Impedance across AC voltage source

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SUMMARY

The discussion centers on calculating the power dissipated in an impedance of 1000(1 + i) Ω connected to an AC voltage source with an amplitude of 10 V and a frequency of 60 Hz. The power is determined using the formula P = Re(V*I), where I is calculated as I = V/Z. The participant correctly computes the RMS voltage and applies the formulas to find the real part of the product V*I, ultimately concluding that the power dissipated is 1/40 W.

PREREQUISITES
  • Understanding of complex impedance in AC circuits
  • Familiarity with RMS voltage calculations
  • Knowledge of power calculations in AC circuits
  • Basic grasp of complex numbers and their operations
NEXT STEPS
  • Study the concept of complex impedance in AC circuit analysis
  • Learn about RMS voltage and its significance in AC power calculations
  • Explore the use of complex conjugates in power calculations
  • Investigate the implications of frequency on impedance and power dissipation
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in AC circuit analysis and power calculations will benefit from this discussion.

Simfish
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Homework Statement



An impedance 1000(1 + i) Ω (and note it contains an imaginary
part) is connected across an AC voltage source of amplitude 10 V
and frequency 60 Hz. Whatʼs the power dissipated during one cycle
within the impedance?

Homework Equations



P = Re(V*I), where V* is complex conjugate of voltage
I=V/Z

The Attempt at a Solution



So it's alternating current, so I first take the RMS V:10/\sqrt{2}. Then I = V/Z, so I get

V*I = V*V/Z = 100/2 \frac{1}{1000(1+i)} = \frac{1-i}{20*2}. Then I take the real part of this, which is 1/40. Am I doing this right?

Thanks!
 
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Correct!

ehild
 

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