No, the receptacle's rating is 240 volt and that is the maximum voltage allow to be applied to that device. To prevent this condition from happening 277 volt receptacles and switches have a larger box that they fit into. The retaining screws are set apart further that a 240 volt device which makes it impossible to install a 240 volt device in a 277 volt junction box.
A standard 115 rated bathroom circuit is enough power. You are supposed to have a GFCI on the line, either attached to your blow dryuer plug or as a receptacle.
Each voltage has a range and devices rated in that range will all work . The 110 to 120 volts is one range of voltages. The 220 to 240 is another range of voltages. The reason there is a range of voltages is to help the utility company out. They are mandated to keep the voltages within a + or - 5% range of a set voltage. Because there are loads coming on line and loads dropping off line the voltage fluctuates. Using a base voltage of 115 and 230 volts means that the voltage can rise by 115 +5% = 121 volts and drop by 115 - 5% = 109 volts. In the 230 volt range the voltage can rise by 230 + 5% = 241 and drop by 230 - 5% = 219 volts. As you see they are still in the usable voltage ranges.
The voltage rating of the cord is actually an insulation rating of the wire in the cord set. Your cord has a maximum voltage rating of 300 volts or less. The other standard maximum voltage cord ratings are 600 volts and 1000 volts. 480 volts would drop into the category of 600 volts or less.
To calculate watts, you need to multiply the voltage (in volts) by the current (in amps). For the 208 volts, 8 amps heating element: Watts = 208 volts * 8 amps = 1664 watts For the 110 volts, 8 amps heating element: Watts = 110 volts * 8 amps = 880 watts
dear all, 460/600 = 2/3 =0.66... you mean operating the motor at 2/3rd the volts.... 1. you may run the motor in star connection, speed will be little less then the synchronous speed, but as long as starting torque & running torque matches the load requirements, you may operate the same on lesser volts. In fact, starting line currents will be less. 2. Use VFD drive - to keep v/f ratio same, then, you get the full load rated torque even at the 460 volts.
Run a fused power line from the receptacle to either the battery for constant Hot or to the fuse box for an ignition controlled receptacle and ground the receptacle either by mounting on a metal surface or run a ground wire from the body of the receptacle to a good chasis ground
A standard 115 rated bathroom circuit is enough power. You are supposed to have a GFCI on the line, either attached to your blow dryuer plug or as a receptacle.
Plug the phone line into the phone line receptacle in the back.
An AC transmission line is typically a set of three wires that transports electricity over great distances at potentials exceeding 115,000 Volts. A common intrastate transmission line voltage is 230,000 Volts (230 kV), and interstate lines are normally rated 500,000 Volts (500 kV). When it is desired to transport electricity to a non synchronized area -- like Texas, DC transmission lines are used instead.
Most probably the receptacles downstream from the GFCI would not be protected by the GFCI receptacle.
The conscience is no. There is a work about by using an intermediary relay coil rated at 120 volts to switch the 240 volt feeder line.
It's the RMS value. A 120 volt lamp (light bulb) is rated according to its RMS voltage. Just like appliances in the home are rated at 120 volts (like your fridge, microwave and toaster), or 220 volts (like your clothes dryer). Note that these appliances will have to stand up to the peak voltage on the AC line. Naturally. And the peak voltage on an AC line is 1.414 times the RMS value of voltage. That means in a 120 volt AC line (120 volts RMS), the peak value of the voltage will be 1.414 times the 120 volts, or right at about 170 voltspeak for each cycle.
On a three phase system with a line to line voltage of 13800, a wye connection will give you a voltage of, 13800/1.73 = 7977 volts to ground.
When the alternator on a vehicle is in the charging mode its output is close to 14.5 volts. If a bulb is rated at 12 volts, which is the voltage of a vehicle battery at rest, when the alternator comes on line at a higher voltage the life span of the bulb will be shortened.
Power = voltage times current, and the power loss is the loss in the line, I^2 * R. At 11,000 volts, the current will be (11,000 / 415 = ) 3.77% of what it is at 415 volts. So the power loss in the line at 11,000 volts will be (3.77% ^2 = ) .14% of what it is at 415 volts.
48 VOLTS
I checked with my other meter and get 120 volts!