I Have the following circuit scheme (I have to find the U): $$R_1=10, R_2=20 (\Omega); A_1->i=0.6, A_2->i=0.8(A)$$

enter image description here

I have a problem with calculating the impedance Z since I need it to calculate he U, but I don't know how to calculate it since Xc (capacitor resistance)**** is not given.

What I've done so far: $$i_1=i_2+i_3=1.4 A;\quad U_{r1}=i_1*R_1=14V;\quad U_{r2}=i_1*R_2=12V; $$ and after that I get stuck because I haven't got the Xl (inductor resistance) to solve for Ur3? How should I proceed?

I hope this question is appropriate for this community (my first question).

EDIT: Results that are given: enter image description here

  • \$\begingroup\$ Presumably the point in your course teaches you that you can assume the amp-meters to represent zero resistance? Hopefully that's a decent hint. \$\endgroup\$
    – Asmyldof
    Commented Sep 11, 2016 at 10:11
  • \$\begingroup\$ But I wouldn't even take them into account in my problem, I just use the facet they give me the value of the current through that branch. I think I didn't quite understand... \$\endgroup\$ Commented Sep 11, 2016 at 10:25
  • \$\begingroup\$ check my answers \$\endgroup\$ Commented Sep 13, 2016 at 14:00
  • \$\begingroup\$ check my improved answer \$\endgroup\$
    – Decapod
    Commented Sep 13, 2016 at 20:06

1 Answer 1


First remember current of capacitor and resistor not same, like this picture : enter image description here

  • \$\begingroup\$ wow, that's amazing, well deserved 50 points! I m, sorry but I haven to wait 14 hours to award you with it. \$\endgroup\$ Commented Sep 13, 2016 at 19:21
  • \$\begingroup\$ Could you just explain why did you use the cos theorem, I've never seen that before begin used in such examples? \$\endgroup\$ Commented Sep 13, 2016 at 19:26
  • \$\begingroup\$ The calculations are almost correct. However the currents I1 and I2 are swapped This brings the voltage over R2 at 12V (20*0,6) Entering this correction brings Vu to 19.7 V \$\endgroup\$
    – Decapod
    Commented Sep 13, 2016 at 20:22
  • \$\begingroup\$ O I see. tnx for the correction. Without you guy I would have never solved this. \$\endgroup\$ Commented Sep 13, 2016 at 20:43
  • 2
    \$\begingroup\$ Uh what a complicated way. The point is we already have I1 as resistive and capacitive componets I1=I2+I3, then you find modulus, angle and then back to components. There's no need for this round back trip. Circuit can be solved in one single line. \$U_1=U_\text{R2}+R_1(I_2+I_3)=R_2 I_2+ +R_1(I_2+I_3)= 20\,\Omega \times 0.6\,\text{A}+10\,\Omega\times (0.6+\text{j} 0.8\,\text{A})=18+\text{j}8\,\text{V}\$ which in modulus yields \$|U_1|=\sqrt{(18\,\text{V})^2+(8\,\text{V})^2}\approx 19.7\,\text{V}\$ \$\endgroup\$
    – carloc
    Commented Sep 15, 2016 at 6:32

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