Power in AC Circuit: Can Power Be Made Arbitrarily Small?

AI Thread Summary
In an AC circuit, average power dissipation is calculated using P=VIcos(φ), where φ is the phase angle. In highly inductive or capacitive circuits, φ can approach ±π/2, leading to very low real power, but this does not mean power can be made arbitrarily small if resistance is present. Even with resistance, the power loss remains I²R, ensuring some heat generation. Adding capacitors does not reduce the actual value of real power, but it affects the phase relationship between voltage and current. Thus, while real power can be minimized, it cannot be eliminated entirely in the presence of resistance.
ResonantW
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In an AC circuit, the average power dissipated is given by P=VIcos(\phi). Does that mean that in a highly inductive, or highly capacitave, circuit where \phi approaches \pm \pi/2, the power can be made arbitrarily small? Even if a resistor were present? Does that mean it wouldn't heat up at all?
 
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As a fraction of apparent power, real power can be small, but adding a capacitor doesn't reduce the actual value of the real power.
 
ResonantW said:
In an AC circuit, the average power dissipated is given by P=VIcos(\phi). Does that mean that in a highly inductive, or highly capacitave, circuit where \phi approaches \pm \pi/2, the power can be made arbitrarily small? Even if a resistor were present? Does that mean it wouldn't heat up at all?
In a highly inductive element, there is only a very small component of current that is in phase with the voltage (leaving most to be in phase quadrature). But if resistance is added, then ɸ will no longer be close to Pi/2.

If a current I (RMS) passes through a resistance R, the power loss is I²R. ALWAYS.
 
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