Do alternators obey Ohm’s law? The alternator output voltage is proportional to the rotor excitation current. When reducing rotor excitation, the output voltage drops and the load resistance stays constant. Can why still use ohm law to determine the output current.

Example alternator 480 v / 4 ohm current output is 120 a if why now reduce rotor excitation output voltage drops tu 220 v / 4 ohm output current is 55 a can why calculete like this.

Another example is car alternator have 900 w of pawer tu give that pawer with 13.7 v output current is 65 a and load resistance is 0.21077 this all calculation are based on ohm low.

Can why calculate like this of course every equipment must support rated current

also i understand principle of impedance can why calculate a show i above examples

if why can apply ohm's law tu alternator haw why can determinant output current

i am really confused about this because i am always thought that resistance determine the current same people say they alternator does not follow ohm law ok but haw then calculate current output and load itself

 

The output current of an alternator will depend on the resistance of the windings, the excitation current, the rotation speed, the output voltage, the load resistance, the magnetic saturation of the cores, the battery voltage if the load is a battery....
There are a lot of variables to take in. All electronic devices obey ohms law. The trick is to know all the variables.
Just why do you need to calculate the output current of the alternator anyway?

 

I have a alternator without label plate i understand that if i wanna know exactly alternator output current. That I have to take into account the listed thing and i basically wanna know the if i have 13.7 v output and for example 90 w load that i can use ohm law to determine the load resistance and then current output.

 

yes

Edit: i.e......if the 90 watt load is 13.7VDC rated.

 

The rated output current is the current shown on the rating plate.

In this case it is 80 amps.

The amount of current at any particular time is that described by dendad in post #2.

 

RE:""Example alternator 480 v / 4 ohm current output is 120 a if why now reduce rotor excitation output voltage drops tu 220 v / 4 ohm output current is 55 a can why calculete like this.""

At the specified turns (rpm) the EDS of alternator is the same whatsever the resistivity You have. But, as soon You begin to USE the CURRENT, this current will produce the VOLTAGE LOSSES in that parasithics. Therefore EDS=480V, I=55A, R=4. It means that dU=I*R=55*4=220V of voltage losses. It means that U(in exit)=480-220=260 V only. Be in awake, alternator converts to the heat 220*55=12 kW !!! Run away while the rolling mad network null have cached You and have not divided You with itself. As You know the dividing with null is something always very painful like teeth-pains
More seriously, if Your load will be designed to have 55A at 480V, then picture will be skewed quite much. The R(loa)=U/I=480/55=8.7272727273 Ohms. And in serie with that Yo`ll have the 4 Ohms of parasitic. It means that Amperage in circuit will far not be 55A but I=EDS/R(tot)=480/(4+8.273)=37.7142857143 Amp, and correspondingly, the heat-losses on motor, sorry, alternator will be dU=R(mot)*I(tot)=37.71*4=
150.8571428571Volts or N(thermal)=I*2*R=37.7*37.7*4=5689.4693877551 Watts, therefore in the network will be exerted EDS-dU=480-150.65=329.1428571429 (useful) Volts or N(useful)=U(out)*I(circuit)=329 V * 37A=12413.387755102 Watts.