For the part of the circuit of sinusoidal circuit given below, the following data is given (effective values): $$E=10V,I_g=2\sqrt 3 A,I_2=2A.$$

Impedance of conductor is $$Z_C=5\Omega.$$

Current Ig is leading in phase with regards to E for 2pi/3, and E is running late in phase with regards to I2 for pi/2.

Evaluate effective value of voltage U10.

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I think that the following intuitive solution is completely incorrect:

By using potential of nodes method on branch 1-0, if we set 0 node as referent, we have: $$\underline{V_1}\frac{1}{\underline{Z_C}}=\frac{\underline{E}}{\underline{Z_C}}\Rightarrow \underline{V_1}=\underline{E}$$

Could someone explain why this solution is incorrect?

What is a correct approach?

  • \$\begingroup\$ "Evaluate effective value of voltage" - what voltage? \$\endgroup\$ – Andy aka Jun 29 '17 at 15:41
  • \$\begingroup\$ @Andy aka, Voltage U10. I said that in the new line from the last question. \$\endgroup\$ – user156262 Jun 29 '17 at 15:42
  • \$\begingroup\$ Eh??????????????? \$\endgroup\$ – Andy aka Jun 29 '17 at 15:56
  • \$\begingroup\$ @Andy aka, Just read the last sentence of the problem and include U10. \$\endgroup\$ – user156262 Jun 29 '17 at 15:59
  • \$\begingroup\$ Just fix the question dude!! \$\endgroup\$ – Andy aka Jun 29 '17 at 16:05

First apply Kirchhoff's current rule at junction 0. That gives you the current in the 1-0 branch, because you already know the other two currents. Now just trace a path from 0 to 1 algebraically adding voltages. Then you have the answer. That's basically what the first person said.

I think the mistake you're making is that you're assuming zero current in the 1-0 branch. You cannot arbitrarily do so. The branch is not open, it's just that any other connection, it might have, isn't shown.


I hope this helps: Capacitors transmit displacement current, they do not allow actual flow of electrons.

So the initial equality: $$\frac{V_1}{Z_c} = \frac{E}{Z_c}$$ is invalid.

The voltage between 1 and 0 will be E+ the voltage across the capacitor. If the capacitor voltage is opposite in polarity to E, then the magnitude of V_1 could be less than the magnitude of E.

The trick is to find the displacement current though the capacitor, the following should help.

$$I_g=I_2 +\frac{V_1-E}{Z_c}$$

Back To Phase Shifting

For the voltage source E, it is treated as a short circuit for current flow.

   Ig is leading E by 120 degrees.

   I2 is leading E by 90 degees.

Converting from polar to rectangular coordinates: $$i_g = 3+j1.732$$ $$i_2 = 0+j2$$

$$i_c = i_g-i_2$$ I got $$3-j0.268=3.0\angle-5.1^o$$

So v across c: $$V_c = I_c \times Z_c$$ I got $$15\angle-5.1^o$$

So V1: $$V_1= V_c+E$$ I got $$24.98\angle-3.1^o$$

  • \$\begingroup\$ @A Gern, Do you mean that the solution from my original post, $$\underline{V_1}=\underline{E}$$ is correct? \$\endgroup\$ – user156262 Jun 29 '17 at 16:23
  • \$\begingroup\$ @A Gern, Could you clarify? Please give detailed solution if possible. \$\endgroup\$ – user156262 Jun 29 '17 at 16:26
  • \$\begingroup\$ @A Gern, How to determine the voltage across the capacitor? \$\endgroup\$ – user156262 Jun 29 '17 at 16:59
  • \$\begingroup\$ Updated my first answer, using the algebra you can ignore the difference between displacement and 'real' current, that just describes the physics of the current flow. \$\endgroup\$ – A Gern Jun 29 '17 at 17:06
  • \$\begingroup\$ @A Gern, I am really stuck at this problem. Could you give a detailed solution? \$\endgroup\$ – user156262 Jun 29 '17 at 17:17

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