Complex Power

Thread Starter

xz4chx

Joined Sep 17, 2012
71
So I have the circuit below and this is the information given:

Load 1 absorbs 4203.22-2835.11j [VA]
Voltage sources delivers 7500 [W]
v_s(t) = 400cos(100[rad/s]t+38°) [V]
Find all possible values for L_A

So my idea was to put it into phasor domain.

So V_s=400<38°
Since the voltage source only delivers power the angle of voltage and current have to be the same since the power factor is 1 so I came up with

S_L=.5(V_s)(I_L*) (Complex Power Formula)

So I_L=(7500/400*2)<38° = 37.5<38°

Then I could find the voltage of the Load and that comes out to be
Using S_L=.5(V_L)(I_L*)

V_L= 2*(4203.2-2835.1j)/(37.5<38°)

V_L= 269.744+18.86j

After that i thought I could just do a KVL to get the value for L_A (the inductor)

So KVL
-V_s+(L_A+R_1)(I_L)+V_L=0

So I get
-400<38°+37.5<38°(100LjΩ+3.3Ω)+ 269.744+18.86j = 0

But this does not give me a real value for L_A, I am getting a complex value
For reference the answer is 77.1 [mH]
 

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panic mode

Joined Oct 10, 2011
2,759
you did not post original question (please post it exactly "as is") and reading though your comment gives me a headache ;-)

7500W is P not S. S is measured in VA not W.

you have there series circuit so current is the same in L, R and load. note that:
Psource=Presistor+Pload
7500 = Presistor + 4203.22
Presistor = 3296.78W

I=sqrt(Presistor/R)=31.607A
but this is only magnitude of RMS current, we don't know the angle of it yet.

similarly, Rload=Pload/I^2=4.2074 Ohm

Re(Z)=R+Rload = 3.3+4.2074=7.5074 Ohm
 
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