Basic Electrical Question

Ohms law is only true with DC circuitry. You must consider the reactance and therefore the phase angle of the current created by the reactive load.

You are confusing yourself. First of all, u have to understand what is constant and what is variable. Remember the generator V(voltage) is constant and will not change as the load increases or decreases.

The reduced voltage on increasing the load, apart from other reasons can also be due to IR drop on the wires where R is constant.

Gajanan Phadte

Voltage is the difference between two potentials and the resistance restricts the current that is trying to equalise the two.

Therefore if there is more current flowing the voltage is lower. If there is no resistance then the potentials will equalise on contact and there will be no voltage to measure.

Hasantha, the system is not just a passive resistance, there is also a voltage regulator controlling the voltage out of the generator. As this is dynamic you cannot use Ohm's law.

Your generator has some internal resistance and the voltage drop across that internal resistance (I X R) subtracts from the voltage the generator supplies, so you measure a lower voltage at the output of the generator when the current is higher.

When you have a higher load, the load resistance is smaller and hence it draws more current with a constant supplied voltage.

See the basics of the A.C electric circuit for more details .

V=I*R this Ohm's low is used for D.C circuit only for A.C Ohm's low is V=I*impedance. in the generators if a load draws a current the voltage over the load should be stabile that is an A.V.R's working .

Ohm's law applies, but if AC generator it must be written as follows: V = IZ, where Z is the total impedance (includes resistive load, inductive load, and capacitative load). Also your voltage droop is due to the internal source (Thevinin source) impedance to the load. Generally, all systems will experience droop of voltage output with high loads near the actual machine limits, however most modern machines control the extent of the voltage droop with the use of automatic voltage control circuitry that affects the excitation of the generator. Inductive impedance is related to inductance by the following: Z_L = iωL. Capacitative impedance is related to capactiance by the following: Z_C = 1/iωC. Resistive impedance is simply the resistance R. The total impedance is simply the sum of the parts.

The total impedance is.

Ztotal=Zc+R. for capacitive load .----> Ζtotal=1/2∏fC +R.

Ztotal=Zl+R . for inductance load.---->Ztotal=2∏fL+R.

because Z is a vector quantum not scaler

Yes, it is a vector impedance. You do not include anything that states or even implies a vector. You should add the imaginary unit vector of the square root of -1 (i) to these equations. As James Stewart already wrote.

Thanks for the smiley, Fred.

please write the complete equation of the impedance for each of the capacitive and inductance load that has the imaginary part and why you didn't .?

I guess you do not recognize that James Stewart already wrote what you asked for by using the radians per second symbol (ω) for frequency instead of the cycles per second symbol (f). Now you should remember that these equations are only precise for ideal components. In reality the there are parasitic resistors and reactance that effect the impedance depending on the component construction and frequency of interest.

Oops! I forgot about all the parasites. Are there any Jebusites, too?

Yes i do ( ω) is some thing new. on me

D.C. Generator also exists and no reactance is involved here[sorry!]. There are also generators without voltage regulation [ may be without rpm regulation, also].The phenomenon is simple :more load=more current that means less voltage at generator terminal due to generator internal voltage drop-as Rixter exposed. If the drive unit could overcome the load the rpm will be constant and if the excitation current will be constant-no voltage regulation-then the generator EMF will be constant so more current will be more voltage drop then less voltage at the terminals.

In order to compensate the drop one has to raise the excitation current [manual or automatic] and so to amplify the EMF.

If the Generator [synchronous a.c. generator- or induction generator] is connected with the Grid then it is provided with all necessary, of course.

For record only, the impedance [written in complex-real and imaginary] it is not a "vector" but a "phasor" since vector it is a physical notion [force, field and so on] and phasor it is only a symbolic representation. See: http://en.wikipedia.org/wiki/Phasor

You are correct, James. A phasor is not a spatial vector but a vector none the less.

I find it funny that the cited Wikipedia link says this in the very first sentence:

In physics and engineering, a phase vector, or phasor, is a representation of a sinusoidal function whose amplitude (A), frequency (ω), and phase (θ) are time-invariant.

A phasor is one type of vector.