4
You can reduce induced voltage in control cable by applying the following methods. 1. While laying control cable keep a minimum distance 300mm from the power cable. 2. If possible try to lay control cable on the separate cable tray. 3. Earth the Armour/ screen of the control cable on both side of the cable or at-least at one side. Some times it is not important to reduce the the induced voltage in control cable, but it's effect on relay or contactor as the contactor or relay does not pick-down even after the removal of the control supply. In that case it is recommended to use two relay/contactor in parallel and use the contacts in series or parallel for control circuit interlocking.
8000 watts divided by 240 volts equals 33.3 amps. If using NM cable indoors, Table 310.16 of the NEC says #8 wire is good to 40 amps. However, since this is a heating application which would qualify under "continuous duty" classifications(3hours or more of run time), then you figure it as 125%. This would in turn bump you up a wire size to #6. 33 amps times 125% equals 41.6 amps. Voltage drop would not be an issue even with #8 copper.
Not a good idea. Install a back board. A backboard of 4' x 4' is of ample size to install the panel and anything else that needs to be mounted. A thickness of 1/2" is OK but 3/4" is perfect. Some times telephone and cable companies need mounting points and this board is just perfect for this.
AS PER THUMB RULE THE CURRENT CARRYING CAPACITY OF WIRE IS 4 TIMES OF ITS CROSSECTIONAL AREA OF WIRE i.e. 10 sq.mm wire carrying maximum current 40 amp , 16 sq.mm carrying maximum current 64 amp , now amp convert in watt by multiplying by volt i.e. 240 16 sq.mm carrying max load 64x240= 15360 watt
At least two different methods. One is to run power from panel to one light, then to the next, then next, then next. Run separate cable from each light to the switch location where you want the control. Other way is to run power from panel (hot, neutral and ground) to first switch box, then an identical cable to the light #1, contine always hot wire from switch box 1 to switch box 2, then identical cable to light 2, repeat 2 moe times. If you are talking about 4 switches in one box, use pigtails from incoming (from panel) hot to each switch--then apply that concept to the options listed above for switches in different locations. Watch and anticipate/plan how many conductors will be in each box and check code for maximum wires in various size boxes.
Type ACT armored cable should generally be bent with a minimum radius of 8 times the overall diameter of the cable. This ensures that the cable maintains its integrity and performance without risking damage to the armor or internal conductors. Always refer to the manufacturer's specifications for precise bending requirements specific to the cable type and application.
The radius of the curve of the inner edge of the bends shall be at least 6 times the external diameter for armoured cable.
4 times the diameter of the cable
Four times the outer jacket diameter.
The bending radius for MICS (Microphone Integrated Circuit) cable typically refers to the minimum radius that the cable can be bent without causing damage or degradation to its performance. This value can vary depending on the specific cable design and manufacturer, but it is generally recommended to maintain a bending radius of at least 10 times the cable's diameter. Adhering to this guideline helps prevent issues like signal loss or cable breakage. Always consult the manufacturer's specifications for the most accurate information.
The minimum bending radius for a 1.5-inch diameter non-shielded cable typically ranges from 6 to 10 times the cable's diameter, depending on the specific type and application of the cable. For a 1.5-inch diameter cable, this translates to a minimum bending radius of approximately 9 to 15 inches. It's important to consult the manufacturer's specifications for the exact bending radius to ensure optimal performance and prevent damage.
The minimum bending radius for a nonshielded cable is typically 8 to 10 times the diameter of the cable. For a cable with a 1.5-inch diameter, this would mean a minimum bending radius of 12 to 15 inches. Therefore, the closest correct answer is B, 12.0 inches.
Typically, wires have a bend radius of 10 times the diameter of the wire. Measuring the inside. This is basically to prevent injury or changes in impedance at the bend. Also note that the tighter the bend can damage the insulation in coaxial or triaxial cables and cause lower dielectric strength between layers. Best to install cables where ripples don't form on the surface and moderate force is recommended. MIL-W-5088 has more information.
Pi times Radius times Radius times height
(radius+radius) times pi
4 times Pi times radius times radius
Radius times 2. then, that times Pi