To find the voltage drop over a resistor, you measure it with a voltmeter connected across the resistor. You also need to make sure the impedance of the voltmeter is high enough to not shift the effective resistance more than the required accuracy of the measurement.
frist off all calculate current through that resistance then use ohms law and remember one thing resistance connected parallel acros Voltage source.
A voltmeter is used to measure voltage. To calculate the voltage across a resistor you must first know the current through it, and then use Ohm's law.
Ohm's law states that voltage = current x resistance.
A potentiometer.
Due to the physical construction and size of the resistor, at a certain voltage, the insulation will break down and the applied voltage will arc over. This is generally bad. Operating the resistor within its voltage rating will prevent this failure mode.
This is a voltage drop question. To answer this question a voltage must be given.
AC card have over/under voltage adjustment for fault. If voltage drop to that range it will trip the fault. Also the unit have voltage adjustment in the bottom right. Voltage coming out of the cable vs voltage coming out Underwood can be different.
Zener impedance is calculated, like anything else, with Ohm's law. Take voltage and divide by current, and you get resistance (or impedance). The only thing special about a zener is that it is not linear, and you can not expect the impedance to remain the same if you change the voltage or current. In fact, the primary characteristic of a zener is that the voltage is relatively constant over the specified current range so, in effect, the zener is a dynamic resistor or, more correctly, a mostly-independent voltage load.
Generally, yes. It would depend on the device you are talking about. In a resistor a change in current will result in an instant change in voltage. Inductors and capacitors do not change instantaneously but will over time.
A potentiometer.
Due to the physical construction and size of the resistor, at a certain voltage, the insulation will break down and the applied voltage will arc over. This is generally bad. Operating the resistor within its voltage rating will prevent this failure mode.
If the voltage is supplying any current through the cable, i.e. if there is any 'load' at the end, then the voltage will drop through the cable.
Voltage drop due to the resistance present in the series circuit causes voltage split over a series circuit.
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.
The voltage drop depends on the current through the cable.For DC current in cable of 16 mm diameter, at 68° F, the voltage drop is(0.00857) x (current, Amperes) volts.
What happens to the current in a circuit as a capacitor charges depends on the circuit. As a capacitor charges, the voltage drop across it increases. In a typical circuit with a constant voltage source and a resistor charging the capacitor, then the current in the circuit will decrease logarithmically over time as the capacitor charges, with the end result that the current is zero, and the voltage across the capacitor is the same as the voltage source.
There is voltage drop over any sysetm that does not have infinitely low resistance, but with reasonable cable size there would be very little drop over 200 metres. The number of phases makes little difference.
If the input resistor is 3k ohms and the feedback resistor is 33k ohms in an inverting amplifier the voltage gain is -11, the ratio of 33 over 3. Consider this... The negative input terminal is a virtual ground, assuming that the positive terminal is tied to ground (usually) through a resistor. That means you have a simple voltage divider between output and input, and the output will be whatever it takes to get the common point (the negative input) to be zero. From there, its all just Ohm's law.
A rheostat is a mechanically operated variable resistor, consisting of a wire coiled around a rod (or ring) the contact slides over the coil, varying the length of wire that the current passes through, giving an adjustable voltage drop.
A voltmeter would measure the voltage. If you measure the voltage drop over a known low resistance you get a kinda-sorta idea of the power available.