1
\$\begingroup\$

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

Results given: enter image description here

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

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

\$\endgroup\$
  • \$\begingroup\$ 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. \$\endgroup\$ – winny Sep 14 '16 at 13:51
  • \$\begingroup\$ you mean tangent fi?--> whit extracti fi out of it... \$\endgroup\$ – eugene_sunic Sep 14 '16 at 13:52
  • 1
    \$\begingroup\$ +1 for a clear question and a clear attempt of a solution. \$\endgroup\$ – Arsenal Sep 14 '16 at 13:55
  • 1
    \$\begingroup\$ 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. \$\endgroup\$ – winny Sep 14 '16 at 14:07
  • 2
    \$\begingroup\$ 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. \$\endgroup\$ – ilkkachu Sep 14 '16 at 14:39
3
\$\begingroup\$

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.

\$\endgroup\$
  • \$\begingroup\$ You Sir are a genius, I' ll folow your instructions and hopefully get the result in the list. Tnx \$\endgroup\$ – eugene_sunic Sep 14 '16 at 14:04
  • \$\begingroup\$ One more question, do I have to calculate 4 powers for each element r1,xc,r2,xl and then add them up? \$\endgroup\$ – eugene_sunic Sep 14 '16 at 14:25
  • 1
    \$\begingroup\$ Once you know the current (V/Z) the power is just I^2R. There is no power dissipated as a result of the current through an ideal capacitor or ideal inductor. So there are only two powers to add up- that in R1 and that in R2. \$\endgroup\$ – Spehro Pefhany Sep 14 '16 at 14:27
  • \$\begingroup\$ Answer is c) I,ve updated my question with the answer tnx to your explanation! \$\endgroup\$ – eugene_sunic Sep 14 '16 at 17:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.