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

Question

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: $$I_2=\frac{U}{Z_2}=\frac{50}{5}=10A$$ $$P_1=I^2*R_1=4*20=80W$$ $$P_2=I^2*R_2=100*3=300W$$

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

Answer 1

You have calculated the impedances correctly. Now since you know the voltage across R1 (given to be 40V) you can easily calculate the current in that branch from V=I/R. Since you know the impedance, you can then calculate U in volts RMS.

From U in volts RMS and the impedance of the second branch, you can calculate the current in that branch.

Since you now know both currents, the power is just I^2R in each branch, since the reactances are lossless. No need to consider the angles at all (directly). The total power is just the sum of the powers dissipated in each branch by the respective resistor.

The correct answer is in the listed numbers, so you should be fine.