Understanding AC Power: Maximum and Minimum Currents on Resistor and Inductance

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SUMMARY

This discussion focuses on calculating maximum and minimum currents in AC circuits involving resistors and inductors. The key equations presented include the relationships for maximum power (prmax = Um * Im) and average power (Pav = Vrms * Irms * cos(φ)). The confusion arises from the notation used, particularly regarding the term V2, which is interpreted as the square root of 2. The discussion clarifies that instantaneous power can be expressed as a combination of average power and a varying component influenced by the circuit frequency.

PREREQUISITES
  • Understanding of AC circuit fundamentals
  • Familiarity with power calculations in electrical engineering
  • Knowledge of impedance and its role in AC circuits
  • Proficiency in trigonometric functions as they apply to electrical equations
NEXT STEPS
  • Study the derivation of power formulas in AC circuits
  • Learn about the significance of the power factor (cos(φ)) in AC systems
  • Explore the concept of impedance and its calculation in AC circuits
  • Investigate the effects of inductance on current and voltage in AC circuits
USEFUL FOR

Electrical engineering students, circuit designers, and professionals working with AC power systems will benefit from this discussion, particularly those focused on power calculations and circuit analysis.

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Homework Statement



I have some questions about powers in AC circuit. What is the maximum and minimum current power on resistor and inductance.

Homework Equations



Um = U / V2
Im = I / V2

The Attempt at a Solution


prmax = U*I=(Um/V2)*(Im/V2)=(Um*Im)/2. But answer seems to actually be that prmax = Um*Im, how can that be? And for plmax = 1/2 Um * I am plmin=- 1/2 Um *Im
 
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hi 0fibonacci1! :smile:

your notation is difficult to understand :redface:

here's an extract from the pf library on impedance
Power:

Power = work per time = voltage times charge per time = voltage times current:

P = VI =\ V_{max}I_{max}\cos(\omega t + \phi/2)\cos(\omega t - \phi/2)
=\ V_{max}I_{max}(\cos\phi + \cos2\omega t)/2​
(because 2cosAcosB = cos(A-B) + cos(A+B)))
=\ V_{rms}I_{rms}(\cos\phi + \cos2\omega t)
=\ \left(V_{rms}^2\left/|Z|\right.\right)(\cos\phi + \cos2\omega t)​

So (instantaneous) power is the constant part, P_{av} = V_{rms}I_{rms}\cos\phi (the average power), plus a component varying with double the circuit frequency, V_{rms}I_{rms}\cos2\omega t (so a graph of the whole power is a sine wave shifted by a ratio \cos\phi above the x-axis).
 
0fibonacci1 said:

The Attempt at a Solution


prmax = U*I=(Um/V2)*(Im/V2)=(Um*Im)/2.
It almost seems that you might be using V2 as denoting √[/size]2 ?
 

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