Want this question answered?
Nothing will happen to the possible output power of the power source i.e it will not increase. Each power source has its maximum possible output power. Adding more lamps in parallel will result in a drop in the circuit's total resistance which causes the total current drawn by the lamps to increase. Your wires might be burnt as a result so be careful.
Nothing will happen to the possible output power of the power source i.e it will not increase. Each power source has its maximum possible output power. Adding more lamps in parallel will result in a drop in the circuit's total resistance which causes the total current drawn by the lamps to increase. Your wires might be burnt as a result so be careful.
the actual power
usually power source where power comes from logic gates outputs are commonly called source or sink say the power source to the package is positive (VCC is +5) to call the output a source we would use an NPN transistor with the collector connected to +5V the base goes to the logic the emitter is the output terminal the output is the source of +5v sink output would have the emitter grounded base to logic the collector is the output so the output will sink positive voltage to ground A BUFFER usually has two output transistors one for source one for sink and can do either or both
The lamps will get dimmer. In a parallel circuit, voltage is constant. Whereas, in a series circuit, amps are constant.
You measure it with a clamp-on ammeter.
take the power it needs and the current it pushes, and divide.V=P/IAnswerLook at the nameplate of the voltage converter -it will specify the output voltage to the lamps. Alternatively, look at one of the lamps, it should have its rated voltage printed on it together with its rated power.
here, the power required by the receiver is the output power and that required from the source is input power. Gain in dB=10 log(output power/input power) we have, loss in dB = -gain in dB = 10 log(input power/output power) or, 50 = 10 log(input power/10nW) or, anti-log(5) = input power/10 nW so the power required from the source is antilog(5)*10nW = 1 mW
72kVA
Neither, it is a power source providing the energy to make the device work.
The Purpose of an output shaft is to Transmit Motion From A power source to other shafts With the Help of Helical Or spur Gear arrangement.
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.