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I have the following problem. Consider a circuit node where 3 sinusoidal currents with the same frequency converge, i1 i2 and i3. Knowing that the effective values of i1 and i2 are I1ef=1A and I2ef=2A. What can we say about I3ef:

Options: $$(a)1A \leq I_{3ef} \leq 3A$$ $$(b)0 \leq I_{3ef} \leq 3A$$ $$(c)2A \leq I_{3ef} \leq 3A$$

My attempt: So using KCL we have: $$i_1+i_2+i_3=0$$

Using phasors $$\overline{I_1}+\overline{I_2}+\overline{I_3}=0$$

where $$\overline{I_i}=I_ie^{j\phi_i}$$

Then $$I_1e^{j\phi_1}+I_2e^{j\phi_2}+I_3e^{j\phi_3}=0 $$

Because $$I_i=I_{efi}\sqrt{2}$$ then:

$$I_{ef1}\sqrt{2}e^{j\phi_1}+I_{ef2}\sqrt{2}e^{j\phi_2}+I_{ef3}\sqrt{2}e^{j\phi_3}=0 $$

Now I'm stuck in this. I don't know how should I proceed from this to obtain the interval of values for I3ef. I think the complex exponentials are what is bothering me. Can someone help me?

Thanks!

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  • \$\begingroup\$ Then math is not needed. If \$I_1\$ & \$I_2\$ are opposite, result is 1A. If they are in phase, result is 3A. Something else, between 1 and 3. \$\endgroup\$ Nov 28, 2017 at 19:37

1 Answer 1

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No full solutions for homeworks, only quidance. See the following image:

enter image description here

The phasor of -I3 is as long as the phasor of I3. Think different possible directions for the phasor of the 2A current. Decide the maximum and minimum lengths for phasor -I3 when the angle beta varies.

There's no need to think varying directions for the phasor of the 1A current. Angle alpha can be fixed because the image can allways be rotated.One angle in the system can be arbitary.

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  • \$\begingroup\$ That was the hint I needed! I didn't even think of using phasor diagrams. Thank you very much! \$\endgroup\$ Nov 28, 2017 at 21:44
  • \$\begingroup\$ Aren't we missing a $\sqrt{2]$ factor in the absolute value of the phasors? \$\endgroup\$ Nov 28, 2017 at 22:15
  • \$\begingroup\$ @GrangerObliviate The phasors work ok also with the effective values (= peak/sqrt(2)) of voltages and currents. The equation I wrote (and you too) stays equivalent if all terms on both sides are divided or multipliplied by sqrt(2) \$\endgroup\$
    – user136077
    Nov 28, 2017 at 22:19
  • \$\begingroup\$ Yes I just verified that! Now I got it! \$\endgroup\$ Nov 28, 2017 at 22:24
  • \$\begingroup\$ @GrangerObliviate, If the answer is right, why do not you accept it? \$\endgroup\$
    – Hazem
    Nov 29, 2017 at 8:48

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