E12 ( 10%): 10 12 15 18 22 27 33 39 47 56 68 82
E24 ( 5%): 10 11 12 13 15 16 18 20 22 24 27 30
33 36 39 43 47 51 56 62 68 75 82 91
The basic difference is that the E24 have "extra" values between those of the E12. It's roughly midway between the E12 points. They're more expensive too.
There's also an E6- it runs:
10 15 22 33 47 68
At the top end there's an E192- wildly expensive but extremely accurate. They're used in life-critical systems or super accurate industrial machines.
Usually you'll need 100ohm resistor for each led. In this kind of setting, you will get the LED intensity nonuniform because of its differences due tolerance value. the best thing is to use LED driver IC. Another benefit of using LED driver is, you can mix red led with other color. Now, if we wanted to connect a red LED with a forward voltage of 1.8V and a current of 20mA directly to the transformer output then use the following equation: R=V/I = 9 / 0.02 = 450 Ohms. You could use 470 ohms from the E12 resistor series.
To figure this out, you need to know the expected forward voltage and current of the LED. Lets assume 2ma and 2V. (Actually, 2ma is small, but I intend to make a point.) By Kirchoff's Voltage Law, you know that the signed sum of the voltage drops going around a series circuit must add up to zero. This means that the voltage across the resistor must be 228 volts. (-230 + 228 + 2 = 0) By Kirchoff's Current Law, you know that the signed sum of the currents entering a node is zero. As a consequence, you also know that the current at every point in a series circuit is the same. Therefore, the current through the resistor is also 2ma. By Ohm's Law, you know that resistance is voltage divided by current, so you know that the resistor is 228V divided by 2ma, which is 114K. The nearest standard value in the E12 scale is 100K. Recalculate the current for 100K, and you get about 2.25ma. (You could also use 120K, and I'll let you run the calculations yourself.) Don't stop here. There are some issues... By the power law, you know that power is voltage times current, so you know that the power dissipated by the resistor is 228V times 2.25ma, which is 513mW. I would put a one watt resistor in there. However, consider this. 2ma is a low current LED. Some of them pull 25ma. The power in the resistor in that case is about 6.5W, which is getting pretty high. Secondly, you need to consider the reverse breakdown voltage on the LED. I assume that when you said 230V, you meant AC, not DC, which means that there is going to be 230V (actually, a peak value of 325) across that LED for one half the line cycle. You need to check the datasheet and make sure the LED can handle that. If not, you need to put an ordinary signal diode, such as a 1N4148, in parallel with the LED, in the reverse direction, so that it clamps the reverse voltage at about 0.7 volts. (Don't worry about the reverse breakdown on the 1N4148, because the LED will protect it, on opposite half line cycles.) Last, but not least, you need to consider the safety of the operator. 230V is a high voltage, and LED's are not the most rugged thing around. If the LED breaks, you need to consider if its internal wiring could come into contact with the operator. I would certainly demand a UL listed device in this application.
10% tolerance.
The main difference between e11 and e12 light bulb sockets is their size. E11 sockets are smaller and have a diameter of 11mm, while E12 sockets are slightly larger with a diameter of 12mm. This means that E11 bulbs are not compatible with E12 sockets and vice versa. It is important to check the socket size before purchasing a light bulb to ensure it will fit properly.
Series 2 E12 Dead to Rights
If it is a standard resistance (E12 series, with 10% tolerance) it might do. However, note that electrical designs must be prepared to face the tolerances of the components, so unless it is a specific and important resistor (for the gain of an amplifier, current limiter, etc.), there should be no problem. Anyway, resistors being a cheap component, you may as well go to the shop and get a 1% tolerance 30ohm resistor!
The compatibility options for an e12 lamp socket include e12 bulbs, candelabra bulbs, and some LED bulbs designed to fit the smaller socket size.
cross polarization = xpd so, xpd = 20 log E11/E12 where E11 and E12 are the two waves.
Usually you'll need 100ohm resistor for each led. In this kind of setting, you will get the LED intensity nonuniform because of its differences due tolerance value. the best thing is to use LED driver IC. Another benefit of using LED driver is, you can mix red led with other color. Now, if we wanted to connect a red LED with a forward voltage of 1.8V and a current of 20mA directly to the transformer output then use the following equation: R=V/I = 9 / 0.02 = 450 Ohms. You could use 470 ohms from the E12 resistor series.
Yes It Is.
It is 2.3476*10^12, almost the same as in the question!
This can be used to blow up enemy ships (e.g E12)
See Provided Link
It it a formula in a spreadsheet requiring the summing of the values in cells E9 to E12, both inclusive.