Question

Prove that the power consumed by an inductor is zero.

Answer 1

To prove that the power consumed by an inductor is zero, let's start by understanding how an inductor behaves in an AC circuit.

1. **Voltage and Current Relationship**: For an inductor, the voltage across it is given by \( V(t) = L \frac{dI(t)}{dt} \), where \( L \) is the inductance and \( I(t) \) is the current through the inductor.

2. **Instantaneous Power**: The instantaneous power \( p(t) \) consumed by the inductor is given by:
   \[
   p(t) = V(t) \cdot I(t)
   \]
   Substituting \( V(t) = L \frac{dI(t)}{dt} \):
   \[
   p(t) = L \frac{dI(t)}{dt} \cdot I(t)
   \]

3. **Power Average Over Time**: To find the average power over a complete cycle, we calculate:
   \[
   \text{Average Power} = \frac{1}{T} \int_{0}^{T} p(t) \, dt
   \]
   where \( T \) is the period of the AC cycle.

4. **Substitute \( p(t) \)**:
   \[
   \text{Average Power} = \frac{1}{T} \int_{0}^{T} L \frac{dI(t)}{dt} \cdot I(t) \, dt
   \]
   Using the product rule:
   \[
   \frac{d}{dt}(I(t)^2) = 2 I(t) \frac{dI(t)}{dt}
   \]
   Therefore:
   \[
   \frac{dI(t)}{dt} \cdot I(t) = \frac{1}{2} \frac{d}{dt} (I(t)^2)
   \]
   Substituting into the integral:
   \[
   \text{Average Power} = \frac{L}{T} \int_{0}^{T} \frac{1}{2} \frac{d}{dt} (I(t)^2) \, dt
   \]
   The integral of a derivative over a period \( T \) is just the difference between the values at the end and start of the period. Since \( I(t) \) is periodic, \( I(T) = I(0) \):
   \[
   \int_{0}^{T} \frac{d}{dt} (I(t)^2) \, dt = I(T)^2 - I(0)^2 = 0
   \]
   Thus:
   \[
   \text{Average Power} = \frac{L}{T} \cdot \frac{1}{2} \cdot 0 = 0
   \]

Therefore, the average power consumed by an inductor over a complete cycle is zero. The inductor only stores and releases energy, but does not consume it in the long run.

Answer 2

To prove that the power consumed by an ideal inductor is zero, consider the following:

1. **Voltage and Current Relationship**: For an ideal inductor, the voltage \( V(t) \) and current \( I(t) \) are related by \( V(t) = L \frac{dI(t)}{dt} \), where \( L \) is the inductance.

2. **Instantaneous Power**: The instantaneous power \( p(t) \) consumed by the inductor is given by:
   \[
   p(t) = V(t) \cdot I(t)
   \]
   Substituting the voltage:
   \[
   p(t) = \left( L \frac{dI(t)}{dt} \right) \cdot I(t)
   \]
   \[
   p(t) = L I(t) \frac{dI(t)}{dt}
   \]

3. **Average Power**: To find the average power over one period \( T \) for a sinusoidal current \( I(t) = I_0 \sin(\omega t) \), compute the average value of \( p(t) \):
   \[
   P_{\text{avg}} = \frac{1}{T} \int_{0}^{T} p(t) \, dt
   \]
   Substituting \( p(t) \):
   \[
   P_{\text{avg}} = \frac{1}{T} \int_{0}^{T} L I(t) \frac{dI(t)}{dt} \, dt
   \]

4. **Using Chain Rule**:
   Note that \( \frac{dI(t)}{dt} \, dt = dI(t) \):
   \[
   P_{\text{avg}} = \frac{L}{T} \int_{I(0)}^{I(T)} I(t) \, dI(t)
   \]
   Since \( I(t) \) returns to its initial value after one period, the integral of \( I(t) \, dI(t) \) over a full period is zero. Thus:
   \[
   P_{\text{avg}} = 0
   \]

Therefore, the average power consumed by an ideal inductor over a complete cycle is zero.