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!

## Best Answer

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

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.