Current phase angle difference (AC circuit)

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SUMMARY

The discussion focuses on calculating the phase angle difference between two sinusoidal current generators, I_{g1} and I_{g2}, connected to three receivers with complex impedances Z_1, Z_2, and Z_3. The impedances are Z_1=(125+j375)Ω, Z_2=(700+j100)Ω, and Z_3=(500-j500)Ω. When the switch is closed, the active power of all three receivers is halved, leading to the conclusion that the total active power is reduced to approximately 0.4235W. The final calculations yield I_{g1} as approximately 0.032A, with the phase angle θ determined to be -π/4 or 7π/4.

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  • #31
gneill said:
Note that there should be two candidate angles that will yield the required cosine. One in quadrant 2 and one in quadrant 3.

You are right: \theta=\frac{3\pi}{4} or \theta=\frac{5\pi}{4}.
 
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  • #32
Yup, or ##\pm \frac{3}{4} \pi## . Phase angles are conventionally given in the range -180° to +180°.

You should confirm that the resulting total current yields half the power as the original current, say by determining the active power dissipated in Z1.
 
  • #33
Thanks for the help.
 
  • #34
You're welcome!
 

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