Dielectric loss in a lossy capacitor under sinusoidal excitation A parallel-plate capacitor has C = 200 × 10^-12 F and dielectric loss tangent tanδ = 4 × 10^-4. It is energized at frequency f = 10^6 Hz with a sinusoidal voltage of peak value 2 V. Calculate the dielectric power loss (in watts), using V_rms and the relation P ≈ V_rms^2 * ω * C * tanδ.

Introduction / Context:
Lossy dielectrics are modeled by a nearly ideal capacitor with a small loss tangent tanδ. Under AC excitation, a small in-phase current component causes real power dissipation. Estimating this dielectric loss is crucial for RF design, high-frequency filters, and reliability of capacitors operated at MHz ranges.

Given Data / Assumptions:

• Capacitance C = 200 × 10^-12 F.
• Loss tangent tanδ = 4 × 10^-4 (small angle approximation applies).
• Frequency f = 10^6 Hz, so ω = 2πf.
• Applied voltage peak Vp = 2 V → V_rms = Vp / √2.
• Power formula for small tanδ: P ≈ V_rms^2 * ω * C * tanδ.

Concept / Approach:
For a near-ideal capacitor, the reactive current leads the voltage by ~90°. A lossy dielectric introduces a small phase angle δ, represented by tanδ. The in-phase component is responsible for real power loss. Using V_rms and angular frequency, we compute P directly.

Step-by-Step Solution:
Compute V_rms: V_rms = 2 / √2 = 1.4142 V → V_rms^2 ≈ 2.0.Compute ω: ω = 2πf = 2π × 10^6 ≈ 6.2832 × 10^6 rad/s.Evaluate ω * C: 6.2832 × 10^6 * 200 × 10^-12 ≈ 1.2566 × 10^-3.Multiply by tanδ: 1.2566 × 10^-3 * 4 × 10^-4 ≈ 5.0264 × 10^-7.Compute P: P ≈ V_rms^2 * (ω C tanδ) ≈ 2.0 * 5.0264 × 10^-7 ≈ 1.0 × 10^-6 W.

Verification / Alternative check:
Using the conductance model, the equivalent series loss yields the same order of magnitude. Because tanδ is much less than 1, linear approximations are valid and results agree.

Why Other Options Are Wrong:

• 10^-8 W and 4 × 10^-7 W underestimate the loss by factors of ~2.5 to 10.
• 10^-4 W and 10^-2 W are two to four orders of magnitude too high for these values.

Common Pitfalls:

• Using V_peak instead of V_rms in power formulas leads to a 2× error.
• Confusing f with ω; forgetting the 2π factor changes results by ~6.28×.

Final Answer:
10^-6 W