A battery is rated to supply a certain number of volts. However, it actually supplies less, because they are "lost" as the current has to get out of the battery in the first place.(The battery has internal resistance)
The amount of lost volts depends on the current being drawn:
The less resistance a circuit has, the more current is drawn, because it's easier to flow.
Example:
If the circuit has little resistance, it draws a large current and the battery's internal resistance causes more lost volts.
If the circuit has high resistance, it draws a small current and there are fewer lost volts.
This is why when you short-circuit a battery (give it hardly any resistance to go through) it heats up and may explode. A large current is drawn and all the volts are used by the battery's internal resistance.
There's a couple ways. You could count the number of semiconductor junctions, resistances and other voltage attenuators, do all your Ohm's Law and drop calculations to come up with a reasonably good number...but if the circuit has been constructed, you could always just power it up, read the voltages at the supply and at whatever point you're trying to calculate the voltage loss for, and subtract one from the other.
The conversion factor from electron volts to volts is 1 eV 1.602 x 10-19 volts.
(1,000,000,000) One billion volts.
Electron volts (eV) and volts (V) are both units of energy measurement, but they are used to measure different types of energy. Volts measure the electrical potential difference between two points, while electron volts measure the energy of particles, such as electrons, in an electric field. In simpler terms, volts measure electrical potential, while electron volts measure the energy of particles in that potential.
Divide Watts by Volts ; this gives you Amps.
Internal resistance
The voltage is lost at the slice.
There's a couple ways. You could count the number of semiconductor junctions, resistances and other voltage attenuators, do all your Ohm's Law and drop calculations to come up with a reasonably good number...but if the circuit has been constructed, you could always just power it up, read the voltages at the supply and at whatever point you're trying to calculate the voltage loss for, and subtract one from the other.
A; The best way to describe is this way the load requires 10 volts but due to wiring and bad connections it gets to be 9v 1 volt is lost on IR drop so to compensate the input voltage needs to be boost up to 11 volts to insure 10 volts across the load
1/3 rd lost
volts abbr (V)
900,000 Volts. That is Nine Hundred Thousand volts.
150 volts is 125% of 120 volts.
The conversion factor from electron volts to volts is 1 eV 1.602 x 10-19 volts.
5000 volts
45 volts
12 volts