# Can two different power factors cos(fi) be added together?

Image of the problem (I need to calculate the overall power for these 4 elements):

Results given:

My main problem is that I'm unsure whether I can add cosfi1 + cosfi2=cosfi. I calculated them both, one for RC and the second for RL.

Here is my overall (exhaustive) attempt on this problem (please correct if you think something is wrong). Data given: $$R_1=20\Omega;R_2=3\Omega;L=12.73*10^{-3}H;C=212.2*10^{-6} F; f=50Hz, U_{R1}=40V$$

Solution: $$X_L=2\pi fL=4\Omega$$ $$X_C=(2\pi fC)^{-1}=15\Omega$$ $$Z_1=\sqrt {R_1^2+X_C^2}=25\Omega$$ $$Z_2=\sqrt {R_1^2+X_L^2}=5\Omega$$ $$I_1=\frac{U_{R1}}{R_1}=2A$$ $$cos_1\phi=\frac{R_1}{Z_1}=0.8$$ $$cos_2\phi=\frac{R_2}{Z_2}=0.6$$ $$U_{XC}=I1*X_C=30V$$ $$U_1=\sqrt {U_{R1}^2+U_{XC}^2}=50V$$ $$U_1=U_2=50V \quad (see \quad image)$$

$$I_2=\frac{U_2}{R_2+X_L}=7.14$$

And now calculating total I with total U being 50V

$$I=\sqrt {I_1^2+I_2^2}=7.41A$$ $$P=U*I*(cos_1\phi+cos_2\phi)= 50*7.14*(0.8+0.6)=499.8W$$

I get 500 W, which is not listed in the results. Can anyone pinpoint my error?

EDIT: Tnx to Spehro Pefhany (answer below) I've found my answer: $$I_2=\frac{U}{Z_2}=\frac{50}{5}=10A$$ $$P_1=I^2*R_1=4*20=80W$$ $$P_2=I^2*R_2=100*3=300W$$

## $$P=P_1+P_2=380W \quad answer: \quad c)$$

• It's not that simple. If one has say 1 kVA and 0.8 leading cos phi and the other 1 kVA and 0.8 tailing cos phi, yes, they will cancel out. In any case, draw visar diagrams and/or calculate all currents and their phases and add them up. – winny Sep 14 '16 at 13:51
• you mean tangent fi?--> whit extracti fi out of it... – eugene_sunic Sep 14 '16 at 13:52
• +1 for a clear question and a clear attempt of a solution. – Arsenal Sep 14 '16 at 13:55
• If you just want to find the power, you only need to calculate the Z for each element and you will have the current with ohms law. With the current though each resistor and its resistance, you have the power. Spice will also solve it for you. – winny Sep 14 '16 at 14:07
• As the power factor is basically the ratio of the apparent power that does real work, it can't be higher than 1. But if you sum up the power factors like that, you get 0.8 + 0.6 = 1.4, which immediately tell you that something is wrong with that approach. – ilkkachu Sep 14 '16 at 14:39