I'am trying to calculate v2 from the circuit below, but my answer, 9.81cos(1000t+166.64°) mV, has a different angle from the correct answer, 9.81cos(1000t-13.36°) mV.

My approach

I tried to solve it with node analysis doing the following steps.

  1. Found the impedance for the two capacitors, -j1.67Ω and -j2, and for the inductor, j2Ω.
  2. I stated my current convention, currents going out of the node will be positive and currents going into the node will be negative.
  3. Then, using Kirchhoff's Current Law or LCK in spanish, I made an equation for each node.
  4. Solved the system of equations using a calculator.
  5. Changed v2 from polar coordinates to phasor.

Here's the process: enter image description here

  • \$\begingroup\$ Hint: you are off by 180 degrees, exactly. Look for a sign change. \$\endgroup\$ – placeholder Jun 12 '16 at 3:43
  • \$\begingroup\$ I already checked it several times, but couldn't find the mistake. \$\endgroup\$ – laco13 Jun 12 '16 at 4:47
  • \$\begingroup\$ Are you taking numerator and denominator signs into account when doing arctan? \$\endgroup\$ – Chu Jun 12 '16 at 6:59
  • \$\begingroup\$ I calculated the angle by first solving the system of equations and then changing from rectangular to polar, all with the calculator. \$\endgroup\$ – laco13 Jun 12 '16 at 21:47

Your solution is correct, the given "correct" answer is wrong. The voltage \$v_2\$ in complex form turns out to be

$$v_2=-0.0095443 + 0.0022665i$$

which is in the second quadrant (negative real part, positive imaginary part). If you just take the arctangent of the imaginary part divided by the real part, then you actually get the angle of \$-v_2\$. This is what the one who computed the "correct" solution must have done. What you (or your software) did, is correct. The angle of \$v_2\$ is given by


The addition of \$\pi\$ in this case is done automatically by the function atan2.


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