P=v*i i=p/v=0.5/3=1/6 (a)
A chemical reaction generates free electrons which build up a potential called a voltage and then the electrons can be drawn from the battery as a current to supply a load like a flashlight light bulb or your cell phone. Some batteries can be recharged to provide additional free electrons in conjunction with the chemical reaction.
The 7805 is a standard 5 volt regulator with a voltage input, a regulated 5V output and an earth reference. With no load on the 5V output, the current drawn by the regulator is nor more than a few milliamps. The actual figure varies from one type to another. Once a load is added to the 5V output, the current drawn from the battery is the same as the current drawn by the load itself (plus the few milliamps drawn by the regulator itself). As an example, if the load current is measured at 1 amp, then the current drawn from the battery will also be 1 amp. If current draw is a problem, then consider a switching regulator instead of a linear type. Switching regulators will be more efficient and for a 1 amp load, it is possible to see only a 0.6 - 0.7A current from the battery itself. as well as conserving battery life, the regulator will also generate less heat than the 7805.
Load!
The maximum current that can be drawn from a voltage source is dependent on the impedance of that source, the impedance of the connections to the source, and the energy available from that source.
p=r*i
A chemical reaction generates free electrons which build up a potential called a voltage and then the electrons can be drawn from the battery as a current to supply a load like a flashlight light bulb or your cell phone. Some batteries can be recharged to provide additional free electrons in conjunction with the chemical reaction.
Inverter efficiencies vary about 50% when a small about of power is being used, to over 90% when the output is approaching the inverters rated output. Your question cannot be answered without knowing the inverters rated output.
The 7805 is a standard 5 volt regulator with a voltage input, a regulated 5V output and an earth reference. With no load on the 5V output, the current drawn by the regulator is nor more than a few milliamps. The actual figure varies from one type to another. Once a load is added to the 5V output, the current drawn from the battery is the same as the current drawn by the load itself (plus the few milliamps drawn by the regulator itself). As an example, if the load current is measured at 1 amp, then the current drawn from the battery will also be 1 amp. If current draw is a problem, then consider a switching regulator instead of a linear type. Switching regulators will be more efficient and for a 1 amp load, it is possible to see only a 0.6 - 0.7A current from the battery itself. as well as conserving battery life, the regulator will also generate less heat than the 7805.
Cells and batteries.A more detailed answerIn general, most power supply sources - including cells and batteries - cannot supply either a truly constant current or voltage.Because most load devices are designed to require a nearly constant voltage - and not a specific constant current. - most power supplies - including cells and batteries - are designed to supply a given output voltage whatever the current that is being drawn, provided of course that the current stays within a pre-determined designed range of output current values.One definition of "constant current source" is that it has zero output resistance, which in practice cannot exist.However, a transistor can act as a constant current source if biased and used in that manner. This can also be done using one of the several standard "voltage regulator" components - which are available from electronics parts suppliers - because they can be hooked-up to act as constant current sources to operate over a small range.
With the batteries in series, the alternator provides the same current, and therefore the same charge to all the batteries. The alternator can take no account of any current being drawn from the 12V take off point. Part of the charge current is being fed to the 12V load and not to the batteries. Therefore, the 12-24V batteries are being fully charged while the 0-12V batteries are being partially charged. Over a period of time, the difference between the charge levels will become more obvious.
mechanical characteristics can be drawn between speed and torque electrical characteristics are 1. Output power vs speed 2. Output power vs input power 3. Output power vs efficiency 4. Output power vs torque 5. Output power vs line current
ANSWER In rectifiers for power supplies, the capacitor size is determined by the allowable ripple on the output. This can be determined by the rate at which the capacitor is drained. Specifically, this rate is the current drawn from the capacitor. Assume a half wave rectifier made from four diodes. For part of the cycle, the output current is supplied by the rectifier diode. This is also when the capacitor is charged. While the rectifier is not supplying current -- when the input waveform has dropped below the output voltage -- the capacitor must supply the current. Then, as the input waveform rises above the capacitor voltage, the rectifier supplies the current to charge the capacitor and the output circuit.
Load!
load
The amperage drawn from batteries is governed by the connected load. The voltage of the batteries can be one of two voltages. in parallel the 8 batteries will give you a voltage of 6 volts. In series the 8 batteries will give you a voltage of 48 volts. The amp/hour capacity of the batteries will give you the amount of current the device can draw over a specific length of time. Equation to fine amperage is I = W/E, Amps = Watts/Volts. Watts = Amps x Volts.
Yes, the output line used to find the losses, usually for smaller machines we take copper losses equal to the iron losses.
The output voltage available at a USB port is controlled by the regulator in the computers power supply. It is fixed, you can not change or control it. The maximum current which should be drawn from a USB2 port is 500mA, from a USB3 port it can (I believe) be up to 850mA.