A megohm is 1,000,000 ohms, so 2.2 megohms is 2,200,000 ohms, often abbreviated to 2.2M, or 2.2Mohm, or 2M2.
Using a resistor color code chart below, you will find that the resistance is calculated this way: Resistance = (1st Digit x 10 + 2nd Digit) x Multiplier Hold the resistor with the gold or silver band to the right and read the resistor from the left to the right. So, in this case, the resistor is: RED RED YELLOW GOLD Using the formula and resistor color chart found in the link below, the resistance would be: R = 1st Digit x 10 + 2nd Digit) x Multiplier R = (RED x 10 + RED) x YELLOW R = (2 x 10 + 2) x 10,000 R = 22 x 10,000 R = 220,000 ohm (220 K ohms) Since the final band is gold, the tolerance is 5%
The time constant of a 4.7 µF capacitor in series with a 22 KΩ resistor is about 103 ms.
Ok so if you have to take the 5 amperes and multuply them by the 22 ohm resistance giving you the answer of 110 5 * 22 = 110 volts
220 ohms with a 10 percent tolerance. Red is 2 and brown is 1. Brown is in the multiplier band so it is 10 times the value in the first two bands. Silver is in the tolerance band. Gold would have been a %5 tolerance device.
If the resistors are in series the voltage can not be divided, as it has to pass first through one then the other. The amount of current that flows through a set of resistors in series will be the same at all points and the total resistance in the circuit must be equal to the sum of all the individual resistors added together. In other words the 22k and 12k Ohm resistors are the sames as a single 34k Ohm resistor.
22Kiloohm is 22,000 ohms 22megohm is 22,000,000 ohms Kilo is a thousand meg/mega is a million
"Sensitivity" is not a word normally applied to resistors. Characteristics of resistors include "resistance", "tolerance", "power rating", and "temperature coefficient". "Inductance" and "capacitance" are also used in describing certain critical performance resistors. A 22 KOhm resistor will require 22 v of voltage to induce a current of 1 ma. This is Ohm's Law: voltage = current times resistance.
You can do it by putting a resistor in series to limit the current, but you have to calculator the resistance needed. First you need to find out the capacity of the battery, e.g. 2000 mAh (milliamp-hours). The charging current should be set at at one tenth the capacity (or less). So in the example, the charging current is 200 mA, or 0.2 amp. The resistance is found by dividing the voltage drop, four volts, by the current. That gives a resistance of 4/0.2 or 20 ohms. You can use a 22-ohm resistor. It dissipates a power of 4 x 0.2 watts, or 0.8 watts, so it should be a 22-ohm 1 watt resistor.
Using a resistor color code chart below, you will find that the resistance is calculated this way: Resistance = (1st Digit x 10 + 2nd Digit) x Multiplier Hold the resistor with the gold or silver band to the right and read the resistor from the left to the right. So, in this case, the resistor is: RED RED YELLOW GOLD Using the formula and resistor color chart found in the link below, the resistance would be: R = 1st Digit x 10 + 2nd Digit) x Multiplier R = (RED x 10 + RED) x YELLOW R = (2 x 10 + 2) x 10,000 R = 22 x 10,000 R = 220,000 ohm (220 K ohms) Since the final band is gold, the tolerance is 5%
The time constant of a 4.7 µF capacitor in series with a 22 KΩ resistor is about 103 ms.
The size of the resistor will depend on the load. Let's look at this a bit to see if we can make sense of it. You want to drop the applied voltage to a device from 12 volts AC to 11 volts AC. That means you want to drop 1/12th of the applied voltage (which is 1 volt) across the resistor so that the remaining 11/12ths of the applied voltage (which is 11 volts) will appear across the load. The only way this is possible is if the resistor has 1/11th of the resistance of the load. Here's some simple math. If you have an 11 ohm load and a 1 ohm resistor in series, you'll have 12 ohms total resistance ('cause they add). If 12 volts is applied, the 1 ohm resistor will drop 1 volt, and the 11 ohm load will drop the other 11 volts. A ratio is set up here in this example, and each ohm of resistance will drop a volt (will "feel" a volt) across it. See how that works? If the resistance of the load is 22 ohms and the resistance of the (series) resistor is 2 ohms, each ohm of resistance will drop 1/2 volt, or, if you prefer, each 2 ohms of resistance will drop 1 volt. The same thing will result, and the load will drop 11 volts and the series resistance will drop 1 volt. That's the math, but that's the way things work. You'll need to know something about the load to select a series resistance to drop 1/12th of the applied voltage (which is 1 volt) so your load can have the 11 volts you want it to have. There is one more bit of news, and it isn't good. If your load is a "dynamic" one, that is, if its resistance changes (it uses more or less power over the time that it is "on"), then a simple series resistor won't allow you to provide a constant 11 volts to that load. What is happening is that the effective resistance of the load in changing over time, and your resistor can't "keep up" with the changes. (The resistor, in point of fact, can't change its resistance at all.) You've got your work cut out for you figuring this one out.
It means the distance from zero. | 22 | (the absolute value of 22) is 22; | -22 | (the absolute value of -22) is also 22. Basically just take out the negative, and that is the absolute value.
Resistance will increase.
" value of the colteer 4-22 rifle ? "
Ok so if you have to take the 5 amperes and multuply them by the 22 ohm resistance giving you the answer of 110 5 * 22 = 110 volts
-22 ( negative 22 )
value of mossberg 800c 22-250