A clock is not a generator so it does not have any power output. If anything the clock could consume 30 watts of power in its operation. The formula you are looking for is I = W/E, Amps = Watts/Volts.
With increasing torque load the armature tends to slow down; the motor draws more current to compensate, and if there is armature resistance the back emf generated by the armature falls to allow the increased current to flow, which causes the motor to settle at a lower speed. The mechanical output power is the speed times the torque, and increasing the torque increases the power output provided the speed does not drop much.
Output power can never be more than input power. With a transformer, it is possible to increase the output current (while decreasing the output voltage), or to decrease the output current (while increasing the output voltage).
probably clock and/or alarm system - draws a lot of power
With increasing torque load the armature tends to slow down; the motor draws more current to compensate, and if there is armature resistance the back emf generated by the armature falls to allow the increased current to flow, which causes the motor to settle at a lower speed. The mechanical output power is the speed times the torque, and increasing the torque increases the power output provided the speed does not drop much.
Current drawn = 5.5 / (1.732 X Voltage X power factor)AnswerWithout knowing its efficiency, you cannot determine the input power to this motor (5.5 kW is its output power), so you cannot determine its current.
When the clock light is on, it typically means that the clock is active and functioning. The light may indicate that the clock is receiving power and displaying the current time.
A power supply unit (PSU) converts mains AC to low-voltage regulated DC power for the internal components of a computer. PSU is a output device because it is outputting power to the rest of the device. Most personal computers can be plugged into standard electrical outlets.
Power limitation transformer circuit structure of power supply, including: an electromagnetic interference filter unit, a rectifying unit, a power factor correction section, a transformer having a primary input terminal and secondary input terminal, a pulse controlling unit connected with the secondary input terminal of the transformer, a primary power limitation circuit and at least one secondary power limitation circuit, an output current controlling unit and an output voltage controlling unit. The input terminal of the output current controlling unit is connected with the secondary output terminal of the transformer. The primary output terminal and secondary output terminal of the output current controlling unit are respectively serially connected with the primary and secondary power limitation circuits. The input terminal of the output voltage controlling unit is connected with the secondary output terminal of the transformer. The output terminal of the output voltage controlling unit is connected with the secondary power limitation circuit for controlling the magnitude of the voltage and current of the primary and secondary power limitation circuits so as to control the output power.
The formula you are looking for is W = I x E.
Your power adapter is rated up to 30 volts at 1000ma. (1 amp.) If you have a device which only draws 1 milliamp, then it should work with this adapter, as long as the voltage is correct.
Power = (voltage) x (current) = (120) x (0.042) = 5.04 watts
the power output increases