Reactive Element Power Absorption Explained

In electrical circuits, elements can be broadly classified into three types: resistive, inductive, and capacitive. A purely reactive element refers to an ideal inductor or an ideal capacitor. Unlike resistive elements, which dissipate energy as heat, reactive elements store and release energy.

Understanding Power in AC Circuits

In an alternating current (AC) circuit, power is not always consumed continuously. We consider different types of power:

• Instantaneous Power \((P(t))\): This is the product of the instantaneous voltage \(V(t)\) and instantaneous current \(I(t)\) at any given moment in time, i.e., \(P(t) = V(t)I(t)\).
• Average Power \((P_{avg})\): Also known as real power or active power, it is the average of the instantaneous power over one complete cycle. This is the power that is actually converted into another form of energy (like heat, light, or mechanical work).
• Reactive Power \((Q)\): This is the power that oscillates between the source and the reactive elements (inductors and capacitors). It is not dissipated but is exchanged between the circuit and the magnetic/electric fields of the reactive components.

Instantaneous Power in Purely Reactive Elements

Let's consider a purely reactive element (either an ideal inductor or an ideal capacitor) connected to an AC voltage source. Assume the voltage across the element is given by:

\[V(t) = V_m \sin(\omega t)\]

Where \(V_m\) is the peak voltage and \(\omega\) is the angular frequency.

For a purely reactive element, the current and voltage are 90 degrees (or \(\frac{\pi}{2}\) radians) out of phase.

• For an ideal inductor: The current \(I_L(t)\) lags the voltage \(V(t)\) by 90 degrees.

  \[I_L(t) = I_m \sin(\omega t - \frac{\pi}{2}) = -I_m \cos(\omega t)\]

  The instantaneous power absorbed by the inductor is:

  \[P_L(t) = V(t)I_L(t) = (V_m \sin(\omega t))(-I_m \cos(\omega t))\] \[P_L(t) = -V_m I_m \sin(\omega t)\cos(\omega t)\]

  Using the trigonometric identity \(\sin(2\theta) = 2\sin(\theta)\cos(\theta)\), we get:

  \[P_L(t) = -\frac{V_m I_m}{2} \sin(2\omega t)\]

• For an ideal capacitor: The current \(I_C(t)\) leads the voltage \(V(t)\) by 90 degrees.

  \[I_C(t) = I_m \sin(\omega t + \frac{\pi}{2}) = I_m \cos(\omega t)\]

  The instantaneous power absorbed by the capacitor is:

  \[P_C(t) = V(t)I_C(t) = (V_m \sin(\omega t))(I_m \cos(\omega t))\] \[P_C(t) = V_m I_m \sin(\omega t)\cos(\omega t)\]

  Using the trigonometric identity \(\sin(2\theta) = 2\sin(\theta)\cos(\theta)\), we get:

  \[P_C(t) = \frac{V_m I_m}{2} \sin(2\omega t)\]

In both cases, the instantaneous power is a sinusoidal waveform that oscillates at twice the supply frequency (\(2\omega\)).

Average Power Calculation for Reactive Elements

The average power absorbed over one complete cycle (T) is calculated as:

\[P_{avg} = \frac{1}{T} \int_{0}^{T} P(t) dt\]

For the instantaneous power in a purely reactive element (either inductor or capacitor), we have a term like \(\sin(2\omega t)\). The integral of a complete cycle of a sine (or cosine) function is always zero.

• For the inductor:

  \[P_{avg,L} = \frac{1}{T} \int_{0}^{T} \left(-\frac{V_m I_m}{2} \sin(2\omega t)\right) dt\] \[P_{avg,L} = -\frac{V_m I_m}{2T} \int_{0}^{T} \sin(2\omega t) dt\]

  Since \(\int_{0}^{T} \sin(2\omega t) dt = 0\) over a full cycle \(T = \frac{2\pi}{\omega}\),

  \[P_{avg,L} = -\frac{V_m I_m}{2T} \times 0 = 0\]

• For the capacitor:

  \[P_{avg,C} = \frac{1}{T} \int_{0}^{T} \left(\frac{V_m I_m}{2} \sin(2\omega t)\right) dt\] \[P_{avg,C} = \frac{V_m I_m}{2T} \int_{0}^{T} \sin(2\omega t) dt\]

  Since \(\int_{0}^{T} \sin(2\omega t) dt = 0\) over a full cycle,

  \[P_{avg,C} = \frac{V_m I_m}{2T} \times 0 = 0\]

Key Takeaway: Zero Average Power

The calculations show that for both ideal inductors and ideal capacitors, the average power absorbed over a complete cycle is zero. This is because reactive elements continuously exchange energy with the source. During one half-cycle, they store energy (e.g., an inductor stores energy in its magnetic field, a capacitor stores energy in its electric field), and during the next half-cycle, they return this stored energy back to the source. There is no net consumption of real power over a complete cycle.

Therefore, the average power absorbed by a purely reactive element is zero.