 * 0,03 = 0,039W\$ (You can use 1/8W or 1/10W)
You have already gotten some direct answers, but here are a few other things to think about.
You didn't state it explicitly, but you implied you want 30mA thru the LED. Is this really correct? Certainly there are LEDs and applications for which this is valid. This is not meant as a offence, but if you're here asking a basic Ohm's law question you probably have a common T1 or T1-3/4 LED which is most likely rated for 20mA.
Despite what the maximum allowed current is, you need to consider what the purpose of the LED is. Is it for illumination? In that case you probably do want the maximum light you can get. However, your picture implies it's a red LED, so I'm guessing not. If this is just a indicator, you don't need to run it anywhere near its maximum current unless there are unusual circumstances like it needs to be visible in sunlight. Full current can make the result uncomfortably bright as a indicator for a indoor device. I usually run 5mA thru a 20mA LED for indicator use.
Is the power coming from a battery? If so, it is likely worth getting a more efficient LED and running it at a lower current.
My second comment is about your equation, "3/0.03 = 100 Ohm". Please be more careful with units. That will not only communicate your intentions better, but make it more likely you think about the underlying physics better. Technically your equation is incorrect since the ratio of two dimensionless numbers can't result in a value in Ohms. From context, you meant 3 Volts divided by 0.03 Amps = 100 Ohms.
While that would at least make it correct, I recommend to write values in electronics in engineering notation. That means use a power of 1000 such that the value is at least 1 but less than 1000, then apply the appropriate prefix. In this case, 0.03 Amps is better said as 30 milliAmps, or 30 mA. I know this can seem arbitrary and inconvenient to beginners, but it's worth learning and getting used to. After a while you'll form a picture in your mind what a milliamp, microvolt, kiloOhm, etc, is. Those skilled in the art whom you hope will answer your question likely had to look at 0.03 Amps and then think to themselves "Ah, he means 30 mA".
I'm new to this web site and don't yet have a good understanding of answers versus comments. I tried to answer your comment as a comment, but apparently they have a small character limit.
It is not only OK to put units in equations, but it is wrong not to. Your "3V / 30mA = 100 Ohms" is exactly how it should be written.
You bring up a good issue with a calculator though. Calculators generally don't handle units. They basically operate as if everything is a dimensionless quantity. This is where using the scientific notation capability of your calculator becomes very useful. To compute the resistance you do 3 / 30e-3. A value with a small exponents like 30e-3 might be easy to convert in your head to .030, but consider doing a computation with 22pF. 22e-12 is easy enough to enter, but what's the chance of getting .000000000022 right?
