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Every component in a series circuit must have a potential difference across its terminals if current is to flow through that component. We call each of these individual potential differences a 'voltage drop', which is actually a bit of a misnomer as it's not really a 'drop' at all; it would be better termed a 'distribution'. The sum of these 'voltage drops' will then equal the supply voltage applied to the entire circuit.
It is another way of saying "Apply a voltage" or "supply (something) with a voltage"
12 volts...! The voltage drop across a 2 ohm resistor depends on the current flowing through it. As voltage (E) equals current (I) times resistance (R), if 1/2 amp is flowing through your 2 ohm resistor, 1/2 times 2 = 1 volt. If 1 amp is flowing through your 2 ohm resistor, 1 times 2 = 2 amps. Piece of cake. If the two ohm resistor is the only component in the circuit, it will drop whatever the applied voltage is. Put a 2 ohm resistor across a 6 volt battery, it drops 6 volts. If you put your 2 ohm resistor across a 9 volt battery, it drops 9 volts. Another way to say voltage drop may help. The voltage drop across a resistor is the voltage it "feels" when in a circuit. And that last couple of examples says that very well. In a circuit where a given resistor is the only component, it drops all the voltage in the circuit. It "feels" all the voltage in the circuit. In a circuit where there are 2 resistors of equal value in series, each one drops or "feels" half of the applied voltage. (The sum of the voltage drops equals the applied voltage.) As you work more with simple circuits using resistors in different arrangements with a given voltage source, try thinking of the voltage drop of a resistor as the voltage it "feels" when the circuit is energized.
It isn't. It's the other way around. The line voltage is 1.732 times the phase voltage. The figure results when you vectorially add the relevant phase voltages.
A: No matter how many resistor of different value are inserted the current will remain the same for each. The voltage drop will vary with the difference in resistors and i a parallel path is found along the way the current will divide according to the resistors values
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Are resistor drops the voltage or current? ANSWER: The voltage DROP is a way to imply that the voltage no matter of the value is what it must be whether it is measured or calculated
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
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IF USING A COPPER AT 105 DEG. CELSIUS - 70 FT. ONE-WAY LENGTH: TO GET AT VOLTAGE DROP = 3.72% AT 25 AMP LOAD ---- USE #1 AWG TO GET AT VOLTAGE DROP = 2.94% AT 25 AMP LOAD ---- USE #1/0 AWG TO GET AT VOLTAGE DROP = 1.85% AT 25 AMP LOAD ---- USE #3/0 AWG
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120 volt wall outlets. Their could me a small voltage between neutral and ground,up to around 1.5 volts. What you are measuring is the voltage drop on the system at that point in the system. You see the neutral and the ground are at the same level some where up stream (service panel). The neutral is under the same load as the phase conductor, and the neutral will drop voltage same as the phase. This is in fact the way I measure voltage drop, (neutral to ground.) However don't get fooled by high imped. meter, They have a way of ghosting a voltage. Or floating high, You get a reading but the voltage is really not their.
Every component in a series circuit must have a potential difference across its terminals if current is to flow through that component. We call each of these individual potential differences a 'voltage drop', which is actually a bit of a misnomer as it's not really a 'drop' at all; it would be better termed a 'distribution'. The sum of these 'voltage drops' will then equal the supply voltage applied to the entire circuit.