10. Given that the total instantaneous power for a Y load
= 3VmIm*cos(theta)=P then P=sqrt(3)VllIl *Cos(Theta)
find Q,S,S*
Show that P,Q,S* is same for y and delta loads.
P Q and S are Active power, Reactive Power, and Apparent Power respectively.
please tell me what these powers are telling us
(addendum- what do they say about the circuit, load, and
why are these powers there (necessary)). What
information do these convey
Avg Power, Ptotal, Q, S,S*.
Choose a value for Vm to your liking, not 120V.
Power consumed by a load is given by the product of voltage and current. P = VI. (Considering only magnitude of current in below calculations, so I = I*).
Assuming Vm and Im to be the rms voltage and rms current in a single phase of a 3 phase load, then instantaneous power in a phase is given by P = VmIm and total power consumed by the load is P = 3VmIm.
But inan ac system, there are 3 forms of power,
1. Active Power : It is the actual power
consumed by the load to perform useful work or the heat dissipated.
Active Power, P, is given by P =
3VmImcos
. (
Units is Watts,W).
2. Reactive Power : This is an average zero
power proportional to sin
. So the power
contributed in performing useful work is zero. Reactive Power, Q,
is given by Q = 3VmImsin
. It is
the power equivalent measure of the reactive component of load
current, which flows from source to load in the positive half cycle
and from load to source in the negative half cycle. (hence average
is zero). This generates additional current. But reactive power is
useful for generating and maintaining magnetic fields of various ac
machines and other equipments in the power system so it cannot be
nullified. In general, inductive loads consume reactive power and
capacitive loads generate reactive power. (Units is
Volt-Ampere-Reactive,VAr).
3. Apparent Power : It is the complex
combination of both active and reactive powers, S = P +
jQ. So the magnitude of Apparent Power,
. Since both active power and reactivepower
contributes to power generated by the source, apparent power can be
considered as the net power generated by source. (Units is
Volt-Ampere, VA). (S* is explained below).



.
P,Q and S for Y and delta :
Let line to line Voltage be Vll and line current be Il .
1.For Y load,
and
.




For Delta load,
and
.




So for both Y and delta loads P,Q and S are the same.
Other powers mentioned in question :
1. Average Power : It is the average work done or energy transferred or heat dissipated in unit time.
2. Total Power : It is the total work done or energy transferred or heat dissipated over a given period of time.
3. Reactive Power,Q : Explained above.
4. Apparent Power,S : Explained above.
5. Apparent Power,S* : S* is same as S, the only difference is that while using the complex power for calculations the sign of reactive power Q is reversed. (S = P + jQ. So, S* = P - jQ). This is done to satisfy the general load flow convention that power flowing from source to load is positive power.
10. Given that the total instantaneous power for a Y load = 3VmIm*cos(theta)=P then P=sqrt(3)VllIl *Cos(Theta)...
Need help solving the circuit with 120V 60Hz power network
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