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
The voltage drop should not exceed 3% on a feeder or branch circuit.
You usually use V for volts. There can also be a subscript. Say for the voltage drop of a resistor could be called VR.
Not enough information. To answer this question the amount of current in AMPS the circuit is drawing is needed.
Voltage across all parallel capacitor's is same i.e. it is equal to supply voltage, it can be measured using digital volt meter (any high input impedance volt meter). When capacitors are in series; voltage drop depends on charge stored in the capacitor. it can be given by the formula V x V = 2 / (joules x capacitance). This voltage can also be measured using digital volt meter.
A wire size of 250 MCM will limit the voltage drop to 3% over a distance of 200 feet.
The voltage drop should not exceed 3% on a feeder or branch circuit.
If there is nothing else in the circuit, then the voltage drop across the resistor will be the full supply voltage of 5 volts. The size of the resistor does not matter in this case - it will always be 5 volts.
Must also know current load to determine.
A 500 MCM copper conductor will limit the voltage drop to 3% or less when supplying 350 amps for 150 feet on a 208 volt system.
You usually use V for volts. There can also be a subscript. Say for the voltage drop of a resistor could be called VR.
It depends on the use it is being put to. It is sufficient for a 24 volt circuit. Too much for a 12 volt circuit and too little for a 240 volt circuit.
Yes, the voltage listed on the bulb is the nominal voltage and it will work perfectly on a 120 volt circuit.
A volt meter is used to detect the presence of voltage, and it also measure the amount of voltage (electrical pressure) in a circuit.
Not enough information. To answer this question the amount of current in AMPS the circuit is drawing is needed.
Volt difference causes a short circuit! ChaCha
For voltage drop checking on the wire from the battery positive post to the starter main terminal, you place the voltmeter reading on the 1 to 3 volt range, (or what lower range you have). You then place one voltmeter lead on the positive battery post, the other lead you place on the starter positive terminal and have someone crank the engine. While the engine is cranking, you observe the voltage drop. It should be less than a volt. This seems strange because you would think the cable would not drop voltage but it sure can. If the battery post is dirty or bad connection, you could see a voltage drop all the way to the maximum 12 volts battery voltage depending upon the connection. Also you can check the return path the same way--- Negative case of the starter to the negative post on the battery for voltage drop. If it is more than a volt or two, you have a bad ground return circuit.
In this situation, to calibrate a transmitter you need a power circuit and communicator circuit. The Hart communicator used in the calibration process is connected to the power source circuit in parallel. The power source circuit is the one that has ammeter, 250 Ohm resistor, and power source all connected in series. As the transmitter sends output mA, it creates volt drop across the 250 Ohm resister. Let's say the volt drop across the resistor was 1 Volt. Now, back to the Hart communicator. It is a load, meaning there will be a volt drop across the Hart communicator. Since it is in parallel with the power circuit, it is also parallel with the resistor. So, the 1 volt drop across the 250 Ohm resistor will also make 1 volt drop across the Hart communicator. Technically speaking, the 1 volt drop across the Hart communicator is only true if its resistor is also 250 Ohm. However, it does NOT matter what voltage drop is in the Hart communcator. It only sees the "relative" voltage drop changes to measure the changes in transmitter outputs.