In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.
In both Exercise 7.3 (inductor circuit) and Exercise 7.4 (capacitor circuit), the net power absorbed over a complete cycle is zero.
Explanation: In a pure inductor circuit, the current lags the voltage by 90°, whereas in a pure capacitor circuit, the current leads the voltage by 90°. In both cases, the voltage and current are out of phase by π/2 radians. The instantaneous power is given by p = vi. When averaged over a complete cycle, the power becomes zero because the average of sin(2ωt) over a complete cycle is zero. As derived in the text for an inductor: P_L = ⟨-i_m v_m/2 × sin(2ωt)⟩ = 0. Similarly, for a capacitor, the average power consumed is also zero. Hence, neither inductors nor capacitors dissipate energy; they only store energy temporarily and return it to the circuit.
Explanation
This question tests understanding of power consumption in reactive circuits. The textbook context clearly shows that for an inductor, the average power over a complete cycle is zero because the average of sin(2ωt) over a complete cycle is zero. The same principle applies to a capacitor - in both cases, the phase difference between voltage and current is 90°, resulting in zero average power consumption. Students should understand that reactive components (inductors and capacitors) only store energy temporarily during one part of the cycle and return it during another part, resulting in zero net energy absorption.
Solution Steps
Step 1: Identify that Exercise 7.3 involves a pure inductor and Exercise 7.4 involves a pure capacitor.
Step 2: Recall that in a pure inductor, current lags voltage by 90°, and in a pure capacitor, current leads voltage by 90°.
Step 3: Apply the power formula: Average power P = VI cos(φ), where φ is the phase difference.
Step 4: Since φ = 90° for both circuits, cos(90°) = 0, therefore net power absorbed = 0.