In AC circuits, the average power dissipated is calculated using the formula P=VIcos(φ). In highly inductive or capacitive circuits where φ approaches ±π/2, the real power can be made arbitrarily small, but this does not eliminate power dissipation entirely, especially when resistors are present. The presence of resistance alters the phase angle φ, ensuring that power loss is always calculated as I²R, where I is the RMS current and R is the resistance. Thus, while real power can be minimized, it cannot be completely negated.
PREREQUISITESElectrical engineers, students studying circuit theory, and professionals involved in power system design and analysis will benefit from this discussion.
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.ResonantW said:In an AC circuit, the average power dissipated is given by [itex]P=VIcos(\phi)[/itex]. Does that mean that in a highly inductive, or highly capacitave, circuit where [itex]\phi[/itex] approaches [itex]\pm \pi/2[/itex], the power can be made arbitrarily small? Even if a resistor were present? Does that mean it wouldn't heat up at all?