Kapanadze and other FE discussion

[Sergh]

Voltmeter on the generator terminals, shows U -10.4V; ammeter in the circuit of the generator winding and resistance shows: I -1.3A; R in the circuit -3 Ohm! Here you calculate, all other missing parameters of this combination: r -?; E -? Ui -?

Another simple problem: In order to get an active load current of one kilowatt, to the connected AC network 220V/50Hz what EMF should be in the phase of the generator, with a conventional resistance of the generator winding of 0 Ohm.

If the voltage of 10.4 volts is the open circuit voltage of the generator, then this must be reported so as not to be misleading.
If the voltmeter shows 10.4 volts and the 3 ohm resistor is disconnected, then this means that the current in the circuit is zero.
The output power in such a circuit is also zero.
To determine the power, it is necessary to measure the voltage and current at the same time, with the load connected.

[Sergh]

BUNK 

It seems to me that honest people behave somehow differently. 

[rakarskiy]

If the voltage of 10.4 volts is the open circuit voltage of the generator, then this must be reported so as not to be misleading.
If the voltmeter shows 10.4 volts and the 3 ohm resistor is disconnected, then this means that the current in the circuit is zero.
The output power in such a circuit is also zero.
To determine the power, it is necessary to measure the voltage and current at the same time, with the load connected.

Sadly, things are not going well at all for you.  The suitability of the profession, you could say zeroed out. The measurements were taken with the load connected. Simultaneous Ammeter and Voltmeter on the alternator creams. Doesn't the figure explain that. You have a problem with elementary knowledge, elementary power circuitry. You don't know how the current in a circuit is formed. For you as well as for most (radio masters) current is electrons. But Ohm's Law itself denies it by dividing the voltage and current into different components. Ask another question, if you turn off the load, open the circuit between the ammeter and the resistor, what voltage will show the voltmeter. I would simply condition the school principal to remove such a teacher from the school.

[Sergh]

The measurements were taken with the load connected. Simultaneous Ammeter and Voltmeter on the alternator creams. Doesn't the figure explain that. You have a problem with elementary knowledge, elementary power circuitry. You don't know how the current in a circuit is formed. For you as well as for most (radio masters) current is electrons. But Ohm's Law itself denies it by dividing the voltage and current into different components. Ask another question, if you turn off the load, open the circuit between the ammeter and the resistor, what voltage will show the voltmeter. I would simply condition the school principal to remove such a teacher from the school.

rakarskiy, please stop writing this. You disgrace yourself.
A voltmeter cannot show 10.4 volts across a 3 ohm resistor at a current of 1.3 amperes. Impossible. Since I have exhausted all efforts to help you, please consult your local physics teacher.

[rakarskiy]

 

The condition of effective power of the motor and the generator must be met: P=IU   

You take an ordinary generator with permanent magnets, with a 1 kW / 220V heater connected, you achieve at the generator terminals - 220V and current for 1 kW. Then open the circuit and watch the voltage on the terminals.

To get the appropriate current in the circuit, at the appropriate voltage in the circuit, the voltage drop across the load resistance must be taken into account. The EMF of the generator will always be greater by the value of the effective mains voltage. So when you measure 220V on the socket terminals and a current of 4.55A for a 1kW load, the EMF of the generator will be greater to the extent that the voltage drop on the load resistance and the circuit elements is greater.

The question is did you not know that?

This condition is mandatory, if you need to charge the battery your EMF voltage must be greater by the amount that will provide the appropriate charging current. Because if the operating voltage falls below the voltage of the battery, it will not be charging, but discharging, taking into account the resistance of the circuit. Then yourself, do radio engineers innovators, ohm's law for the circuit section, so that they do not work out.