For a 180-meter distance with a 160 ampere current in a three-phase supply, you would typically use a larger cable size to minimize voltage drop and ensure safety. A common choice would be a 35 mm² or 50 mm² copper cable, depending on the acceptable voltage drop and specific installation conditions. It's essential to consult local electrical codes and standards to confirm the appropriate size based on factors like insulation type and installation method. Always consider factors such as ambient temperature and grouping of cables as well.
Cable tray is a system of wiring in a building in which insulated electrical wires are used to distribute power and communication to various access points. Cable trunks are power supply lines that are used to distribute power and communication services to large areas such as industrialized cities.
Cable sheath current is a current induced in the sheath by radio, magnetic or other interference, or a fault condition. The cable sheath should be grounded at one end so that this current flows to ground safely.
For appliances that need a lot of power, the advantage of a higher voltage is that less current is needed, because power = voltage x current. That means that the cable doesn't need to be so thick. It also means that for a given percentage of voltage drop in the cable, the allowable volt-drop in the cable is higher when the supply voltage is higher. This also allows the use of a thinner cable especially when the cable is 100 ft long or more.
445 amps. For example Mulberry AAAC cable.
The cable size is : 4*25 mm2
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.
If it's a 380 v 3-phase supply, the current is up to 300 amps which could be supplied over a short distance by XLPE cable with 70 sq. mm cross-section conductors.
Assuming you're taking about 'twin core and earth' cable - it's commonly used to supply power to high-current devices such as cookers and immersion heaters.
That depends on waht the instrument is and is being used for and what is being carried in the electrical cable. If the electrical cable is carrying DC, there is not much to worry about unless the current is very high. AC/modulated current in the electrical current will induce currents in the instrument cable but how great the problem will be depends on current and frequency. Normally, a few inches separation would be enough. If there is likely to be a significant probelm it would be normal to used screened cable for one or both.
· A power cord, line cord or mains cable is known as the supply cable, which temporarily connects an appliance to the mains electricity supply to a wall socket. · A supply cable connects the refrigerator to the mains electricity supply via a wall socket.
yes
what should be the distance between instrument cable and electrical cable
· A power cord, line cord or mains cable is known as the supply cable, which temporarily connects an appliance to the mains electricity supply to a wall socket. · A supply cable connects the refrigerator to the mains electricity supply via a wall socket.
what is current capacity of 95 sqmm cable?
Current in the single core cable would induce a magnetic current in the steel cable, though a transformer effect. This would heat the steel armored strands, and the circuit would increase more electrical power from the load supply point. The earthing of the cable glands would complete the circuit and the return current would flow in the earth bonding cable between the two points. This is called Eddy currents generated in the cable by the twist of the steel armored around the central core, current flowing in one direction
Yes, you need a cable to connect the power supply to the amplifier, but the specific type of cable required depends on the model and power requirements of the amplifier. In most cases, a standard power cord with the appropriate voltage and current rating will suffice. It's always recommended to refer to the amplifier's user manual for the correct power supply requirements.
The formula for a simple DC voltage drop across a cable is:VDrop = Vmeasured at the input of cable - Vmeasured at the output of cableThis formula may seem simplistic however keep in mind that the sensitivity, accuracy and resolution of measuring instrument instrument is what is really important.If you know the DC current flowing through the cable and the impedance of the cable then you can use Ohm's law. Vdrop = I R where Vdrop is the voltage drop across the cable in volts I is the current flowing through the cable in amperes R is the resistance of the cable in ohms.ADDITIONALCalculation of the voltage drop is given by the formula belowVoltage Drop (Volt) = [(mV/Am)/1000] * I * lWhere mV/Am = millivolt drop per meter per ampere of the cable(this information is given in the table based on IEE Wiring Regulation)I = Current in the cable (in Ampere)l = Distance of cable (in Metre)See related links below