If the loads are wired in series, add the three values to get the total R.
If they are in parallel then the formula is 1/R = 1/R1 + 1/R2 + 1/R3
The total resistane when resistors are in parallel is ALWAYS LESS than the lowest individual value.
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
We can do this by using Ohm's Law: V = I x R, where V is the voltage, I is the current and R is the resistance. Re-arranging Ohm's Law, R = V / I so 110 / 5 = 22 ohms. Again re-arranging Ohm's Law, I = V / R so 230 / 22 = 10.45 amps.
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.
"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.
15% of 22= 15% * 22= 0.15 * 22= 3.3
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.
15% of 22 = 15% * 22 = 0.15 * 22 = 3.3
1.4667
The relationship between Volts, Amps and resistance is expressed by the following formulas. V = A*R, A = V/R, R = V/A, So to calculate the resistance of the lightbulb use R=220/10. Resistance equals 22 ohms. If you were to measure the resistance of the bulb with an ohm-meter you would get a lower reading. The resistance of the filement goes up as the bulb heats up. So the current (amps) are higher than 10 amps when the bulb is first turned on.
15 % of £22 = 15/100 × £22 = £3.30
"Deep, Dark, and Dangerous" by Mary Downing Hahn has a total of 15 chapters.
Resistance will increase.